Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 1 Section 1.2 1.1 A matrix is an orthogonal matrix if XTX = I Is the following matrix an orthogonal matrix?  −1 −1   1 X =  1 −1  2  −1 1   1 1  Solution: x={{-1.,-1},{1,-1},{-1,1},{1,1}}/2; Transpose[x].x//MatrixForm yields  1 0   0 1    Therefore, X is an orthogonal matrix.

1.2 If  1 −1  A=   2 −1 

B=

 1

1   4 −1 

does (A + B)2 = A 2 + B 2?

Solution: a={{1,-1},{2,-1}}; b={{1,1},{4,-1}}; ((a+b).(a+b)-a.a-b.b)//MatrixForm yields  0 0   0 0    Therefore, the expressions are equal.

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

1.3 Given the two matrices  1

A=  2

−3   5 4 

4

and

 4 B= 2

1  6 

 0

3 

Find the matrix products AB and BA. Solution:  1

4 −3 



4 1   12 16   =  AB =   2 6  2 5 4   0 3   18 44     4 1   6 21 −8   1 4 −3     BA = 2 6  =  14 38 18   0 3   2 5 4   6 15 12  Aa={{1,4,-3},{2,5,4}}; Bb={{4,1},{2,6},{0,3}}; Aa.Bb//MatrixForm Bb.Aa//MatrixForm 1.4 Given the following matrices and their respective orders: A (nm), B (pm), and C (ns).

Show one way in which these three matrices can be multiplied. What is the order of the resulting matrix?

Solution: CT ABT → (n  s)T (n  m)( p  m)T → (s  n)(n  m)(m  p) → (s  p)

1.5 Given  ab A= 2  −a

b2   −ab 

Determine A2.

Solution: From Eq. (1.13)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 a11 a   a 12 11 AA =   a21 a22   a21

 2 a12   a11 + a12a21 a12 (a11 + a22 ) a22  =  a (a + a ) a a + a2   21 11 22 21 12 22

 a2b2 − a2b2 b2 (ab − ab)  =  =0 2 ( 2 2 2 2 )  −a ab − ab −a b + a b   Aa={{a b, b^2},{-a^2,-a b}}; Aa.Aa//MatrixForm 1.6 Given the matrix  −4 A= 2   4

−3 −1  1 1   −2 4 

Determine the value of 4I − 4A − A2 + A3.

Solution: 11 −3  −7 3   −4 −22 10   −14 −1 −7  3  A = 6 −7 7 

 6 2 A =  −2

 12

−30

22 

Then,  1 0 0   −4 −3  − 4  0 4I − 4 A − A + A = 4 0 1 2 1   4 −2  0 0 1  2

3

 6 −  −2

11 −3   −14 −7 3  + 6  −4 −22 10   12

−1  1  4  −1 −7  −7 7  −30 22 

 0 0 0  = 0 0 0   0 0 0  Mathematica verification Aa={{-4,-3,-1},{2,1,1},{4,-2,4}};

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

A2=Aa.Aa; MatrixForm[A2] A3=A2.Aa; MatrixForm[A3] (4 IdentityMatrix[3]-4 Aa-A2+A3)//MatrixForm Section 1.3 1.7 Given the following matrices:  1  x= ,  2 

 2 a  A=  ,  3 4 

 6 4  B=  7 5 

What is the value of a that satisfies the following equation?

xT Ax = det B Solution: x Ax = T

1 2

 2 a  1   3 4   2  =  

1 2 

 2 + 2a   11   

= 24 + 2a  6 4  = 65−47= 2 detB = det  7 5   Therefore, 24 + 2a = 2 a = −11 Mathematica verification Solve[{1,2}.{{2,a},{3,4}}.{{1},{2}}==Det[{{6,4},{7,5}}],a]

1.8 Show that  a b+c 1   det b a + c 1  = 0    c a+b 1 

Solution:  a b+c 1  b 1 det b a + c 1  = a a + c 1 − (b + c) + b a+c   a+b 1 c a+b c 1  c a+b 1  = a(c − b) − (b + c)(b − c) + b(a + b) − c(a + c) =0

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Mathematica verification Det[{{a,b+c,1},{b,a+c,1},{c,a+b,1}}] 1.9 Expand the following determinants and reduce them to their simplest terms. a)  1+ a a a   det b 1+ b b    b 1+ b   b

Solution:  1+ a a a  det  b 1+ b b  = (1 + a)(1 + b)2 − b2  − a b(1 + b) − b2  + a b2 − b(1 + b)   b 1+ b   b = (1 + a)1 + 2b  − ab − ab = 1 + a + 2b + 2ab − 2ab = 1 + a + 2b Mathematica verification Det[{{1+a,a,a},{b,1+b,b},{b,b,1+b}}]

b)  x3 + 1 1 1    det 1 x3 + 1 1   x3 + 1  1 1 

Solution:  x3 + 1 det  1   1

1 x3 + 1 1

  = x3 + 1  x3 + 1 2 − 1 −  x3 + 1 − 1  +  1 − x3 − 1 ( )( )        x3 + 1  2 = ( x 3 + 1 )  ( x 3 + 1) − 1 − 2x3   1 1

= (x3 + 1) x6 + 2x3  − 2x3

= x3 (x6 + 2x3 )+ x6 + 2x3 − 2x3 = x9 + 2x 6 + x6 = x6 ( x 3 + 3) Mathematica verification

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Det[{{x^3+1,1,1},{1,1+x^3,1},{1,1,1+x^3}}]

1.10 Determine if the following determinant a function of a  ex det  ex   1

sin x cos x 1− a

cos x   sin x  a 

a function of a?

Solution:  ex  det  ex  1

 ex sin x cos x   cos x sin x  x e − det cos x sin x  = det    1 − a a    1 1− a a   x + det  e  1

sin x   sin x a 

cos x  cos x 1− a 

= ex (a cos x − (1 − a)sin x) − sin x (aex − sin x) + cos x (ex (1 − a) − cos x)

= ex (a cos x − sin x + a sin x − a sin x + cos x − a cos x) + sin2 x − cos2 x = ex (cos x − sin x) + sin2 x − cos2 x which is not a function of a.

Mathematica verification Det[{{Exp[x],Sin[x],Cos[x]},{Exp[x],Cos[x],Sin[x]},{1,1-a,a}}] 1.11 Show that

⎛ 2 ⎜x 1 det⎜⎜ x2 ⎜ 2 ⎜ 2 ⎜⎝ x3

x

⎞⎟ ⎟ 1 ⎟ = ( x − x )(x − x )(x — x ) ⎟ 1 2 1 3 2 3 ⎟ 1 ⎠⎟ 1

1

x

2

x3

Solution:

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

⎛ 2 ⎜x 1 det⎜⎜ x2 ⎜ 2 ⎜ 2 ⎜⎝ x3

⎞⎟ ⎟ 1 ⎟ = x2 (x — x )− x ( x 2 − x 2 ) + ( x 2 x — x 2 x ) 3 ⎟ 1 2 1 2 3 2 3 3 2 ⎟ 1 ⎠ ⎟ = x2 (x − x )− x (x − x )(x + x )+ x x (x − x ) 1 2 3 1 2 3 2 3 2 3 2 3 = ( x − x )⎡ x2 − x (x + x )+ x x ⎤ 2 3 ⎢⎣ 1 1 2 3 2 3 ⎥⎦ = ( x 2 − x3 )⎡ x1 (x1 − x2 )− x3 (x1 − x2 )⎤ ⎢⎣ ⎥⎦ =(x1 − x2 )(x1 − x3 )(x2 − x3 )

x

1

1

x

2

x3

Mathematica verification Simplify[Det[{{x1^2,x1,1},{x2^2,x2,1},{x3^2,x3,1}}]]

Section 1.4 1.12 Given A=

 17 7   19 9 

Determine A−1 and verify your result.

Solution: From Eq. (1.27)  a 11   a21

−1  1 a12  a  22  = a11a22 − a12 a21  −a21 a22 

−a12   a11 

−1  17  9 −7  1 7  =  19  17  9 − 19  7  9 −19 17   9 / 20 −7 / 20  =   −19 / 20 17 / 20 

Since, from Eq. (1.13)  a 11

  a21

 b b12   a11b11 + a12b21 11   = a22   b21 b22   a21b11 + a22b21 a

12

a b +a b 11 12



12 22

 a21b12 + a22b22 

then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 17 7   9 / 20 −7 / 20   17  9 / 20 − 7  19 / 20 −17  7 / 20 + 7  17 / 20   =    19 9  −19 / 20 17 / 20   19  9 / 20 − 9  19 / 20 −19  7 / 20 + 9  17 / 20    = 1 0  0 1  Mathematica verification Aa={{17,7},{19,9}}; Inverse[Aa]//MatrixForm Inverse[Aa].Aa//MatrixForm

Section 1.5 1.13 Given the two matrices  3 1 4  A= 2 1 2   4 2 3 

and

 2 1 3  B= 1 2 5   0 2 1 

Show that (AB)T = BTAT. Solution: T

T

 3 1 4   2 1 3   7 13 18   T   ( AB) = 2 1 2   1 2 5  =  5 8 13     10 14 25    4 2 3  0 2 1   7 5 10  =  13 8 14   18 13 25  and

 2 1 3 T  3 1 4 T  2 1 0   3 2 4  BT AT =  1 2 5   2 1 2  =  1 2 2   1 1 2   0 2 1   4 2 3   3 5 1   4 2 3   7 =  13

5 10  8 14   18 13 25 

Mathematica verification

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Aa={{3,1,4},{2,1,2},{4,2,3}}; Bb={{2,1,3},{1,2,5},{0,2,1}}; Aa.Bb//MatrixForm Transpose[Aa.Bb]//MatrixForm Transpose[Bb]//MatrixForm Transpose[Aa]//MatrixForm Transpose[Bb].Transpose[Aa]//MatrixForm Section 1.7 1.14 Does the following system of equations have a solution?

100 y1 + 420 y2 + 486 y3 = 17 400 y1 +1050 y2 + 972 y3 = 18 700 y1 +1680 y2 +1458 y3 =−3

Solution: It is seen from Eq. (1.30) that a = 300, b = 630, c = 486, and k = 2. Therefore, the determinant equals zero.

Mathematica verification Det[{{100,420,486},{400,1050,972},{700,1680,1458}}] 1.15 Without solving, determine whether the following system of equations has a solution. a1 + 2a2 + 3a3 = 1 4a1 + 5a2 + 6a3 = 0 7a1 + 8a2 + 9a3 =−7

Solution: It is seen from Eq. (1.28) that c = 3 and d = 6. Therefore, the determinant equals zero.

Mathematica verification Det[{{1,2,3},{4,5,6},{7,8,9}}] 1.16 Given the following system of equations

7 2  x1  1 0  x1  0 5 1 x  −  0 1 x  =      2  0   2 When  = 4, what are the values of x1 and x2? Justify your answer.

Solution: When

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 2   x1  0 3 2   x1  0 3 2  7 − → 5 1−   x =  x  =   → det   = −19 0 5 −3 0 5 −3   2     2     Since determinant is not equal to zero, x1 and x2 must equal zero.

1.17 Under what conditions does the following system of equations have a solution?

(1−  ) y1 + 2 y2 + 3y3 = 0 4 y1 +(6 −  ) y2 + 5y3 = 0 7 y1 + 8 y2 +(9 −  ) y3 = 0

Solution: ⎛ 1−  ⎜ ⎜ det 4 ⎜ ⎜ ⎜⎝ 7

2 6− 8

⎞⎟ ⎛ 6−  ⎛ 4 5 ⎞ 5 ⎞ ⎜ ⎜ ⎟ ⎟ = (1−  )det⎜ ⎟ − 2det⎜ ⎟⎟ ⎜⎝ 7 9 −  ⎠ ⎜⎝ 8 9− ⎟ ⎠ ⎟⎟ 9− ⎠ ⎛ ⎞ + 3det ⎜ 4 6 −  ⎟ ⎟ 8 ⎠⎟ ⎜⎜⎝ 7 = (1−  )((6 −  )(9 −  ) − 40)− 2(4(9 −  ) −35) 3 5

+ 3(32 −7(6 −  )) =− 3 +16 2 − 29 +14 + 8 − 2 + 21 −30 =− 3 +16 2 −18 Thus, this system of equation has a solution for those values of  that satisfy − 3 +16 2 −18 = 0 That is, when  = −1.02814,  = 1.09908, and  = 15.9291.

Mathematica verification N[Eigenvalues[{{1,2,3},{4,6,5},{7,8,9}}]] 1.18 Given the following system of equations (5 −  )x1 + x2 = 2 −2x1 + (2 −  )x2 = 1 Under what specific conditions can one solve for x1 and x2.

Solution The determinant of the coefficients must not equal zero; that is, we can solve for x1 and x2 provided that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

det

 5−  

−2

 1 0 2− 

the values of x1 and x2 are zero. This will be true for all values of  except when  has the following values   1 =0 det 5 −    −2 2 −   (5 −  )(2 −  ) + 2 = 0  2 − 7 + 12 = 0 1,2 = 3,4 Therefore, when  = 4, values of x1 and x2 are unknown.

Mathematica verification Eigenvalues[{{5,1},{-2,2}}]

1.19 Determine the eigenvalues and eigenvectors for the following matrix and verify that the eigenvectors are not orthogonal.  1 0 −1  A =  −2 −1 2     −1 2 1 

Solution:  1−   det −2   −1

0 −1 −  2

 −2 −1 −    −1 −  −1  2   = (1 −  )det  2  − det    2 1 −  −1 2     1−   = −(1 −  )(1 +  )(1 −  ) + 4  − (−4 − 1 −  ) = −4 (1 −  ) − (1 −  )(1 −  2 ) + 5 +  = −4 + 4 − 1 +  +  2 −  3 + 5 +  = 6 +  2 −  3

Therefore, the eigenvalues are: 1 = −2, 2 = 0, 3 = 3. The eigenvector corresponding to 1 = −2 is obtained as

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 1 − (−2)  −2   −1

  c1    =0 c −1 − (−2) 2  2   2 1 − (−2)   c3   3 0 −1   c1   −2 1 2  c   1 2   c2   2  = →       2   2 3   c3   1     −1 2 3  c    3  0

−1

 2 2    1 3 

4

Therefore, det c2 =

=

  −1 det  1 2   2 3 

det = −4

c3 =

 1 2   2 1 

=

−3

  −1 det  1 2   2 3 

=3

and x(1) = {1, −4,3}T. For 2 = 0,  1 0 −1   c1    −2 −1 2  c  = 0     2   −1 2 1  c3     1 0 −1   c1   −2 −1 2  c   −1 2   c2   2  = →       2   2 1   c3   1      −1 2 1  c   3  Therefore, det c2 =

 2 2    1 1 

=

0

  −5 det −1 2   2 1 

=0

c3 =

 −1 2  det  2 1  

=

−5

  −5 det  −1 2   2 1 

=1

and x(2) = {1, 0,1}T. For 3 = 3,  1 − (3) 0 −1   c1     =0 c −2 −1 − (3) 2  2     −1 2 1 − (3)   c3     −2 0 −1   c1   −2 −4 2  c   −4 2   c2   2  = →       2   2 −2   c3   1     −1 2 −2   c   3 

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Therefore,  2

 det  −4  2 c3 =  det  −4  2

2   1 −2  −6 −3 = c2 = =  −4 2  4 2 det   2 −2  det

2  1  −8 = −2 = 4 2  −2 

and x(3) = {1, −3/2, −2}T. Thus,  1   = −4   3

1 1  0 −3 / 2   1 −2 

( ) x = 0, i  j; therefore,

For the eigenvectors are linearly independent x (i)

(x ) x =  1 −4 3  (1)

T

(2 )

T

( j)

 1     0  = 1+ 3 = 4  0  1   

has to be zero, which is not the case. Therefore, the eigenvectors are not orthogonal. Without actually solving, we also know that because A is not a symmetric matrix the eigenvectors are not orthogonal.

Mathematica verification {lam,xn}=Eigensystem[{{1,0,-1},{-2,-1,2},{-1,2,1}}]; (Transpose[Transpose[{xn[[All,3]]}]].xn[[All,1]])[[1]]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 2 Section 2.1 2.1 Show that a + jb =1 b + ja Solution:

(

a + jb a + jb b − ja 2ab + j b2 − a2  = = b + ja b + ja b − ja a2 + b2

)

Therefore, 1 a + jb = 2 2 b + ja a + b 1 = 2 2 a +b

1 a + b2 1

4a2b2 + (b2 − a2 )2 = a4 + b4 + 2a2b2 =

2

4a2b2 + a4 + b4 − 2a2b2 =1

a +b 2

2

Mathematica verification z=(a+I b)/(b+I a); Simplify[ComplexExpand[Re[z]^2+Im[z]^2]] 2.2 If z = x + jy, then verify the following relations (a) z + 5 j = z − 5 j (b) jz = − jz (c) (1+ j)4 = −4 1+ j (d) = j 1− j Solution: (a) z + 5 j = x + j( y + 5) = x − jy − j5 = z − 5 j (b) jz = j(x + jy) = − y + jx = − y − jx = − j(x + y / j) = − jz (c) (1+ j)4 = (d)

( 2e ) = ( 2e ) = 4e = −4 j tan−11

4

j /4

4

j

1+ j (1+ j) (1+ j) 1+ 2 j − 1 = = = j 1− j (1− j) (1+ j) 2

Mathematica verification z=x+I y; Conjugate[z+5 I] Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Conjugate[I z] ComplexExpand[(1+I)^4] ComplexExpand[(1+I)/(1-I)] Section 2.2 2.3 Show that j=

(a)

1+ j

2 3/2  −1+ j 3  (b)   = 1 2   Solution: Since

(

)

zn =rn exp  jn  + 2m 

m = 0, 1, 2,...

then j = e j( /2+2m )/2 = e j /4e jm = (cos / 4 + j sin / 4) =

1+ j 2

 −1+ j 3    2  

3/2

 1 3 = + e j tan  4 4

−1

 3/( −1)  

3/2

(

= e j ( 2 /3+2m )

) =e 3/2

j +3m

= 1

Mathematica verification ComplexExpand[Sqrt[I]] ((-1+I Sqrt[3])/2)^(3/2) 2.4 Show that (−1+ j)10 = −32 j Solution: We note that −1+ j = 2e j where  = tan−1

1 −1

=

3 4

Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(−1+ j)10 = 25 e j (15 /2) = 25 e j (3 /2+6 ) = 25 e j 3 /2 = 25 cos(3 / 2) + j sin(3 / 2) = − j25 Mathematica verification ComplexExpand[(-1+I)^10] 2.5 Use Euler’s formula to obtain expressions for cos( + ) and sin( + ) in terms of sin(), sin(), cos(), and cos(). Hint: note that ej( + ) = ej ej .

Solution: e j(+ ) = cos( +  ) + j sin( +  )

= e j e j = (cos + j sin )(cos  + j sin  )

= cos cos  − sin sin  + j (sin cos  + cos sin  ) Therefore, upon equating the real and imaginary parts, we obtain cos( +  ) = cos cos  − sin sin  sin( +  ) = sin cos  + cos sin  Mathematica verification ComplexExpand[Exp[I (a+b)]] ComplexExpand[ComplexExpand[Exp[I a]] ComplexExpand[ Exp[I b]]] 2.6 Show that the four roots of z4 − 2cos(2 )z 2 + 1 = 0 are ej and e−j.

Solution:

(

)

1 2cos(2 )  4cos2 (2 ) − 4 = cos(2 )  cos2 (2 ) − 1 2 = cos(2 )  j sin(2 ) = e2 j

z2 =

Therefore, z = e j Mathematica verification PowerExpand[FullSimplify[Solve[ z^4-2 Cos[2 th] z^2+1==0,z],Assumptions->th  Reals]]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2.7 If 

1

 qn = 1 − q

q 1

n=0

then show that 1 −  cost



  cos(nt ) = 1 +  2 − 2 cost n

 1

n=0

Hint: Let q = ejt.

Solution: We note that q =  e jt → qn =  ne jnt Then 



 q =  e n

n

n=0



jnt

n=0

1 =   n (cos nt + j sin nt ) = 1 − ejt n=0

Therefore, 1



n   (cos nt + j sin nt ) =

1

1 −  cost + j sin t

 = 1 − ejt 1 −  cost − j sin t 1 −  cost + j sin t 1 −  cost + j sin t 1 −  cost + j sin t = = 2 2 2 (1 −  cost ) +  sin t 1 − 2 cost +  2 cost +  2 sin2 t

n=0

=

1 −  cost + j sin t 1 +  2 − 2 cost

Equating the real and imaginary parts gives 

1 −  cost

  cos nt = 1 +  2 − 2 cost n

n=0 



 n sin nt =

n=0

 sin t 1 +  2 − 2 cost

Mathematica verification q=e Exp[I t]; cq=ComplexExpand[1/(1-q)]; Simplify[Re[cq],Assumptions->{e,t}  Reals] (*Imag part = 0*) Simplify[Im[cq],Assumptions->{e,t}  Reals] (*Imag part = 0*) 2.8 Using the appropriate results of Exercise 2.7, show that 

2 e− n=0

n

sin(nt ) =

sin t cosh − cost

 0

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution: 

 sin t

  sin nt = 1 +  2 − 2 cost n

n=0

We let  = e− and obtain 

e

−

n

sin nt =

n=0

sin t sin t e− sin t = = 1 + e−2 − 2e− cost e + e− − 2 cost 2(cosh − cost )

2.9 If z = x + jy, then show that (a)

tanh(z / 2) =

(b)

coth(z / 2) =

sinh x + j sin y cosh x + cos y sinh x − j sin y cosh x − cos y

Solutions: (a) tanh(z / 2) = = =

ez/2 − e−z/2 ez/2 + e−z/2

=

ez − 1 ez + 1

=

ex+ jy − 1 ex+ jy + 1

=

ex (cos y + j sin y) − 1 ex (cos y + j sin y) + 1

ex cos y − 1+ jex sin y ex cos y + 1− jex sin y  ex cos y + 1+ jex sin y ex cos y + 1− jex sin y 2 2 ex cos y − 1+ ex sin y + jex sin y ex cos y + 1− ex cos y + 1

(

)

(

)

(

)

(e cos y + 1) + (e sin y) e − 1+ j2e sin y (e − e ) / 2 + j sin y = = e + 2e cos y + 1 (e + e ) / 2 + cos y 2

x

2x

2x

=

x

x

x

2

−x

x

x

−x

sinh x + j sin y cosh x + cos y

(b)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

coth(z / 2) = = =

ez/2 + e−z/2 ez/2 − e−z/2

=

ez + 1 ez − 1

=

ex+ jy + 1 ex+ jy − 1

=

ex (cos y + j sin y) + 1 ex (cos y + j sin y) − 1

ex cos y + 1+ jex sin y ex cos y − 1− jex sin y  ex cos y − 1+ jex sin y ex cos y − 1− jex sin y 2 2 ex cos y − 1+ ex sin y + jex sin y ex cos y − 1− ex cos y − 1

(

)

(

)

(

)

(e cos y − 1) + (e sin y) e − 1− j2e sin y (e − e ) / 2 − j sin y = = e − 2e cos y + 1 (e + e ) / 2 − cos y 2

x

2x

2x

=

x

x

x

2

−x

x

x

−x

sinh x − j sin y cosh x − cos y

Mathematica verification z=x+I y; ComplexExpand[Tanh[z/2]] ComplexExpand[Coth[z/2]] Section 2.3 2.10 If z = x + jy, which of the following functions are analytic and for those that are analytic, what is their region of analyticity? (a) f (z) = e2x cos y + je2x sin y (b) f (z) = sinh z (c) f (z) =| z | sin z (d) f (z) = z (e) f (z) = jz + z (f) z− j f (z) = z+ j Solutions: (a) Not analytic f (z) = e2x cos y + je2x sin y u v = 2e2x cos y  = e2x cos y x y v u 2x = 2e sin y  − = −e2x sin y x y (b) Analytic everywhere

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

f (z) = sinh z → cos y sinh x + j sin y cosh x u(x, y) = cos y sinh x v(x, y) = sin y cosh x u v = cos y cosh x = = cos y cosh x x y v u = sin y sinh x = − = −sin y sinh x x y (c) Not analytic f (z) =| z | sin z → x2 + y2 (sin x cosh y + j cos x sinh y) → u(x, y) =

x2 + y2 sin x cosh y

→ v(x, y) =

x2 + y2 cos x sinh y

u x sin x cosh y] = x2 + y2 cos x cosh y + x x2 + y2 v y cos x sinh y] = x2 + y2 cos x cosh y + y x2 + y2 v x cos x sinh y = − x2 + y2 sin x sinh y x x2 + y2 u y sin x cosh y − =− − x2 + y2 sin x sinh y y x2 + y2 (d) Not analytic u(x, y) = x, v(x, y) = − y u v =1 = −1 x y v u =0=− x x (e) Not analytic f (z) = jz + z = − y + jx + x2 + y2 u = − y + x2 + y2 , u = x

x x +y 2

2

,

v=x v =0 y

(f) Analytic

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

f (z) =

z − j x + j( y − 1) x + j( y − 1) x − j( y + 1) = = z + j x + j( y + 1) x + j( y + 1) x − j( y + 1)

x2 + y2 − 1 2x −j 2 2 2 x + ( y + 1)2 x + ( y + 1) 2 2 x + y −1 2x u= 2 , v = − x + ( y + 1)2 x2 + ( y + 1)2 =

u 4x( y + 1) = , 2 x 2 2 x + ( y + 1)

(

)

( ) = , x (x + ( y + 1) )

v

2 x − ( y + 1) 2

2

(

2

2

2

v 4x( y + 1) = 2 y x 2 + ( y + 1)2 −

u y

)

(

2 x − ( y + 1)2

=

2

)

(x + ( y + 1) ) 2

2

2

Mathematica verification (* u and v must be determined first *) cr[u_,v_]:=Simplify[{D[u,x]==D[v,y],D[v,x]==-D[u,y]}] cr[Exp[2 x] Cos[y],Exp[2 x] Sin[y]] cr[Sinh[x] Cos[y],Cosh[x] Sin[y]] cr[Sqrt[x^2+y^2] Sin[x] Cosh[y],Sqrt[x^2+y^2] Cos[x] Sinh[y]] cr[x,-y] cr[-y+Sqrt[x^2+y^2],x] cr[(x^2+y^2-1)/(x^2+(y+1)^2),-2 x/(x^2+(y+1)^2)] 2.11 Express the following expressions in terms of trigonometric and hyperbolic functions and them determine the value(s) of the real quantities a and b that make them analytic: (a) cos(ax + jby) (b) sin(ax − jby) Solutions: (a) 1 j (ax+ jby ) − j (ax+ jby ) +e (e ) 2 1 = (e−by (cos ax + j sin ax) + eby (cos ax − j sin ax )) 2 1 = (eby + e−by )cos ax − j (eby − e−by )sin ax 2 = cosh by cos ax − j sinh by sin ax

cos(ax + jby) =

(

)

Therefore, u(x, y) = cosh by cos ax v(x, y) = − sinh by sin ax

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Consequently u x v x

v

= −a cosh by sin ax = −a sinh by cos ax

−

y u y

= −b cosh by sin ax = −b sinh by cos ax

Hence, to be analytic, we see that the C-R conditions are satisfied if a = b. (b) sin(ax − jby) = =

(

1 j ( ax− jby ) − j ( ax− jby ) e −e 2j

)

(

)

1 by e (cos ax + j sin ax)− e−by (cos ax − j sin ax) 2j

((

)

(

)

)

1 eby − e−by cos ax + j eby + e−by sin ax 2j = cosh by sin ax − j sinh by cos ax =

Therefore, u(x, y) = cosh by sin ax v(x, y) = −sinh by cos ax Consequently u x v

= a cosh by cos ax = a sinh by sin ax

x

v y u

= −bcosh by cos ax = bsinh by sin ax

y

Thus, the Cauchy-Riemann conditions are satisfied when a = −b.

Mathematica verification (* u and v must be determined first *) crr[u_,v_]:={D[u,x]==D[v,y],D[v,x]==-D[u,y]} crr[Cosh[b y] Cos[a x],-Sinh[b y] Sin[a x]] crr[Cosh[b y] Sin[a x],-Sinh[b y] Cos[a x]] 2.12 Given the analytic function f (z) = u(x, y) + jv(x, y) Determine f(z) when: (a) u(x,y) = x3 − 3xy2 (b) u(x,y) = excosy Solutions:

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(a) u(x, y) = x3 − 3xy2 → Since the function is analytic, u

v

=

x

u

= 3x2 − 3y2

x

= 3x2 − 3y2 → v = 3x2 y − y3 + c

y

As a check, we note that v

u

= 6xy = −

x

= 6xy

y

(b) u(x, y) = ex cos y →

u

= ex cos y

x

Since the function is analytic, u

=

x

v

= ex cos y → v = ex sin y + c

y

2.13 Determine the value of the following integrals. singularities.

Assume that the path C includes all

e−3 z (a)  2z + j dz C



(b)

C

z

e cos z

(z −  / 2)

2

zn

 z−a

(c)

C

(

)

n+1

dz

dz

Solutions: 1 −3 z −3 j / 2 =  je =  j(− j) =   2z + j dz = 2 j 2 e z =− j /2 C e−3 z

ez cos z

d

z

 ( z −  / 2) dz = 2 j dz (e cos z ) 2

C

2 j d n

zn

 ( z − a)

n+1

dz =

C

n! dz n

n

(z )

z

z = /2

= 2 j (e cos z − sin z) z= /2 = −2 je

 /2

= 2 j z=a

Mathematica verification

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2 Pi I Residue[Exp[-3 Pi z]/(2 z+I),{z,-I/2}] 2 Pi I Residue[Exp[z] Cos[z]/(z-Pi/2)^2,{z,Pi/2}] 2 Pi I Residue[z^3/(z-a)^4,{z,a}] 2.14 Evaluate the following integral 1

 z3 (z + 4)dz C

for (a) inside the circle |z| = 2, and (b) inside the circle |z + 2| = 3.

Solutions: (a) Third order pole at z = 0. Therefore,  d2  1 1  −1  z3  d2  2  3  z3 (z + 4) dz = 2 j 2!lim   =  j lim  dz2 ( z + 4)  z→0 dz  z (z + 4)  z→0     |z|=2  −3 2 j  j   =  j lim 2 ( z + 4) = =  z→0   64 32 (b) The center of the circle is at z = −2 and the radius is 3; therefore, both z = 0 and z = −4 are inside this contour and the residue is 2 3  z3(z1+ 4) dz = 2 j 2!1lim  dzd2 z3(zz+ 4)   + 2 j lim  z3(z(z++4)4)  z→0 z→−4    |z+2|=3   j 2 j  j  j = + = − =0 32 (−4)3 32 32 Mathematica verification 2 Pi I Residue[1/(z+4)/z^3,{z,0}] 2 Pi I Residue[1/(z+4)/z^3,{z,0}]+ 2 Pi I Residue[1/(z+4)/z^3,{z,-4}] 2.15 Determine the value of the following integral when the contour includes all singularities. etzdz 2 z2 + 2z + 2 C z

I=

(

)

Solution: The poles are at z = 0 and at z = 1,2

1 −2  4 − 8  = −1  j  2

Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

  z2etz (z − z )etz  d  =  1 C z2 (z2 + 2z + 2) 2 j  dz  z2 (z2 + 2z + 2)  +  z 2(z − z1 )(z − z2 )    z=0 z=z1   (z − z )etz   2 +  z2 (z − z )( z−z ) 1 2 z=z   2  etzdz

 tz  2 etz  (2z + 2)etz 2   2 te  + = 2 j −   (z + 2z + 2) ( z 2 + 2z + 2) z=0  z (z − z2 ) z=z1    tz +  z2 (ez − z )  1 z=z   2  tz 1  etz2 = 2 j t − 1 + 2 e − 2 z1 (z1 − z2 ) z2 (z1 − z2 )  2 But z1 − z2 = −1+ j − (−1− j) = 2 j

( ) = 2e z = (−1− j) = ( 2e ) = 2e z12 = (−1+ j) = 2

2

2e3 j/4

2

−3 j/4

2

2

(cos(3 / 2) − j sin(3 / 2)) = 2 j

−3 j/2 = 2

2

Therefore,

(cos(3 / 2) + j sin(3 / 2)) = −2 j

3 j/2 =2

1 tz1 tz2    C z2 (z 2 + 2z + 2) 2 j  2 + (−2 j)2 j − (2 j)2 j  =  j t − 1 + 2 (e + e ) 1 1     =  j t − 1 + et(−1+ j) + et(−1− j) =  j t − 1 + e−t (ejt + e− jt )     2     2 etzdz

=

t − 1

etz2

etz1

(

)

=  j t − 1 + e−t cost

Mathematica verification Simplify[ComplexExpand[ 2 Pi I Residue[Exp[t z]/(z^2+2 z +2)/z^2,{z,0}]+ 2 Pi I Residue[Exp[t z]/(z^2+2 z +2)/z^2,{z,-1+I}]+ 2 Pi I Residue[Exp[t z]/(z^2+2 z +2)/z^2,{z,-1-I}]]] 2.16 Evaluate the following integral around the unit circle.

 z e dz 2 1/z

|z|1

Hint: Express the exponential function as a series using Eq. (A.3) of Appendix A.

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution: In this case, we expand the argument in series form to get  1 1 1 1  z2e1/z = 2 z  1+ + 2 + + + ...   z 2!z 3!z3 4!z4 1 1 1 = z2 + z + + + + ...

2 2! 3!z 4!z Upon integrating this expansion term by term, the only nonzero residue is that associated with 1/z.

Therefore, from Eq. (2.50), we see that  1 j z2e1/z dz = 2 j =



  3!

|z|1

3

Mathematica verification 2 Pi I Residue[Expand[x^2 Normal[Series[Exp[z],{z,0,4}]] /.z->1/x],{x,0}] 2.17 Determine the order of the pole for the following expressions: (a) etz z sinh(az) (b) 2

ez z4 (c)

sin2(z) z3 Solution: (a) Order equals 2.

1+ tz + t 2 z 2 / 2 +! etz = z sinh(az) z az + a3z3 / 3!+!

(

)

=

1+ tz + t 2 z 2 / 2 +!

(

)

az2 1+ a 2 z 2 / 3!+!

(b) Order equals 4.

2 ez = 1 1+ z2 + z4 / 2 +! z4 z4

(

)

(c) Simple pole.

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 1 z 2z3 sin2 (z) 1  z2 − z 4 2z6 = − + = + +! + !

3  z3 45 3 z   z 3 45

Mathematica verification Series[Exp[t z]/z/Sinh[a z],{z,0,1}] Series[Exp[z^2]/z^4,{z,0,1}] Series[Sin[z]^2/z^3,{z,0,1}]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 3 Section 3.1 3.1 Referring to Figure E3.1, consider a sine wave of magnitude Fo and frequency b that has a duration tb = Ntp, where tp = 2/b and N is an integer. This sine wave is repeated every period T, where T = Mtp = 2/o, M  N, and M is an integer. Determine the Fourier coefficients ai and bi and express them in terms of i, N, and M. To simplify subsequent calculations, it is suggested that the following notation be introduced prior to performing any integrations: 2  N o 2 N  =  t,  = M ,  t = = , and T = 2 o b o o b o b M Solution: For ai ai =

2F o tb

otb

 sin(bt)cos(iot) dt = 2F o  sin(b /  o )cos(i ) d T 0 oT 0

where we have set ot = . To simplify the subsequent integrations, we introduce the following notation: 2 No 2 N  = M ,  t = = , and T = 2 b o o b o M b Then, for ai and bi, we have 2 N M

F a i = o  sin( M )cos(i ) d  0 F 2 N M bi = o  0 sin( M )sin(i ) d

i M i M

From a table of integrals, we find that for i  M, Fo  cos[( M + i) ] cos[( M − i) ] 2 N/ M a = − −  i   2( M + i) 2( M − i) 0 F  1 1 cos[( M + i)(2 N/ M )] cos[( M − i)(2 N/ M )]  = o + − −   2  M + i M − i M +i M −i  F  2M 2 M cos(2iN/ M )  = o −  2 2 2  M − i M 2 − i2  Fo M 2Fo M = 1− cos(2iN/ M ) = sin2(iN/ M ) i  M  2 2 2 2  M − i  M −i 

(

)

(

)

and for i = M, Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

a = Fo M  =

2 N M



sin( M )cos( M ) d =

0

Fo sin2(2 N ) = 0 2 M

Fo 1  2M

sin2( M )

2 N M 0

i= M

For i = 0, F a = o 0

2 N M

2F

 sin( M ) d == n sin2 (n ) = 0 o

0

Similarly, for i  M, Fo sin[( M − i) ] sin[( M + i) ] 2 N/ M −   i   2( M − i) 2( M + i) 0 F sin[( M − i)(2 N/ M )] sin[( M + i)(2 N/ M )]  = o −  2  M +i M −i  Fo  sin(2iN/ M ) sin(2iN/ M )  = − −  2  M +i M −i  − Fo sin(2iN/ M )  1 1  =  M −i + M +i 2   − F M sin(2iN/ M ) o i M =  M 2 − i2

b=



(

)

and for i = M, Ï  sin( M )cos( M ) 2 N b = Fo  sin2( M ) d = o  −  i  0 2M  2 0 F  2 N  Fo N = o = i= M     2M  M Mathematica verification

M

2 N M

ai=Fo/Pi Simplify[Integrate[Sin[M t] Cos[i t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn}  Integers] aM=Fo/Pi Simplify[Integrate[Sin[M t] Cos[M t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn} Integers] a0=Fo/Pi Simplify[Integrate[Sin[M t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn} Integers] bi=Fo/Pi Simplify[Integrate[Sin[M t] Sin[i t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn} Integers] bM=Fo/Pi Simplify[Integrate[Sin[M t] Sin[M t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn} Integers] (*Visual verification: a plot*)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Ai[Nn_,Mm_,n_]:=If[n==Mm,0, 2 Mm Sin[Pi n Nn/Mm]^2/(Mm^2-n^2)/Pi] Bi[Nn_,Mm_,n_]:=If[n==Mm,Nn/Mm, Mm Sin[2 Pi n Nn/Mm]/(Mm^2-n^2)/Pi] ser[t_,Nn_,Mm_]:=Total[Table[Ai[Nn,Mm,n] Cos[n t]+ Bi[Nn,Mm,n] Sin[n t],{n,1,31}]] Plot[ser[t,4,8],{t,0,2 Pi}] 3.2 Expand the following periodic function in complex notation and use the results to express the series as a real quantity. −  x 

f (t ) = x Solution:

It is seen that the period T = 2. Then the Fourier series coefficients are obtained from Eq. (3.2) as 2    n   n    n  cn = 1  x e − jnx dx = 2  n cos − sin sin n0         n 2 2  2 2 − −2 = 2 n =0

n = odd n = even

Thus, cn is a real quantity. The same results are obtained for n < 0. When n = 0, 

1 x dx =  c0 =  2 − 2

Then the final result in complex notation is f (x) =



−

2

2

−



e jnx

 n=−1,−3,−5, n

2

−

2



e jnx

 n2  n=1,3,5,

To express this result as a real quantity, since f(x) is real, we use Eqs. (3.14) and arrive at 



n=1

n=1,3,5...

f (x) = c0 + 2 cn cos[2n x / T − n ] → c0 + 2  cn cos[nx − n ] where  2 2 cn = 2 n c0 =

 = tan−1 n

n = odd 0

( )

−2 / n 2

=

Therefore, Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

f (x) =

 2

+

4







1 cos(nx −  ) =

 n=1,3,5... n2

2



4 −



1

 n=1,3,5... n2

cos nx

Mathematica verification cn=Simplify[ComplexExpand[Integrate[Abs[x] Exp[-I n x], {x,-Pi,Pi},Assumptions->n  Integers&&n>0]/(2 Pi)]] cn=Simplify[ComplexExpand[Integrate[Abs[x] Exp[-I n x], {x,-Pi,Pi},Assumptions->n  Integers&&n<0]/(2 Pi)]] c0=Integrate[Abs[x],{x,-Pi,Pi}]/(2 Pi) 3.3 (a) Obtain the Fourier series expansion of the following periodic function f ( ) = t 2

− t 

(b) Using the results in (a), chose appropriate values for t and show that 

(−1)n+1  2 = 2 12

 n n=1 

2

n = 6 1

2

n=1

(c) Add the results in part (b) and show that (see also Exercise 3.2) 

2

n =8 1

n=1,3,5

2

Solutions: (a) We note that T = 2 and, therefore, o = 2/T = 1. Then, the Fourier coefficients are  3  3  a0 = 1  t 2 dt = 1 t  = 2 = 2  2  −   3 − 3 3 

an = 1  t 2 cos(nt )dt = 1 4n 3

 −

 n

n

(−1) =

4

(−1)n

n2



1 bn =  t 2 sin(nt )dt = 0  − Therefore, f (t ) = t 2 =

2 3



(−1)n

n=1

n2

+ 4

cos(nt )

(b) If t = 0, then we obtain   2 (−1)n  2 (−1)n = −4 → =  2 2 3 n=1 n 12 n=1 n

+1

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

If t = , then we obtain 2 =

2

(−1)n  4

+



3

2

n

n=1 

cos(n )

2n

(−1) n2 n=1

2 2 = 12

2  1 = 2 6  n=1 n (c) Expanding each series, we find 2

= 1−

1

+

1

−

1

+ ...

12 4 9 16 2 1 1 1 = 1 + + 9 + 16 + ...

6 4

Therefore, adding these two expressions, we obtain  2 2 1 + =2  2 12 6 n=1,3,5 n  2 1 =  2 8 n=1,3,5 n

Mathematica verification FourierCoefficient[t^2,t,n] (* Gives coefficients cn only *) 3.4 Obtain the Fourier series expansion of the following periodic function f ( ) = t 4

− t 

and choose appropriate values for t and the results of Exercise 3.3 to show that (−1)n+1 7 4 = n4 720 n=1 



1 4 =  4 90 n=1 n 

Solutions: (a) We note that T = 2 and, therefore, o = 2/T = 1. Then, the Fourier coefficients are

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 5  5  a0 = 1  t 4 dt = 1 t  = 2 = 2  4  −   5 − 5 5 1 8 a =  t 4 cos(nt )dt = 1 8 (−1)n (n2 2 − 6) = (−1)n (n2 2 − 6)

 − 

n

 n4

n4



1 bn =  t 4 sin(nt )dt = 0  − Therefore,

 (−1)n

1

4 f (t ) = t 4 =  + 8  4 5 n=1 n

(n2 2 − 6)cos(nt )

If t = 0, then we obtain    (−1)n+1 1 4 (−1)n+1 2 2 (−1)n+1 2 −  = 8 8  n −6 =  n2 48 n4 5 n4 n=1 n=1 n=1

(

)

  (−1)n+1 = 8   − 48 n4  12  n=1

2





(−1)n+1 n4

n=1

=

2

1  2 4  4  7 4 − =  48  3 5  720

If t = , then we obtain 

 2n   =  + 8 (−1) (n2 2 − 6) = 2  4 + 8 2  1 − 48 1 2 5 n4 5 n4 n=1 n=1 n 4 n=1    1  4  1 4  2  = 1 1 ( + − ) −4 = + −1  4 =   4 + 8 2  48  5 3  48  15 3 20 15 48  5  6  n=1 n 8 4  4 = = 720 90

4

1

4

Mathematica verification FourierCoefficient[t^4, t, n]

(* Gives coefficients cn only *)

3.5 Determine the Fourier series expansion for the periodic function f(t) = 2Aot/T, −T/2  t  T/2 using Eqs. (3.2) and (3.14) and Eqs. (3.18) and (3.19).

Solutions: From Eq. (3.2)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

cn =

1 T /2 2 A t

c0 =

n

Ao

− jy

 ye dy = 2n2 2 − n 2 A T /2 2 Ao T /2 o dt = =0 t dt = 2o t2  2 −T /2 T T T −T /2



o

T −T /2 T 1 T /2 2 A t

T −T/2

− j 2 nt /T

e

dt =

jAo (−1)n n

and since the real part of cn is zero

n = tan

n −1 −(−1)

= (−1) n+1

0

Then, from Eq. (3.14)



2A



1

 2 n+1

(

)

f (t) = c0 + 2 cn cos[2nt / T − n ] = o cos 2nt / T − (−1)  / 2   n=1 n=1 n n+1  2A (−1) = sin(2nt / T ) o   n n=1 From Eq. (3.19), an =

2 T /2 2 Ao t

n cos(2nt / T ) dt = Ao  y cos y dy = 0 T n2 2 − n



T −T /2 T /2

a0 = bn =

2 A T /2 4 Ao t dt = 2o t2 =0  2 −T /2 T T −T /2 2 T 2A t A

T 0 T

and from Eq. (3.18)

o

sin(2nt / T ) dt =



f (t ) = bn sin(not ) = n=1

n

o

n 2

 y sin y dy =

− n

o

n

 (−1)n+1

2A 

2

2 A (−1)n+1

o

 n=1

sin(2nt / T ) n

Mathematica verification FourierCoefficient[2 t/T,t,n,FourierParameters->{1,2 Pi/T}, Assumptions->T>0] (*Verification for cn only*) 3.6 Determine the Fourier series expansion for f(t) = Aoe−t/T,  > 0, over the interval −T/2  t  T/2.

Solutions: From Eq. (3.2)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Ao −t /T T /2 = 2 sinh( / 2) Ao T /2 −t /T e e dt = − −T /2  T −T/2  2sinh(( + j2n ) / 2) A T /2 A T /2 −( + j 2 n )t /T −t /T − j 2 nt /T o o dt = cn = e e dt = e T −T/2  + j2n T −T /2 c0 =

(−1)n sinh( / 2) (−1)n sinh( / 2) = = ( / 2 − jn )  / 2 + jn ( / 2) 2 + (n )2

Then, sinh( / 2)

n = 0,1,2,3,... ( / 2)2 + (n )2 n  = tan−1 n = 1,2,3,... n /2 c = n

Therefore, 

f (t) = c0 + 2 cn cos[n ot − n ] sinh( / 2) n=1

=

 /2



+ 2sinh( / 2)

(

)

cos 2nt − tan−1 (n / ( / 2) ( / 2)2 + (n )2

n=1

3.7 For the wave forms shown in Figure E3.2, use Eq. (3.2) to determine the magnitude and phase of the Fourier coefficients cn.

Solutions: (a) Case 2: From Eq. (3.2) t=td /2

t=0

je− j 2 pt /T je− j 2 pt /T 1 d − j 2 pt /T − j 2 pt /T cp = 1 − e dt −  e dt = T − t /2 2 p t=− t /2 2 p t=0 T 0 0

t /2

d

(

d

)

(

)

j j 1− e jptd /T + 1− e− jptd /T 2 p 2 p j 2j 2 = 2 − 2cos( pt / T )) = sin ( pt / (2T )) ( d d 2 p p c0 = 0 =

and from Eq. (3.11)  = tan−1 n

sin ( 2

pt d / (2T ) 0

) = 2

Mathematica verification A=Sequence[FourierParameters->{1,wo},

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Assumptions->wo>0&&td>0&&td<Pi/wo]; FourierCoefficient[(UnitStep[t+td/2]-UnitStep[t](UnitStep[t]-UnitStep[t-td/2]))/(td/T),t,n,A]/.wo->2 Pi/T (b) Case 3: From Eq. (3.2) and Mathematica

2t cos( pt / T ) t /2 1 d d j2 pt/T − cos(t / t d)e dt = d cp = 2 T −t/2 T 1− (2 pt / T ) d

d

2t c0= d T

 2t cos( pt / T )  t d d  d 2 = p→T /( 2td ) T 2T 1− (2 pt / T )  d 

cp = lim and from Eq. (3.11)

0  n = tan−1 , g

g=

(

)

cos ptd / T 2 1− (2 ptd / T )

Mathematica verification A=Sequence[FourierParameters->{1,wo}, Assumptions->wo>0&&td>0&&td<Pi/wo]; FourierCoefficient[Sin[Pi t/td+Pi/2] (UnitStep[t+td/2]-UnitStep[t-td/2])/(td/T),t,n,A] /.wo->2 Pi/T (c) Case 4: From Eq. (3.2) and Mathematica

− jt sin( pt / T ) t /2 1 d d d j2 pt/T − cp = sin(2t / t d)e dt = T −t/2 T 1− ( pt / T )2 d

d

c0 = 0

c = lim  − jtd sin( ptd / T )  − jtd = 2   p p→T /(td )  T 1− ( ptd / T )  2T and from Eq. (3.11)  n = tan−1

−g , 0

g=

(

)

sin ptd / T 2 1− ( ptd / T )

Mathematica verification A=Sequence[FourierParameters->{1,wo}, Assumptions->wo>0&&td>0&&td<Pi/wo]; FourierCoefficient[Sin[2 Pi t/td] (UnitStep[t+td/2]-UnitStep[t-td/2])/(td/T),t,n,A] /.wo->2 Pi/T Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(d) Case 5: From Eq. (3.2) and Mathematica 1 − j2 pt/T dt + 1 td /2 1− 2t / t c = 0 (2t / t + 1)e ( )e− j2 pt/T dt



p

=

T − td /2 T

d

(1− cos( pt / T )) =

p  td 2

c = 0

2



tTsin0 ( 2 d

)

d

pt d / (2T )

2T ( pt d / (2T ))

2

d

td 2T

and from Eq. (3.11)  = tan−1 n

0

= 0,

g0

g

Mathematica verification A=Sequence[FourierParameters->{1,wo}, Assumptions->wo>0&&td>0&&td<Pi/wo]; Simplify[ComplexExpand[FourierCoefficient[ Piecewise[{{2 (t/td+1/2),-td/2<=t<0}, {2(1/2-t/td),0<=t<td/2}}]/(td/T),t,n,A]]]/.wo->2 Pi/T Section 3.3 3.8 Find the Fourier transform of the following waveform f (t ) = h(1 + at )

−1/ a  t  0

= h(1 − at ) =0

0  t 1/ a

|t |1/ a

What happens when h = a and a → ? What does this result signify?

Solution: The Fourier transform is obtained from Mathematica as 0

1/a

F( ) = h  (1 + at )e− jt dt + h  (1 − at )e− jt dt −1/a

=

0

2ah (1 − cos( / a)) 2

When h = a, we note that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2 4  2a2   1    1  −!  F( )  2 1 − 1 −   +    2  a  24  a   2 1   1 −   +!

24  a 

Then, lim F( ) → 1 a→

This is the Fourier transform of a delta function.

Mathematica verification sol=Simplify[ComplexExpand[FourierTransform[ h (1+a t) (UnitStep[t+1/a]-UnitStep[t]),t,w, FourierParameters->{1,-1}]+ FourierTransform[h (1-a t) (UnitStep[t]-UnitStep[t-1/a]), t,w,FourierParameters->{1,-1}]]] Limit[sol/.h->a,a->Infinity] 3.9 Determine the Fourier transform of the following expression

 n2e−  n

f ( ) =

d

sin(d )u( )

where u() is the unit step function and

d = n 1 −  2 Present you results in terms of the non-dimensional quantity  = /n.

Solution: The Fourier transform is obtained as F( ) = n2  −  sin(  )e − j d  e d  d 0 n

2 2 =  2 + ( n + j )2 =  2 −  2 +  2n 2 + 2 j d

n

d

n

 =  2 −  2 2 −  2 +  2 2 + 2 j

n

2 n

n

F() =

n

n

n

1 1 − 2 + 2 j

In polar form, we have that 1 − 2 + 2 j = r()ej()

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

where r() = () = tan−1

2 1 − 2

Therefore, F() =

1 − j () e r()

Mathematica verification sol=FourierTransform[ wn^2/wd Exp[-z wn t] Sin[wd t]* UnitStep[t],t,w,FourierParameters->{1,-1}]; den=ComplexExpand[Denominator[sol]]; den=Simplify[den/.wd->wn Sqrt[1-z^2]]; fw=Simplify[(Numerator[sol]/den)/.w->Om wn] 3.10 Find the Fourier transform of the following waveform f ( ) = e−|t|

 0

Solution: The Fourier transform is 



0

F( ) =  e− |t|e− jt dt =  e( − j )t dt +  e−( + j )t dt −

=

−

1

0  ( − j )t 

( − j ) 

e

−

0

−

1



e−( + j )t  0 ( + j ) 

=

1 1  + j +  − j + = 2 +2 ( − j ) ( + j )

=

2  +2 2

Mathematica verification FourierTransform[ Exp[-a Abs[t]],t,w, FourierParameters->{1,-1},Assumptions->a>0] 3.11 Find the Fourier transform of the following waveform f ( ) = cost

−1  t 1

f ( ) = 0 otherwise Solution: The Fourier transform is

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

1

F( ) = 

1

cos(t )e−

jt

dt =

−1

1  (e jt + e− jt )e− jt dt 2 −1

1 1 − j( + )t 1 1 j( − )t = e dt + e dt 2 −1 2 −1 = 1 e j( − )t − e− j( + )t +1 2 j ( −  ) j ( +  )  −1 j( − ) − j( − ) − j( + ) 1 e −e e − e j( + )  = −  2 j   −  +  sin( −  ) sin( +  ) = +  −  + Mathematica verification sol=FourierTransform[Cos[a t] (UnitStep[t+1]-UnitStep[t-1]), t,w,FourierParameters->{1,-1}] 3.12 For the wave forms shown in Figure E3.3, use Eq. (3.34) to determine the magnitude of their Fourier transforms. Compare these results with the corresponding results obtained in Exercise 3.7 Solutions: (a) Case 2: From Eq. (3.34) t=0

t=td /2

e− jt e− jt F ( ) =  e dt − − 0 e− jt dt = − j t=−td /2 − j t=0 −td /2 j 2j = 1− e jtd /2 + 1− e− jtd /2 = 1− cos(t / 2) d   4j = sin2 ( t / 4) d  F (0) = 0 0

− jt

td /2

(

)

(

)

Comparison with Exercise 3.7: F( ) sin2 (td / 4) = td t d / 4

 0

F( )

c0 = 0 td / T

=0

td

 =0

cp td / T

=

sin2 ( ptd / (2T )) p  0 ptd / (2T )

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; FourierTransform[(UnitStep[t+td/2]-UnitStep[t](UnitStep[t]-UnitStep[t-td/2]))/td,t,w,B]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(b) Case 3: From Eq. (3.34) and Mathematica td /2

 cos (t / t )e

F ( ) =

d

− td /2

− jt dt = 2t

cos(td / 2) d

 2 −  2t2d

2

F (0) =



Comparison with Exercise 3.7: F( )

= 2

td

cos(td / 2)  2 −  2t d2

F( ) 2 = td 

 =0

c0 = 2 td / T 

F( )

 =  / td

cp

=

td

1 2

cp

   / td

=

td / T

td / T

1

=

2 cos( ptd / T )  1 − (2 ptd / T )2

t d / T = 1 / (2 p),

2

td / T  1 / (2 p), p  0

p0

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; Simplify[FourierTransform[(Sin[Pi t/td+Pi/2]* (UnitStep[t+td/2]-UnitStep[t-td/2]))/td,t,w,B]] (c) Case 4: From Eq. (3.34) and Mathematica td /2

sin(t  / 2)

−td /2

4 −  2t2d

F ( ) =  sin(2t / td )e− jt dt = −4 jt d

d

2

F (0) = 0 Comparison with Exercise 3.7: F( )

=

sin(td / 2)  2 −  2t d2 / 4

=0

 =0

td F( ) td F( ) td

=

1 2

  2 / td

c0 =0 td / T

 = 2 / t d

cp td / T

cp td / T

=

1 sin( ptd / T )  1 − ( pt d / T )2

td / T  1 / p,

p0

p=0 =

1 2

t d / T = 1 / p,

p0

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; Simplify[FourierTransform[Sin[2 Pi t/td]* (UnitStep[t+td/2]-UnitStep[t-td/2])/td,t,w,B]]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(d) Case 5: From Eq. (3.34) and Mathematica td /2

0

F ( ) =

 (2t / t + 1)e

− jt

d

− td /2

=

dt +  (1− 2t / td )e− jt dt 0

(

)

4 8 1− cos(t d / 2) = 2 sin2 (td / 4) 2 t d t d 2 t  sin(t / 4)

= d 2 t F (0) = d 2

d

td / 4

 

Comparison with Exercise 3.7: F( ) td F( ) td

1  sin(t d / 4 )  =  2  td / 4 

2

=

1

 =0

2

 0

cp td / T

=

1 sin2 ( pt d / (2T )) p  0 2 ( ptd / (2T ))2

c0 = 1 td / T 2

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; Simplify[ComplexExpand[FourierTransform[ Piecewise[{{2 (t/td+1/2),-td/2<=t<0}, {2(1/2-t/td),0<=t<td/2}}]/td,t,w,B,Assumptions->td>0]]] Section 3.4.6 3.13 Consider the pulse f (t ) = Ap

u(t) − u(t − t ) − u(t − t ) − u(t − 2t ) p

p

p

where tp is the duration of the positive and the negative portions of the pulse and Ap is the magnitude of each portion. (a) Determine an expression for the magnitude of the Fourier transform of this pulse when a rectangular window function is used; that is, one with no weighting function. The results can be normalized to 2Aptp. (b) Determine an expression for the magnitude of the Fourier transform of this pulse when a Hamming window function is used.

Solutions: (a) The Fourier transform of the signal with the rectangular window is obtained from

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2tp 2 1  − jt − jt dt − A e dt A t e = p p   − j FR ( ) = Ap  p  − j d − 1 e d   0e 0 tp 4 jA t e− j p p = sin2 ( / 2)  tp

where  = tp. Therefore, 2 2 F () = 2A t sin ( / 2) → FR () = sin ( / 2) R p p 2A p t p /2 /2

Mathematica verification sol=FourierTransform[(UnitStep[t]-UnitStep[t-tp])(UnitStep[t-tp]-UnitStep[t-2 tp]),t,w, FourierParameters->{1,-1}] csol=ComplexExpand[w sol] PowerExpand[Sqrt[Simplify[Re[csol]^2+Im[csol]^2, Assumptions->{tp,w}  Reals]]/w] (b) The Fourier transform of the signal with the Hamming window is obtained from tp



FH ( ) = Ap  a − b cos[2t / (2tp )]



e− jt dt − A

0

2tp p

 a − b cos[2t / (2t )]e− jt dt p

tp

= A t  1 a − b cos[ ]e− j d − 2 a − b cos[ ]e− j d  p p   0  1 Upon carrying out the integrations, we obtain

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2 2 FH ( ) aj bj − j = − (e− j − 1) + 2 e + 1 ( ) Aptp   −2 2 2 j = − A (e− j − 1) + B (e− j + 1)

( ( −  )( (  −  ) 2 2

= = =

)

o 2

2

(− A (−2 j sin( / 2)) + B (2 cos( / 2)) ) 2

je− j

(2 −  2 ) 4 je− j ( −  ) 2 je− j 2

)

o

− A e− j (e− j/2 − e j/2 ) + B e− j (e− j/2 + e j/2 )

j

= =

2 2

o o

2

2

o

o

(A sin ( / 2) + B cos ( / 2)) 2

2

o

o

(Ao (1 − cos) + Bo (1 + cos))

(2 −  2 )

2 je− j (Ao + Bo + (Bo − Ao )cos) (2 −  2 )

where

(

Ao = a 2 −  2

)

Bo = b2

Therefore,

(

)

Ao + Bo + (Bo − Ao )cos FH ( ) 2A p t p =  2 −  2

(

)

Mathematica verification (*Integrations only*) a Simplify[Together[(Integrate[Exp[-I Om t],{t,0,1}]Integrate[Exp[-I Om t],{t,1,2}])]] -b (Integrate[Exp[-I Om t] (Exp[I Pi t]+Exp[-I Pi t]), {t,0,1}]-Integrate[Exp[-I Om t] (Exp[I Pi t]+ Exp[-I Pi t]),{t,1,2}])

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 4 Section 4.1.1 4.1 Obtain the solution to the following first-order differential equations: (a) 2x2

dy

= x2 + y2

dx (b) dy

( e + x sin y ) dx + ye − cos y = 0 x

x

(c) dy dx

=

y

1/2

  − k 1− y  x  x

(d)

(3y + 6xe + sin(2x))dy + (4x + 6e + 2 y cos(2x))dx = 0 2

y

3

y

(a) 2x2

dy

= x2 + y2

dx Solution Rearranging terms, we find 2

1 y = +   dx 2 2  x  dy

1

If we let v = y/x, dy dv =v+x dx dx then v+ x

dv

=

1

=

dx

dx

1 dv 1 1 1 + v2 → x = + v2 − v = (v −1)2 2 2 dx 2 2 2

or

2dv (v −1)

2

x

→

2dv

=

dx

 (v −1)2  x

→

−2

= ln x + C

v −1

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solving for v gives v = 1−

2 ln x + C

y = x−

2x ln x + C

and, therefore,

Mathematica verification DSolveValue[2 x^2 y'[x]==x^2+y[x]^2,y[x],x] (b) dy

( e + x sin y ) dx + ye − cos y = 0 x

x

Solution Rearranging terms, we find

(ex + x sin y)dy + ( yex −cos y)dx = 0 Therefore, M (x, y) = yex − cos y N (x, y) = ex + x sin y To determine if this equation is exact, we note that M = ex + sin y y N = ex + sin y x Therefore, the equation is exact and from Eq. (4.12), u(x, y) =  M (x, y)dx + f ( y)

=  ( yex − cos y ) dx + f ( y)

= yex − x cos y + f ( y) = c o To determine f(y), we note that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

u(x, y) = N (x, y) y df ( y) x ex + x sin y + = e + x sin y dy df ( y) = 0 → f ( y) = c 1 dy Then, yex − x cos y + c1 = co yex − x cos y = C which is an implicit form.

Mathematica verification DSolve[(Exp[x]+x Sin[y[x]]) y'[x]+(y[x] Exp[x]-Cos[y[x]])==0 ,y,x]

(c) dy

=

dx

y

  y  2  1/2 − k 1+  

x

 x  

Solution Setting v = y/x, we find that dy dv =v+x dx dx Then, the differential equation becomes v+x 1

dv dx dv

= v − k 1+ v2 =−

k

1+ v2 dx dv

x k = − dx x 1+ v2

Therefore,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

dv

dx

 1+ v2 = −k  x sinh−1 v = −k ln x + ln C

( ) = −k ln x + ln C x v + 1+ v ) = 1 C( ln v + 1+ v2 k

2

and 1  −k xk  v + 1+ v2 = Cx−k → v =  Cx − 2 C  Then, y=

x

k 1+k Cx−k − x  = 1  Cx1−k − x  2  C  2  C 

Mathematica verification TrigToExp[DSolveValue[y'[x]==y[x]/x-k (1+(y[x]/x)^2)^(1/2), y[x],x]/.C[1]->Log[c]]

(d)

(3y + 6xe + sin(2x))dy + (4x + 6e + 2 y cos(2x))dx = 0 2

y

3

y

Solution We see that N (x, y) = 3y2 + 6xey + sin(2x) M (x, y) = 4x3 + 6ey + 2 y cos(2x)

and, therefore, N x

= 6ey + 2cos(2x) =

M y

Thus, we have an exact equation. Then, from Eq. (4.12), we have that

(

)

u(x, y) =  4x 3 + 6ey + 2 y cos(2x) dx + f ( y) = x4 + 6xey + y sin(2x) + f ( y) From Eq. (4.13),

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

df ( y)

6xey + sin(2x) +

= 3y2 + 6xey + sin(2x)

dy df ( y) = 3y2 dy Upon integration, we obtain f ( y) = y3 + C and the solution is x4 + 6xey + y sin(2x) + y3 + C = 0 As a check, we differentiate this equation with respect to x and arrive at dy dy dy 4x3 + 6ey + 6xey + sin(2x) + 2 y cos(2x) + 3y2 =0 dx dx dx

(4x + 6e + 2 y cos(2x))dx + (6xe + sin(2x) + 3y )dy = 0 3

y

y

2

Mathematica verification DSolve[(2 y[x]^2+6 x Exp[y[x]]+Sin[2 x])y'[x]+ 4 x^3+6 Exp[y[x]]+2 y[x] Cos[2 x]==0,y[x],x] 4.2 For   0, find the solution to d d

+ 1+ = 0

when the initial condition is (0) = 1.

Solution: We rewrite the equation as d

 =−   d  → − = − + C →  =  ( − C ) 1+   −

−1/

Using the initial conditions gives −1/

 (0) = 1 = −C  Thus,

 =  ( − C )  

−1/

→ 1 = −C → C = −

→ ( + 1)

−1/

1 

 0

Mathematica verification Simplify[DSolveValue[{y'[t]+y[t]^(1+e)==0, y[0]==1},y[t],t]]//Quiet

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

4.3 Given

 2  − w cot = k cos  −  1 + cos  d 

dw

with the boundary condition w(o) = 0. The parameter k is a constant. Show that  1+ cos  1 1 w( ) = k ln − + sin  1+ cos 1+ cos 1+ cos   o o Solution: The solution is given by 1   ( )q( ) d + C   ( )    ( ) = exp  p( ) d  

w( ) =

where p( ) = − cot   2 q( ) = k  cos  −  1 + cos  Then 1 ( ) = exp  p( ) d  = exp − cot d  = exp − lnsin  =     sin and

 w( ) = sin k

1 

2

cos −

    sin 



d + C



1+ cos 

 

sin

1

Performing the integration yields 1 

2







 sin  cos − 1+ cos  d = ln  tan( / 2)  − 1+ cos However, sin tan( / 2) = 1+ cos Therefore,

 1  w( ) = k ln(1+ cos )− sin + C sin  1+ cos 

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Using the boundary condition, it is found that

 1 sin + C sin = 0 w( ) = k  ln(1+ cos ) − o o o o  1+ cos  o

 1 C = −k  ln(1+ cos − ) o  1+ cos  o

Then, the final result is w( ) =



 1+ cos 1 1 − + sin k ln  1+ cos 1+ cos 1+ cos   o o

Mathematica verification Simplify[DSolveValue[{y'[x]-y[x] Cot[x]== k (Cos[x]-2/(1+Cos[x])),y[xo]==0},y[x],x]]

Section 4.1.2 4.4 Obtain the solution to the following equations: (a) y2

dy

= ex − y3

dx (b) dy

− 2xy = 2x3 y2

dx (a) y2

dy

= ex − y3

dx Solution When the terms are rearranged, this is a Bernoulli equation of the form dy + y = ex y−2 dx Therefore, n = −2 and

p(x) = 1

and

q(x) = ex

Consequently,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

P(x) = (1− n) p(x) = 3 Q(x) = (1− n)q(x) = 3ex and the equation to be solved is du

+ 3u = 3ex

dx where u = y1−n = y3.

The solution to this equation is obtained with the use of an integrating factor as determined by Eq. (4.22). Thus,

(

)

( )

(x) = exp  P(x) dx = exp  3dx = exp(3x) Then from Eq. (4.29), u = y3 =

1  1  3x x (x)Q(x)dx + C  = e 3e dx + C 

  (x)  3 3   = e−3x e4x + C = ex + Ce−3x

e3x 



4  4 1/3 3  y =  ex + Ce−3x   4 Mathematica verification DSolveValue[y[x]^2 y'[x]==Exp[x]-y[x]^3,y[x],x]//Quiet

(b) dy

− 2xy = 2x3 y2

dx for the boundary condition y(0) = 1.

Solution This is a Bernoulli equation with n = 2 and with p(x) = −2x q(x) = 2x3 Then, from Eq. (4.29),

(x) = exp (1− n) p(x) dx  = exp 2 x dx  = ex

2

and

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

1  (1− n)  (x)q(x) dx + C   (x)  2 2 x2  x2 3 x x − − = e −2 e x dx + C  = e e 1− x2 + C     

u=

(

)

= 1− x2 + Ce−x

2

Then, since y = u1/(1-n), y = 1/u. Therefore, y=

1 1− x2 + Ce−x

2

When x = 0, then y = 1 so that y=

1

→ 1+ C = 1  C = 0

1− x2 + Ce−x

2

Therefore, 1

y=

1− x2

x 1

Mathematica verification Simplify[DSolveValue[{y'[x]-2 x y[x]==2 x^3 y[x]^2,y[0]==1}, y[x],x]]

Section 4.2.2 4.5 Find the general solution to d 2  −  d 2W  − 4e− W = 0 e d2  d2  in terms of trigonometric, hyperbolic, and exponential functions when 2 > 2/4.

Solution: Expand the equation and cancel e− to obtain

d 4W d3W +  2 d 2W − 4W = 0 − 2 3 d4 d d2 Assume a solution of the form W = e. Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 4 − 2 3 +  2 2 − 4 = 0

(

)

 2  2 − 2 +  2 − 4 = 0  2 ( −  ) − 4 = 0 →  ( −  ) = 2 2

and, therefore,  2 −  − 2 = 0 →  =



2     2  2  2

With the following definitions =

 2

 p = 2 +  2  m = 2 −  2

2   2

the four solutions are 1,2 =    p 3,4 =   jm Therefore,

( + ) + A e2( − ) + A e3 ( + j ) + A 4e( − j )

W () = Ae1

p

(

p

 

−  p

p = e Ae + A2 e 1

(

p

p

j 

− j p

+ A3e p + A4e

)

= e C cosh   + C sinh   + C cos  + C sin   1

p

2

p

3

m

4

m

)

Mathematica verification DSolveValue[D[Exp[-b x] D[w[x],x,x],x,x]-Om^4 Exp[-b x]* w[x]==0,w[x],x]/.{Sqrt[b^2-4 Om^2]->2 I em, Sqrt[b^2+4 Om^2]->2 I ep} Section 4.2.3 4.6 Find the solution to 2 (x −1) 2 d y + 3(x −1) dy + y = 0 dx2 dx

Hint: Assume a solution of the form y = (x − 1)r to find the first solution and then use the reduction of order to find the complete solution.

Solution:

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Assume a solution of the form y = (x − 1)r. Then dy = r(x −1)r−1 dx d2y = r(r −1)(x −1)r −2 2 dx Using these relations in the original equation gives (x −1)2 r(r −1)(x −1)r−2 + 3(x −1)r(x −1)r−1 + (x −1)r = 0 r(r −1) + 3r +1 = 0 r 2 − r + 3r +1 = 0 (r +1)2 = 0 Thus, we have two equal roots: r = −1. Thus, the first solution is y1 = C1 (x − 1)−1 To find the second solution, we use reduction of order given by Eq. (5.66). For our case, y1 = (x −1)−1 a1 = 3(x −1) a2 = (x −1)2 Noting that

 a (x)   (x −1)  1 exp − 1 dx = exp −3 dx = exp−3ln(x −1) =      (x −1)2  a (x) (x −1)3    2 

Eq. (5.66) becomes

 a (x)  y = C y (x) y−2 (x) exp − 1 + C y (x) 1 1 dx dx  1    2 1 a2 (x)   2 −1 −1 (x −1) = C1 (x −1)  dx + C2 (x −1) (x −1)3 = C1 (x −1)−1 ln(x −1) + C 2 (x −1)−1

Mathematica verification DSolveValue[(x-1)^2 y''[x]+3 (x-1) y'[x]+y[x]==0,y[x],x] 4.7 Find the solution to 4x2

d y 2

dx2

− 4x

dy

(

)

+ 3 − 16x2 y = 0

dx

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

when given that y1(x) = x1/2e2x .

Solution: Since a1(x) = −4x and a2(x) = 4x2, from Eq. (4.51) find that the second independent solution is  we 1 −4x  1/2 2x y (x) = C x e exp − 2 2  1/2 2 x 2   4x2 dx dx x e 1  e−4 x  1/2 2 x ln x 1/2 2 x −4 x 1/2 2 x = C2 x e  4 x e dx = C2 x e  e dx = C2 x e  − xe  4  = C3 x1/2e−2 x

(

)

Mathematica verification DSolveValue[4 x^2 y''[x]-4 x y'[x]+(-16 x^2+3) y[x]==0,y[x],x] 4.8 Find the solution to (2x + 1)x

d 2 y − 2(2x2 − 1) dy − 4(x + 1) y = 0 dx2 dx

when given that y1(x) = x−1.

Solution: Since a1(x) = −2(2x2 − 1) and a2(x) = (2x+1)x, from Eq. (4.51) we find that the second independent solution is  (2x2 − 1)  − dx  dx y2 (x) = C2 x 1  x2 exp2 x(2x2x + 1)    + 1 = C x−1 2 + ln dx  2  x exp2x x2   2x + 1 1 2 2x = C x− x e dx = C x−1 (2x + 1)e2x dx 2

= C x−1 2



(xe ) 2x

 x2  = C e2x

2



2

Mathematica verification DSolveValue[(2 x+1) x y''[x]-2 (2x^2-1) y'[x]-4 (x+1) y[x]==0, y[x],x]

Section 4.2.4

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

4.9 Find the solution to 2

x

d2y dx2

− 2 y = −2

with the boundary conditions y() = 0 and y(1) = 1 when   0.

Solution: The equation is a Cauchy-Euler equation; therefore, we assume a solution of the form yh = xr Substituting this expression into the homogeneous differential equation, we obtain x2r(r − 1)xr−2 − 2xr = 0 → r 2 − r − 2 = 0 → r = −1,2 The particular solution can be determined by inspection to give yp = 1. Then, the total solution is y = C x−1 + C x2 + 1 1

2

To determine the constants, we substitute this expression into the boundary conditions to arrive at y(1) = 1 = C1 + C2 + 1→ C1 = −C2

−1 − y( ) = 0 = C  −1 + C  2 + 1→ C = = 1 2 1  −1 −  2 1−  3

Then y = C1x−1 + C2 x2 + 1

(

)

= C 1 x−1 − x2 + 1 =

 x2 − x−1 + 1 1−  3

(

)

Mathematica verification DSolveValue[{x^2 y''[x]-2 y[x]==-2,y[e]==0,y[1]==1},y[x],x] 4.10 Find the solution to x2

d 2u du − 3x + 13u = 0 2 dx dx

when u(1) = −1 and du(1)/dx = 1.

Solution: We see that this is a Cauchy-Euler equation with a2 = 1, a1 = −3, and a0 = 13. Then, from Eq. (4.80)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

1 4  16 − 4  13 = 2  = 23j −9  2

r = 1,2

Using Eq. (4.60), the solution is u = x2  D cos(3ln x) + D sin(3ln x)  1 2  Substituting this result into the boundary conditions u(1) = −1 = D1 and du dx

= 2x  D cos(3ln x) + D sin(3ln x) + 3x  D cos(3ln x) − D sin(3ln x)  1 2   2 1  =x

(2D + 3D )cos(3ln x) + (2D − 3D )sin(3ln x) 1

2

2

1

Therefore, du(1) dx

= 2D + 3D = 1 → D = 1 1

2

2

The final result is u = x2 sin(ln x3) − cos(ln x3) Mathematica verification DSolveValue[{x^2 u''[x]-3 x u'[x]+13 u[x]==0,u[1]==-1, u'[1]==1},u[x],x] 4.11 Find the solution to 2   y = 0 d2 y dy + 1+ +2 dx2 dx  (1+ 3x)2    [Hint: Assume a solution of the form y = ue−x and then use the change of variables 1 + 3x = et.]

Solution: We note that when y = ue−x.  du  du − x e − ue− x = e− x −u  dx  dx dx d 2 y d  du − −x  −x d 2u du −x du − x = x e − ue = − e − e + ue   dx2 dx dx dx2 dx  dx 2   d u du = −x e  2 − 2 + u dx  dx  dy

=

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Then   2  ue−x = 0  d 2u du + u e−x + 2  du − u e−x + 1+ −2   dx2 dx  (1+ 3x)2   dx   2  d u 2 + u=0 2 dx (1+ 3x )2 Using the change of variable, we find that dt 3 t = ln(1+ 3x) → = dx 1+ 3x du du dt 3 du = = dx dt dx 1+ 3x dt d 2u d  du  dt du d 2t  d 2u dt  dt du d 2t = + = +   dx2 dx  dt  dx dt dx2  dt2 dx  dx dt dx2 d 2u  dt  2 du d 2t  3  2 d 2u  3  2 du = 2   + = − dt  dx  dt dx2  1+ 3x  dt2  1+ 3x  dt Then, d 2u

+

dx2 3  2 d 2u  3  2 du − +  1+ 3x  dt2  1+ 3x  dt

2

(1+ 3x )

2



9

2

(1+ 3x)

2

d 2u

du

dt

dt

2 −9

u=0 u=0

+ 2u = 0

To solve this equation, we assume a solution of the form ert, and obtain 9r 2 − 9r + 2 = 0 The roots of this equation are 1 1 9  81− 4  9  2  = (9  3) 1,2  18 18  1 2 r = , r = 1 2 3 3

r =

Therefore, u = C er1t + C er2t = C et/3 + C e2t/3 1

2

1

2

Converting back to our original variables, we arrive at

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(

)

y = ue− x = C1et /3 + C 2e2t /3 e− x

(

= C1 e

ln(1+3x)/3

+ C2 e

2/3ln(1+3x)

)e

−x

= C1eln(1+3x) + C2eln(1+3x) e− x   1/3

(

2/3

= C1 (1+ 3x) + C 2 (1+ 3x) 1/3

2/3

)e

−x

Mathematica verification DSolveValue[y''[x]+2 y'[x]+(1 + 2/(1+3 x)^2) y[x]==0,y[x],x]

Section 4.2.5 4.12 Obtain a solution to d 2w d

2

+

w

= − − k cos k1 2

and the boundary conditions w(0) = w(/2) = 0.

Solution: The homogeneous solution is wh = C1 cos + C2 sin For the particular solution, we use Eq. (4.65) and assume a solution of the form wp = A + B sin where we have chosen s = 1 since the homogeneous trigonometric function is also a solution to the homogeneous equation. Substituting this solution into the differential equation gives B ( 2 cos −  sin ) + A + B sin = −k1 − k2 cos 2B cos + A = −k1 − k2 cos A + k1 + (2B + k2 )cos = 0 Therefore, A = −k1 and B = −k2/2 and the solution is w = C1 cos + C2 sin − k1 − (k2 / 2) sin Substituting this result into the boundary conditions gives

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

dw

= −C sin + C cos − (k / 2)(sin +  cos ) 1 2 2 d dw(0) = C2 = 0 d dw( / 2) k k = −C1 − 2 = 0 → C1 = − 2 d 2 2 Then, the final result is w( ) = −k1 −

k2

(cos +  sin )

2

Mathematica verification Simplify[DSolveValue[{y''[x]+y[x]==-k1-k2 Cos[x],y'[0]==0, y'[Pi/2]==0},y[x],x]] 4.13 Find the solution to the following boundary value problem. d2y dy + 2 +y=x dx dx2 subject to y(0) = 0 and y(2) = 3.

Solution: We assume that the solution to the homogeneous equation is ex. Thus,

 2 + 2 + 1 = 0 → 1,2 =

(

)

1 −2  4 − 4 = −1 2

Therefore, since we have repeating roots, y = C e−x + C xe−x h

1

2

The particular solution is determined from inspection as yp = x − 2. Then y = y + y = x − 2 + C e−x + C xe−x h

p

1

2

From the boundary conditions at x = 0, we obtain y(0) = −2 + C1 = 0 → C1 = 2 and from the boundary conditions at x = 2, we obtain 3 y(2) = C e−2 + 2C e−2 = 3 → 2e−2 (1+ C ) = 3 → C = e2 − 1 2 2 1 2 2 Then the solution is

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 3 2  −x e − 1 xe  2      3 = x − 2 + e−x 2 + e2 −  2 1 x    

y = x − 2 + 2e−x +

Mathematica verification Collect[Expand[DSolveValue[{y''[x]+2 y'[x]+y[x]==x, y[0]==0,y[2]==3},y[x],x]],{Exp[-x],x}] 4.14 Find the solution to

d 2w

(

+  2w = −  − 2

2

d

)

subject to the boundary conditions w(0) = w(1) = 0.

Solution: The solution to the homogeneous equation is wh = Acos()+ B sin() To determine the particular solution, we use Eq. (4.64) and assume that w = c + c  + c 2 p

0

1

2

Upon substitution into the differential equation, we arrive at

( ) 2c +  c + ( c +  ) + ( c −  ) = 0

2c2 +  2 c0 + c1 + c 22 +  − 2 = 0 2

2

2

0

2

1

2

2

Equating the coefficients of the powers of  to zero, we obtain  2c2 −  = 0 → c2 =



2   2c1 +  = 0 → c1 = − 2 2c +  2c = 0 → c = − 2c2 = − 2 2 0 0 2 4 Then wp = −

2   − 2  + 2 2 4   

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

The solution is w = w + w = Acos ( ) + B sin ( ) − h

p

2

−

4



+

2



2

2

Substituting the solution in to the boundary condition at  = 0 gives 2 2 w(0) = A − =0→ A= 4 4 and that at  = 1 gives 2   w(1) = Acos( ) + B sin( ) − − + =0 4  2 2 2 2 2 B sin( ) = − cos( ) + = 1− cos( )

(

  2 1− cos( ) 4

B= The final result is

2  w=

4

4



)

4

sin( ) 1− cos( )



2





sin() − 4 − 2  + 2   cos () + 4  sin( )    

2

Mathematica verification Simplify[DSolveValue[{y''[x]+b^2 y[x]==-a (x-x^2),y[0]==0, y[1]==0},y[x],x]] 4.15 Determine the solution to

d2y

2 +

dx

6

dy dx

+ 9 y = e3x

when the initial conditions are y(0) = 0 and y(0) = 6.

Solution: Assume a solution to the homogeneous equation as ex. Then, 1  2 + 6 + 9 = 0 →  = −6  36 − 4  9  = −3 1,2  2 Since we have repeated roots, the solution to the homogeneous equation is yh = C1e−3x + C2 xe−3x For the inhomogeneous solution, we assume a solution of the form

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

y p = Ae3x Thus, from Eq. (a) A(9 + 3 6 + 9 ) e 3x = e3x → A =

1 36

Then the solution is

y = yh + yp = C1e

−3x

+ C2 xe

−3x

e3x + 36

Substituting this solution into the boundary conditions, we obtain 1 1 y(0) = C1 + = 0 → C1 = − 36 36 1 y(0) = −3C + C + = 6 → C = 6 − 1 − 1 = 35 1 2 2 12 12 12 6 Therefore, the final solution is

y=−

e−3x 36

−

35xe−3x 6

+

e3x 36

Mathematica verification DSolveValue[{y''[x]+6 y'[x]+9 y[x]==Exp[3 x],y[0]==0, y'[0]==6},y[x],x] 4.16 Find the solution to 1 d  du  r  = −r k  r dr  dr 

0r

Solution: The homogeneous solution is d  duh  r =0 dr  dr  duh c1 = dr r uh = c1 ln r + c2 The particular solution is obtained by assuming a solution of the form up = Ar. substitution yields

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

d  du  r = −r k +1   dr  dr  d

(r Ar ) = −r dr  −1



d

k+1

( Ar ) = −r dr 

k+1

A 2r −1 = −r k +1 Thus, we that for these quantities to be equal A 2 = −1 and  − 1 = k + Therefore,

 = k + 2 and A =

−1 2

=

−1 (k + 2)2

The complete solution is u = uh + up = c2 + c1 ln r +

−r k +2 (k + 2)2

Mathematica verification DSolveValue[D[r u'[r],r]==-r^(k+1),u[r],r] Section 4.2.6 4.17 Use the variation of parameters to find the particular solution to 2 −x d2y dy +2 + y = e dx x2 dx2 Solution: We shall use the variation of parameters method. The homogeneous solution is obtained by assuming a solution of the form yh = ex. Then substitution yields

2 + 2 +1 = 0 → ( +1)2 = 0 Thus, we have repeating roots so that y1 = e−x y2 = xe−x y (x) = C e−x + C xe−x h

1

2

and

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

y1 W[ y1, y2 ] = dy1 dx

y2 e− x dy2 = − x −e dx

xe− x e−x (1− x)

= e−2 x

Then from Eq. (68) with a2 = 1 y2 (x)Q(x) y1 (x)Q(x) y(x) = C1 y1 (x) + C2 y2 (x) − y1 (x)  a (x)W[ y , y ] dx + y2 (x)  a (x)W[ y , y ] dx 2

1

2

2

1

2

−x −x

−x −x

e e xe e dx + 2xe−x  dx 2 −2 x xe x2e−2 x 1 −x 1 = C1e−x + C2 xe−x − 2e−x  dx + 2xe  2 dx x x  −1  = C e−x + C xe−x − 2e−x ln x + 2xe−x 1 2  x    −x −x −x = C1e + C 2 xe − 2e (ln x +1) = C1 e−x + C2 xe − x − 2e−x 

Mathematica verification DSolveValue[y''[x]+2 y'[x]+y[x]==2 Exp[-x]/x^2,y[x],x] 4.18 Use the variation of parameters to find the particular solution to dy d2y − 2 + y = 14x3/2ex 2 dx dx Solution: Assuming a solution of the y = ex, we find that  is determined from 1  = 2  4 − 4  = 1 1,2  2 Since we have repeating roots, the homogeneous solution is y = C ex + C xex h

1

2

and therefore, y (x) = ex , 1

y (x) = xex 2

We shall use Eq (4.79) to find the particular solution. From the differential equation, we see that a2(x) = a0(x) = 1 and a1(x) = −2. First, we compute W using Eq. (4.77). Then, W [ y1, y2 ] =

ex ex

xex = e2 x (1+ x) − xe2x = e2 x x e (1+ x)

From Eq. (4.79), we have

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

ex x3/2ex xex x3/2ex x dx + 14xe  e2 x dx e2 x = −14ex  x5/2 dx + 14xex  x3/2 d

yp (x) = −14ex 

= −4x7/2ex +

28 7/2 x 8 7/2 x x e = x e 5 5

Mathematica verification DSolveValue[y''[x]-2 y'[x]+y[x]==14 x^(3/2) Exp[x],y[x],x] 4.19 Use the variation of parameters to find the particular solution to x

d2y

+ 2(1− x)

2

dy dx

dx

+ (x − 2) y = e2x

where the independent solutions to the homogeneous equation are y (x) = ex ,

y (x) = x−1ex

1

1

Solution: We shall use Eq (4.79) to find the particular solution. From the differential equation, we see that a2(x) = x, a1(x) = 2(1−x), and a0(x) = x−2. First, we compute W using Eq. (4.77). Then, W [ y1, y 2 ] =

ex

x−1ex

ex

ex − x−2 + x−1

(

)

(

)

= e2 x − x−2 + x−1 − x−1e2 x = − x−2e2 x

From Eq. (4.79), we have ex e2 x x−1exe2 x 1 x − yp (x) = e  −2 2 x dx − x e  −2 2 x dx xx e xx e x

= ex  ex dx − x−1ex  xex dx = e2x − x−1e2x (x − 1) = x−1e2 x Mathematica verification DSolveValue[x y''[x]+2(1-x)y'[x]+(x-2) y[x]==Exp[2 x],y[x],x] 4.20 Use the variation of parameters to find the particular solution to d2y dy 4x2 − 4x + 4x2 + 3 y = x7/2 dx dx2

(

)

where the independent solutions to the homogeneous equation are y1(x) = x sin x,

y2 (x) =

x cos x

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution: We shall use Eq (4.79) to find the particular solution. From the differential equation, we see that a2(x) = 4x2, a1(x) = −4x, and a0(x) = 4x2+3. First, we compute W using Eq. (4.77). Then, x sin x x cos x sin x cos x W [ y1, y2 ] = x cos x + − x sin x + 2 x 2 x   cos x  sin x  = x sin x − x sin x + − x cos x x cos x +    2 x  2 x  = −x From Eq. (4.79), we have xx7/2 cos x xx7/2 sin x dx + x cos x  4x2 (− x) dx 4x2 (− x) x x = sin x  x cos x dx − cos x  x sin x dx 4 4 x x = sin x (cos x + x sin x ) − cos x (sin x − x cos x) 4 4 3/2 x = 4

yp (x) = − x sin x 

Mathematica verification DSolveValue[4 x^2 y''[x]-4 x y'[x]+(4 x^2+3) y[x]== x^(7/2),y[x],x]

Section 4.2.7 4.21 Convert the following system of nonlinear equations and boundary conditions to a system of first-order differential equations.  df  2 d3 f d2 f − 2 +3f +T* = 0  d  d3 d2 dT * d 2*T * + 3Pr f =0 d2 d The initial conditions at  = 0 are

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

d2 f

f =0

df =0 d

T* =1

dT * = −b d

d2

=a

where a and b are constants.

Solution: We introduce the following set of dependent variables y1 = f df y2 = d y3 =

y4 = T * dT * y5 = d

d2 f d2

Then, the system of first-order equations for these new variables is dy1 dy4 = y2 = y5 d d dy2 dy5 = y3 = −3Pr y1y 5 d d dy3 = 2y2 − 3y y − y 1 3 4 2 d and the corresponding initial conditions become y1(0) = 0

y4 (0) = 1

y2 (0) = 0 y3(0) = a

y5 (0) = −b

Section 4.2.10 4.22 Determine a solution to the nonlinear equation  d2y 2 2 −1 x2 x ( )  dx2  − 1 = 0 Solution: We use Eq. (4.115) to obtain 2

 dp  x2 − 1 x2  −1 = 0  dx 

(

)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

or p=

dx

 x x2 − 1 = − tan

−1

1 x2 − 1

+ c1

From Eq. (4.111), we find that y = −  tan−1 Upon integrating, we obtain



1 x −1 2

1  1  + ln 1 − y = c2 + c1x − x tan  2  x − 1  2  −1

dx + c1x

x



 x2 − 1 

−

 ln 1 + 2  1

x



 x2 − 1 

Mathematica verification Simplify[PowerExpand[DSolveValue[(x^2-1) x^2 y''[x]^2-1==0, y[x],x]//Quiet]]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 5 Section 5.1.2 5.1 Use the power series method to obtain a solution to 2 dy 2 d y 1− x − 5x − 4 y = 0 dx dx2

(

)

Solution: We see that this differential equation is of the form given by Eq. (5.22) with x0 = 0,  = −1,  = −5, and  = −4. Then, from Eq. (5.26) p(n) = −n(n − 1) − 5n − 4 = −(n + 2)2 Then, Eq. (5.31) yields c 2 n+2 =

2n + 2 c 2n + 1 2n

n = 0,1,2,... )

c

2n + 3 c 2n + 2 2 n+1

n = 0,1,2,...

and Eq. (5.32) yields 2 n+3

=

Expanding these quantities for a few of the even coefficients, we get c2 = 2c0 4 8 c = c = c 4 3 2 3 0 6 68 16 c = c = c = c 6 5 4 3 5 0 5 0 8 16  8 128 c8 = c6 = c0 = c 7 5 7 35 0 and expanding those for the odd coefficients, we get 3 c = c 3 2 1 5 3 5 15 c = c = c = c 5 4 3 24 1 8 1 7 7  15 35 c7 = c5 = c1 = c1 6 68 16 9 9  35 315 c9 = c7 = c1 = c 8 8  16 128 1

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Using Eq. (5.24), the solution is     3 3 15 5 35 7 y(x) = 8 4 16 6 128 8 2 x +! + c1  x + x + x + x +! c0 1+ 2x + x + x + 3 5 35 2 8 16     Mathematica verification Use procedure M5.1 5.2 Use the power series method to obtain a solution to d2y dy 2 + 4x − 2x2 − 12(x − 1) − 12 y = 0 dx dx2

(

)

Show that the solution can be written as 

 n+1 n 2n + 1 n +1 x − 1 ( ) y(x) = c0  n + c1 n ( x − 1) 2 n=0 n=0 2

Note that 2 + 4x − 2x2 = 4 − 2(x − 1)2 Solution: We see that this differential equation is of the form given by Eq. (5.22) with x0 = 1,  = −1/2,  = −3, and  = −3. Then, from Eq. (5.26) 1 1 p(n) = −n(n − 1) / 2 − 3n − 3 = − n2 + 5n + 6 = − (n + 3)(n + 2) 2 2

(

)

Then, Eq. (5.31) yields (2n + 3)(2n + 2) c 2(2n + 2)(2n + 1) 2n 2n + 3 = c n = 0,1,2,...

2n 2(2n + 1)

c2n+2 =

and Eq. (5.32) yields (2n + 4)(2n + 3) n = 0,1,2,... c 2(2n + 3)(2n + 2) 2n+1 n+2 = c2n 1 n = 0,1,2,...

2(n + 1) +

c2n 3 = +

Expanding these quantities for a few of the even coefficients, we get

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

3 c = c 2 2 0 5 5 c = c = 5 3 c = c 4 6 2 62 0 4 0 7 7 c = c = 75 c = c 6 10 2 10  4 0 8 0 ⁝ 2n + 1 c2 n = n c0 n = 0,1,2,...

2 and expanding those for the odd coefficients, we get c3 = c1 3 3 c = c = c 5 4 3 4 1 4 4 c = c = 43c = c 7 6 5 64 1 8 1 5 5 c = c = 5 4 c = c 9 8 7 8  8 1 16 1 ⁝ n +1 c2 n+1 = n c1 n = 0,1,2,...

2 Using Eq. (5.24), the solution can be written as 

 n+1 n 2n + 1 n +1 x − 1 ) y(x) = c0  n ( + c1 n ( x − 1) 2 n=0 n=0 2

or in expanded form

y(x) =



3

5

2

7

3



c0 1+ (x − 1)+ (x − 1) + (x − 1) +!   2 4 8  2 3 3   3 4 + c ( x − 1)+ ( x − 1) + ( x − 1) + ( x − 1) + !

1   4 8

Mathematica verification coe=ConstantArray[0,12]; coe[[1]]=1; coe[[2]]=3/2; coe[[3]]=1; Do[coe[[2 n+2]]=(2 n+3)/(2 (2 n+1)) coe[[2 n]],{n,1,4}]; Do[coe[[2 n+3]]=( n+2)/(2 (n+1)) coe[[2 n+1]],{n,1,4}]; y=c0 (1+Total[Table[coe[[2 n]] (x-1)^(2 n),{n,1,4}]]) +c1 Total[Table[coe[[2n -1]] (x-1)^(2 n-1),{n,1,4}]]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Section 5.1.3 5.3 Use the method of Frobenius to show that the solution to d2y dy 4x 2 + 2 + y = 0 dx dx is y = C1 cos

( x ) + C sin( x ) 2

Solution: We assume a solution of the form 

y(x) =  c n xn+r n=0

Substitution leads to 

4x  cn (n + r)(n + r −1)x n=0





+ 2  c n (n + r)x

+  c nxn+r = 0

 c 4(n + r)(n + r −1) + 2(n + r) x

+  c xn+r = 0

n+r−2



n+r−1

n=0

n+r−1

n

n=0

n=0 

n

n=0 

 c 2(n + r)2(n + r) −1 xn+r−1 +  c xn+r = 0 

n n=0 





n n=0 

2r(2r −1)c xr−1 +  c 2(n + r)2(n + r) −1 xn+r−1 +  c xn+r = 0 0 n n   n=1

 2r(2r −1)c xr−1 +  c 0 n=0

n=0

2(n +1+ r)2(n +1+ r) −1 + c xn+r = 0 n+1   n

Hence, the indicial equation is

2r(2r −1) = 0 → r = 0 and r = 1/ 2 The roots are unequal and their difference in not equal to an integer.

The solution for r = 0 is

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

cn+1 = − c1 = −

cn

n = 0,1,2...

2(n + 1) 2(n + 1) − 1 c0 2(1) c1

c0 2 (4)(3) (4)(3)(2)(1) c0 c c =− 2 =− 3 (6)(5)(4)(3)(2)(1) (6)(5)

c =−

=

⁝ cn =

(−1)n c0 (2n)!

The solution for r = 1/2 is dn+1 = − d1 = −

dn

2(n + 3 / 2) 2(n + 3 / 2) − 1 d0 (3)(2) d1

=−

dn (2n + 3)(2n + 2)

d0 (5)(4) (5)(4)(3)(2) d d0 d =− 2 =− 3 (7)(6) (7)(6)(5)(4)(3)(2) ⁝

d =− 2

dn=

=

(−1)n d0 (2n + 1)!

Therefore, the complete solution is 



y(x) =  c nx +  d n xn+1/2 n

n=0

n=0

 n n = c  (−1) x + d x1/ 2 (−1) x  (2n +1)!

0 0 (2n)! n=0 n=0 

n n

But, 

(−1)n x2 n+1 sin x =  + 1)! n=0 (2n n 2n  (−1) x

cos x =  n=0

(2n)!

and, hence,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab



n n

sin x = x1/2 (−1) x  n=0 (2n + 1)!  (−1)n xn cos x =  (2n)! n=0

and the solution can be written as y(x) = C1 sin x + C2 cos x Mathematica verification DSolveValue[4 x y''[x]+2 y'[x]+y[x]==0,y[x],x] 5.4 Use the method of Frobenius to obtain the solution to 2

d y

dy + (4 − x) y = 0 − 3x dx2 dx To place your solution a compact form, note the following: 2

x

 1 11  1 1 25  1 1 1  = 1+ , = 1+ + , = 1+ + + !

2  2  6  2 3 12  2 3 4   1 1 1 g(n) = 1+ + +!+ (define this function as stated)  2 3  n  3

Solution: Comparing this equation with Eq. (5.78), it is seen that

0 = 1,

1 = 0,

2 = 0

0 = −3,  0 = 4,

1 = 0,  1 = −1,

2 = 0 2 =0

Then, from Eq. (5.89)

p0(r) = r(r − 1) − 3r + 4 = (r − 2)(r − 2) = 0 p1(r) = −1 p2(r) = 0 From the indicial equation, p0(r) = 0, we find that r0 = r1 = r2 = 2 and, therefore, we have two equal roots. Hence, the solutions are given by Eqs. (5.101) and (5.104); that is, y (x) =  a (r )xn+r0 

1

n

n=0

0



y (x) = y (x) ln x +  a(r )xn+r0 2

1

n

0

n=1

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

where, from Eq. (5.97) a0(r) = 1 p1(r)

a (r) = − 1

=

1

(r − 1)2 p (n + r − 1)an−1(r) + p2 (n + r − 2)an−2 (r) an (r) = − 1 p0(n + r) p0 (1+ r)

=

an−1(r) an−1(r) = p0(n + r) (n + r − 2)2

n2

Then a0(2) = 1 a1(2) = 1 a (2) 1 a (2) = 1 = 2 22 22 a2 (2) 1 a (2) = = 3 32 2232 a (2) 1 a (2) = 3 = 4 42 223242 ⁝ 1 a (2) = n (n!)2 and y1(x) is 



xn+2

n=0

(n!)2

y (x) =  a (2)xn+2 = a  n

1

= a x2 I (2 x )

0

n=0

0

0

and we have used Eq. (5.152).

The derivatives are a0(r) = 0 a(r) =

−2

(r − 1)3 2an−1(r) a(r) = − + an−1(r) n (n + r − 2)3 (n + r − 2)2 1 = −2a (r) + (n + r − 2)a (r) n−1 n−1 3 (n + r − 2) 1

(

)

n2

Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

a0(2) = 0 a1(2) = −2 1 a(2) = −2a (2) + 2  a(2) 2 3 1 1 2 1 3 1  1 = −2  1+ 2  (−2) = − = − 1+ 23 22 2  2 

(

)

(

a(2) = 3

)

(−2a (2) + 3  a(2))

1 3

3 1

2

2

1  3  11 1  1 1 −2  + 3 =− =− 1+ +   − 2  3  2 2 3 2   2  3  2 23 2  3  2 3 1 a(2) = −2a (2) + 4  a(2) 4 3 3 3 4  1 1  11   50 1  1 1 1 = −2  +4 =− =− 1+ + +  − 2 2  43   2233   2232 223343  2 3 4  23 4 =

(

⁝ a(2) = − n

=−

)



1 1 1 1+ + +"+ n  12  22  32"n2  2 3 2

2 g(n) (n!)2

where n

g(n) =  k =1

Then, from Eq. (5.104)

1 k 

y2 (x) = y1(x) ln x − 2 n=1

g(n) (n!)2

n+2

x

Mathematica verification DSolveValue[x^2 y''[x]-3 x y'[x]+(4-x) y[x]==0,y[x],x] 5.5 Use the method of Frobenius to obtain the solution to

(5 + x +

10x 2

d2y

dy

)x dx2 + (4 + 3x + 48x 2 ) x dx + ( x + 36x2 ) y = 0 2

Solution: Comparing this equation with Eq. (5.78), it is seen that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

0 = 5,

1 = 1,

 2 = 10

0 = 4,

1 = 3,

2 = 48

 0 = 0,

 1 = 1,

 2 = 36

Then, from Eq. (5.81) p0 (r) = 5r(r − 1) + 4r = r 2 − r = r(5r − 1) p1 (r) = r(r − 1) + 3r + 1 = r 2 + 2r + 1 = (r + 1)2 p2 (r) = 10r(r − 1) + 48r + 36 = 10r 2 + 38r + 36 = 10(r + 2)(r + 9 / 5) From the indicial equation, p0(r) = 0, we find that r1 =1/5 and r2 = 0 and since r1 − r2 = 1/5 is not a positive integer the roots will be distinct. Hence, the solutions are given by Eqs. (5.92) and (5.93); that is, y (x) =  a (1 / 5)xn+1/5 

1

n

n=0 

y (x) =  a (0)xn 2

n

n=0

Using Eq. (5.88), we find that a0(r) = 1 a (r) = − 1

p1(r)

=−

p0(1+ r)

an (r) = − =− =−

(r + 1)2 (r + 1) =− (5r + 4) (r + 1)(5r + 4)

p1(n + r − 1)an−1(r) + p2 (n + r − 2)an−2 (r) p (n + r) 0 2 (n + r) a (r) + 10(n + r)(n + r − 1 / 5)a (r) n−1

(n + r)(5n + 5r − 1)

(n + r)an−1(r) + 10(n + r − 1 / 5)an−2 (r) (5n + 5r − 1)

n−2

n2

Then, for r = 1/5

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

a0(1 / 5) = 1 a (1 / 5) = −

(r + 1)

=−

(5r + 4)

1

a2(1 / 5) = −

6/5 5

=−

6 25

(2 + 1 / 5)(−6 / 25) + 10(2 + 1 / 5 − 1 / 5) (5  2 + 5 / 5 − 1)

(11 / 5)(−6 / 25) + 20 1217 =− 10 625 (3 + 1 / 5)(−1217 / 625) + 10(3)(−6 / 25) a3(1 / 5) = − 15 (16 / 5)(−1217 / 625) − 36 / 5 41972 =− = 15 46875 =−

and for r = 0 a0(0) = 1 a (0) = −

(r + 1)

=−

1

(5r + 4) 4 2(−1 / 4) + 10(9 / 5) 35 a2 (0) = − =− 9 18 3(−35 / 18) + 10(14 / 5)(−1 / 4) −35 / 6 − 7 11 a3(0) = − =− = 14 14 12 1

Using Eq. (c), we find that  6 1217 2 41972 3  y (x) = x1/5 1− x − x + x ... 1  25 625 46875  1 35 2 11 3 y (x) = 1− x − x + x ...

2 4 18 12 Mathematica verification p[j_,r_]:=r (r-1) alp[[j+1]]+r bet[[j+1]]+gam[[j+1]] c[k_,r_]:=(a=Array[0,k]; a[[1]]=1; a[[2]]=-p[1,r]/p[0,r+1]; Do[a[[n+1]]=-(p[1,n+r-1] a[[n]]+p[2,n+r-2] a[[n-1]])/p[0,n+r], {n,2,k-1}];a) alp={5,1,10};

bet={4,3,48};

gam={0,1,36};

rj=Sort[r/.Solve[p[0,r]==0,r],Greater]; k=5;

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

xa=Table[x^n,{n,0,k-1}]; x^rj[[1]] Total[c[k,rj[[1]]] xa] x^rj[[2]] Total[c[k,rj[[2]]] xa] 5.6 Use the method of Frobenius to obtain the solution to

(

d2y

dy

) dx2 + (1 + 3x + 4 x 2 ) x dx + (− x + 2x2 ) y = 0

1 + 2x + x2 x 2

Solution: Comparing this equation with Eq. (5.78), it is seen that

0 = 1,

1 = 2,

2 = 1

0 = 1,  0 = 0,

1 = 3,  1 = −1,

2 = 4 2 =2

(a)

Then, from Eq. (5.81) p0 (r) = r(r − 1) + r = r 2 p1(r) = 2r(r − 1) + 3r − 1 = 2r 2 + r − 1 = (2r − 1)(r + 1)

(b)

p2(r) = r(r − 1) + 4r + 2 = r + 3r + 2 = (r + 1)(r + 2) 2

From the indicial equation, p0(r) = 0, we find that r1 = r2 = 0 and we have the case of equal roots. Hence, the solutions are given by Eqs. (5.92) and (5.95); that is, y (x) =  a (0)xn 

1

n

n=0

(c)



y (x) = y (x) ln x +  a (0)xn 2

1

n

n=1

where the prime denotes the derivative with respect to r. Using Eq. (5.88), we find that a0(r) = 1 a1 (r) = −

p1(r) (2r − 1)(r + 1) (2r − 1) =− =− 2 p0(1+ r) (1+ r) (1+ r)

an (r) = −

(2n + 2r − 3)(n + r)an−1(r) + (n + r − 1)(n + r)an−2 (r) (n + r)2

=−

(2n + 2r − 3)an−1(r) + (n + r − 1)an−2 (r) (n + r)

n2

Then, for r = 0, Eq. (d) gives

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(d)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

a0(0) = 1 a1(0) = 1 a (0) = −

(4 − 3) + 1

= −1

(e)

2

2 (6 − 3)(−1) + (3− 1) −3+ 2 1 a3(0) = − =− = 3 3 3 To determine an, we note that a0 (r) = 0 a(r) = − 1

a (r) = − n

=−

d  (2r − 1)  3 =− dr  (1+ r)  (1+ r)2 d  (2n + 2r − 3)an−1(r) + (n + r − 1)an−2 (r)  dr  1

(n + r) (r) + 3a (r)

(a

)

(f)



(n + r) 1 − (2n + 2r − 3)a (r) + (n + r − 1)a (r) n−1 n−2 n+r 2

n−2

n−1

(

)

Then, from Eq. (f) a0 (0) = 0 a1(0) = −3 1 1 a (r) = − a (0) + 3a (0) − (2n − 3)a (0) + (n − 1)a (0) n−1 n−1 n−2 n n n2 n−2

) (

(

)

Therefore, a0 (0) = 0 a1(0) = −3 1 1 a(0) = − a (0) + 3a (0) − (4 − 3)a(0) + (2 − 1)a(0) 2 1 1 0 2 22 0 1+ 3 1 3 1 =− − (−3) = −1+ = 4 2 2 2 1 1 a(0) = − a (0) + 3a (0) − (6 − 3)a(0) + (3 − 1)a(0) 3 2 2 2 1 3 3 1  2 3 31 1 1  1 = − 1+ 3(−1) − 3 + 2(−3) = + =  9 2 18 3   2 9

(

) (

(

(

) (

)

)

)

Using Eqs. (e) and (g) in Eq. (c), we obtain

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(g)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

x3 ...

3 x 31 2  y (x) = y (x) ln x + x −3 + + x ...

2 1   2 18 y1 (x) = 1+ x − x2 +

Mathematica verification p[j_,r_]:=r (r-1) alp[[j+1]]+r bet[[j+1]]+gam[[j+1]] c[k_,r_]:=(a=Array[0,k]; a[[1]]=1; a[[2]]=-p[1,r]/p[0,r+1]; Do[a[[n+1]]=-(p[1,n+r-1] a[[n]]+p[2,n+r-2]* a[[n-1]])/p[0,n+r],{n,2,k-1}]; da=Array[0,k]; da[[1]]=0; da[[2]]=D[-p[1,v]/p[0,v+1],v]/.v->r; Do[da[[n+1]]=-(p[1,n+r-1] da[[n]]+p[2,n+r-2] da[[n-1]])/ p[0,n+r]+D[-(p[1,n+v-1] a[[n]]+p[2,n+v-2] a[[n-1]])/ p[0,n+v],v]/.v->r,{n,2,k-1}]; {a,da}) alp={1,2,1};bet={1,3,4};gam={0,-1,2}; rj=Sort[r/.Solve[p[0,r]==0,r],Greater]; k=5; (*k term solution*) xa=Table[x^n,{n,0,k-1}]; {a,da}=c[k,rj[[1]]]; x^rj[[1]] Total[a xa] x^rj[[1]] Total[da xa] 5.7 Use the method of Frobenius to obtain the solution to 2

dy

dx2

dx

(1 + x) x2 d y − (3 + 10x) x

+ 30xy = 0

Solution: Comparing this equation with Eq. (5.78), it is seen that

0 = 1, 0 = −3,  0 = 0,

1 = 1, 1 = −10,  1 = 30,

2 = 0 2 = 0 2 =0

(b)

Then, from Eq. (5.81)

p0(r) = r(r − 1) − 3r = r 2 − 4r = r(r − 4) p1(r) = r(r − 1) − 10r + 30 = r 2 − 11r + 30 = (r − 6)(r − 5) p2(r) = 0

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(c)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

From the indicial equation, p0(r) = 0, we find that r1 = 4 and r2 = 0 and r1 − r2 = 4 is a positive integer. In addition, 2 = 2 = 2 = 0; hence, this is Case A of Case 3. The solutions are given by Eqs. (5.92) and (5.101); that is, y (x) =  a (r )xn+r1 

1

n

1

n=0 k −1

y (x) =  a (r )x 2

n

2

n=0

n+r

2

  n+r  + C y (x) ln x +  a(r )x 1 n 1   1 

(e)

n=1

where the prime denotes the derivative with respect to r and from Eqs. (5.100) and (5.102) we have that

a0(r) = 1 p1(n + r − 1)an−1(r) (n + r − 7)(n + r − 6)an−1(r) =− p0(n + r) (n + r)(n + r − 4)

n 1

(f)

p1(r1 − 1) (r − 7)(r1 − 6) (4 − 7)(4 − 6) ak −1 (r2 ) = − 1 a3 (r2 ) = − a3 (0) 4 4 k0 3 = − a (0) 2 3

(g)

an(r) = − and C=−

Then, for r = r1 = 4, a0(4) = 1 a1(4) = −

(1+ 4 − 7)(1+ 4 − 6)a0 (4) (−2)(−1)(1) 2 =− =− (1+ 4)(1+ 4 − 4) 5 5

(2 + 4 − 7)(2 + 4 − 6)a2 (r) =0 a2(4) = − (2 + 4)(2) an (r) = 0

n2

and for r = r2 = 0, a0(0) = 1 an (0) = −

(n − 7)(n − 6)an−1(0) n(n − 4)

1 n  k −1= 3

Therefore,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(h)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

a0(0) = 1 a1(0) = −

(1− 7)(1− 6)a0 (0) (−6)(−5)(1) =− = 10 (1− 4) −3

a2 (0) = −

(2 − 7)(2 − 6)a1(0) (−5)(−4)(10) =− = 50 2(2 − 4) −4

a3 (0) = −

(3 − 7)(3 − 6)a2 (0) (−4)(−3)(50) =− = 200 3(3 − 4) 3(−1)

(i)

Thus, from Eq. (g),

3 3 C = − a (0) = − (200) = −300 2 3 2

(j)

We now use eq. (f) to determine an(4) as follows a0 (4) = 0 d  (n + r − 7)(n + r − 6)an−1(r)  dr  (n + r)(n + r − 4)  r=4

an (4) = −

(k)

(n − n − 14n + 24n)a (4) + 3a (4)(3n − 4n − 8) 4

=−

3

2

2

n−1

n−1

n2(n + 4)2

n 1

Then, using Eqs. (k) and (h) a0 (4) = 0

(1− 1− 14 + 24)(0) + 3(1)(3 − 4 − 8) = 27

a(4) =− 1

(1+ 4)2

25

(2 − 2 − (14)2 + 24(2))(27 / 25) + 3(−2 / 5)(32 − 4(2) − 8) a  (4) = − 4

3

2

2

22(2 + 4)2

2

(16 − 8 − 56 + 48)(27 / 25) − (6 / 5)(1024 − 16) = − 1 =−

30

4(36)

(3 − 3 − (14)3 + 24(3))(−1 / 30) + 3(0)((3)3 − 4(3) − 8) = 0 4

a3(4) = − an (4) = 0

3

2

2

32(3 + 4)2

n3

Using Eqs. (g), (i), (j), and (k), Eq. (e) becomes

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(k)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab



y1(x) =  an (4)x n=0 3

n+4

n

2 5 = x4 − x 5 



n+4



y2 (x) =  an (0)x + C  y1(x) ln x +  an (4)x    n=0 n=1  = 1+ 10x + 50x2 + 200x3 − 27 5 1 6  x − x 300 y1(x) ln x +  25 30  Mathematica verification tes[arg_]:=Expand[(1+x) x^2 D[arg,x,x]-(3+10 x) x D[arg,x] +30 x arg] arg=x^4-2/5 x^5; tes[arg] arg2=1+10 x+50 x^2+200x^3-300(arg Log[x]+27/25 x^5-1/30 x^6); tes[arg2] Section 5.1.8 5.8 A fourth-order differential equation can be factored into the following two second-order differential equations d  n+1 dW     2 nW = 0 d  d  Determine the complete solution; that is, one that contains four independent solution functions.

Solution: Expanding the equation gives n+1

d 2W

+ (n + 1)n

dW

  2nW = 0

2

d 2 dW   2W = 0  2 d W + (n + 1) 2 d d d

Comparing this equation with case 2 of Table 5.2, we see that b 2 k 2 x2b =  2 → b = 1 / 2 and k = 2 or k = 2 j n + 1 = 1 − 2ab → a = −n r = 0 → p2 = a2 → p = n and, therefore, from this table

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

W ( ) =  −n/2 C J (2  ) + C Y (2  ) + 2 n  1 n  W ( ) =  −n/2 C J (2 j  ) + C Y (2 j  ) − 4 n  3 n  =  −n/2 C I (2  ) + C K (2  ) 4 n  3 n  Thus, W ( ) =  −n/2 C J (2  ) + C Y (2  ) + C I (2  ) + C K (2  ) 2 n 3 n 4 n  1 n  Mathematica verification DSolveValue[x^2 w''[x]+(n+1) x w'[x]+lam^2 x w[x]==0,w[x],x] DSolveValue[x^2 w''[x]+(n+1) x w'[x]-lam^2 x w[x]==0,w[x],x] 5.9 Obtain the solution to the following equation that remains finite at x = 0.

2

2

x

d y 2

dx

+ 2x

dy + h2 xy = 0 dx

Solution: Comparing this equation with case 2 of Table 5.2, we see that b2k 2 x 2b = h2x → b = 1 / 2 and k = 2h 1− 2ab = 2 → 1− a = 2 → a = −1 In addition, since r = 0, p = 1. Hence, from the table, the solution that remains finite at x = 0, is y(x) = C (2h)−1 x−1/2 J (2hx1/2 ) 1

1

To show that this solution is finite at x = 0, we use Eq. (5.148). Thus 1/2 lim y(x) = C (2h)−1 x−1/2 1  2hx 

1!

1

x→0

2

1



=

C1 2

Mathematica verification DSolveValue[x^2 w''[x]+2 x w'[x]+h^2 x w[x]==0,w[x],x] 5.10 Given the equation 2d

2

y

x dx2

+x

dy

− ( jx + n ) y = 0 2

2

dx

where n is an integer. Show that one solution to this equation is

y = Bern x + jBein x where

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab



(−1)m+n (x / 2)2m+n

cos( (2m + n) / 4) m!(m + n)! (−1)m+n+1 (x / 2)2m+n sin ( (2m + n) / 4) Bein x =  m!(m + n)! m=0 Bern x =  m=0 

The quantities Ber and Bei, respectively, are known as the real and imaginary parts of the Kelvin function of the first kind. [Hint: The exponential form of −j will prove beneficial.]

Solution: From Case 2 of TA ble 5.2, we see that k2 = −j, a = 0, b = 1, and p = n. Then,

k 2 = − j = e j e j /2 → k = e j /2e j /4 = e j3 /4 and one solution is

y(x) = xab C1 J p(kxb ) + C 2Y p(kxb ) → J n (xe j3 /4 ) From Eq. (5.86), 

J (xe

j3 /4

n

(−1)m (x / 2)2 m+n (e j 3 /4 )2 m+n )= m!(m + n)! m=0

However, e j3 / 4 = e j e j (3 / 4− ) = e j e− j / 4

(e

j3 / 4

)

2m+n

= e j (2m+n)e− j (2m+n)/ 4 = e j 2m e jn e− j (2m+n)/ 4 = (−1)n e− j (2m+n)/ 4

Therefore, 

(−1)m (x / 2)2m+n (−1)n e− j (2m+n)/ 4 m!(m + n)! m=0  (−1)m+n (x / 2)2m+n =  m!(m + n)! cos ( (2m + n) / 4) − j sin ( (2m + n) / 4) m=0

Jn (xe j 3 / 4 ) = 

Hence,



y = Bern x + jBein x = 

m=0

or

(−1)m+n (x / 2)2m+n m!(m + n)! 

cos ( (2m + n) / 4) − j sin ( (2m + n) / 4)

(−1)m+n (x / 2)2m+n

cos( (2m + n) / 4) m!(m + n)! (−1)m+n+1 (x / 2)2m+n sin ( (2m + n) / 4) Bein x =  m!(m + n)! m=0 Bern x =  m=0 

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

5.11 Consider the equation 2 dy 2d y x + a + 2bx x + c +  2x2 + b(a +  − 1)x + b2x2 y = 0 dx dx2

(

)

(

)

where a, b, c, , , and  are real constants. Make the two transformations sequentially   y = x(1−a)/2e−bx /u(x) and z = x  and show that the resulting equation is a Bessel equation. Use Mathematica to obtain dy/dx and d2y/dx2. It will be found that the order p of the Bessel function is 2 1  a − 1 1/2  − c p =    2    Solution: From Mathematica,  du  x(−1−a )/ 2e−bx / 2x + (1− 2bx − a ) u  dx  dx 2 d 2 y 1 (−3−a)/ 2 −bx /  2 d 2u (1 2bx  − a ) 4x du 2 = 2 + x e − 4x dx 4 dx dx  dy



=

1





+(−1+ 4abx + 4b2 x2 + a2 − 4bx )u 

Using this result in the original equation, we obtain   1 d 2u

(

du

)

x2 x( −3−a )/2e− bx / 4x2 2 + 1− 2bx − a 4x 4 dx dx 

(

)

+ −1+ 4abx + 4b2x2 + a2 − 4bx u  1 (−1−a)/2 −bx /  du + x a + 2bx x e 2x + 1− 2bx − a u   dx 2 

(

)

(

)

(

)



+ c +  2x2 + b(a +  − 1)x + b2x2 x(1−a)/2e−bx /u = 0 du  1 −1+ 4abx + 4b2x2 + a2 − 4bx x2 d 2u + a + 2bx + 1− 2bx − a x +  dx  4 dx2   1 + a + 2bx 1− 2bx − a + c +  2x2 + b(a +  − 1)x + b2x2 u = 0  2 

(

(

)(

) (

)

)

Performing further simplification, we arrive at

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2 2 x2 d u + x du + c +  2x2 −  a − 1 u 0  2   = dx  dx2 

However, du du dz = dx dz dx d 2u d 2u  dz  2 du d 2z = + dx2 dz2  dx  dz dx2 Therefore, since dz =  x −1 dx d 2z  −2 = ( − 1)x dx2 we find that du du dz du = =  x −1 dx dz dx dz 2 2 du 2 d 2u d 2u d u  dz  du d 2z =  −1 +  ( − 1)x −2 = +  x   dx2 dz2  dx  dz dx2 dz2 dz

(

)

Hence,   z  2/    z  2−2/ d 2u +  ( − 1)   z  1−2/ du  2            2 dz dz     1/  1−1/ 2       z  a − 1 2  du 2 +   z    z   +    + c −   u = 0        dz       2   Upon simplification, we find that   a − 1 2  2 2  z  − 1+ 1 +  z + c −  2 +  z  u = 0 dz   2   dz 2  2  z2 d u + z du +  z2 − 1   a − 1 − c  u = 0  2 2  dz dz    2    2 du du z2 + z + z2 − p2 u = 0 dz dz2 2 2

d 2u

du

(

)

where

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

  a − 1 2 − c  1 p =    2  2   2

Then the solution is

   y(x) = x(1−a )/2e− bx /  AJ ( x /  ) + AY ( x /  ) p  p  

Mathematica verification FullSimplify[PowerExpand[DSolveValue[ x^2 y''[x]+(a+2 b x^alp) x y'[x]+(c+gam^2 x^(2 d)+ b (a+alp-1) x^alp+b^2 x^(2 alp)) y[x]==0, y[x],x] ]]/.{Sqrt[d^2 ((-1+a)^2-4 c)]->2 p d^2} 5.12 Using the appropriate relationship in Section 5.3.3, determine the solution to d 2 y +  o dy + 2 (o −o ) k x y=0 dx2 x dx Solution: We use Case 6 of Table 5.2; that is, 2 dy 2d y x + a + 2bx x + c +  2x2 + b(a +  − 1)x + b2x2 y = 0 dx dx2

(

)

(

)

Multiplying the original equation by x2, we obtain dy d2 y ( − +2) 2 x +  x + k 2x o o y = 0 o dx dx2 Upon comparing these two equations, we find that a = o ,

b = 0,

c = 0,

 = k,

2 = o − o + 2

Therefore,

=

 o − o + 2 2

Then, from case 6 of Table 5.2, we obtain z = kx /  ,  −1 p= o 2 and, therefore,  y(x) = x(1−o )/2  AJ ( −1)/( 2 ) (kx /  ) + AY( −1)/( 2 ) (kx /  )   o

o

Mathematica verification

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

del=(ao-bo+2)/2; w=x^((1-bo)/2) BesselJ[(bo-1)/(2 del),k x^del/del]; FullSimplify[x^2 D[w,x,x]+bo x D[w,x]+k^2 x^(ao-bo+2) w]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 6 Section 6.1.3 6.1 Place the following equations into Sturm-Liouville form and state whether or not the solutions will be orthogonal functions, subject, of course, to the appropriate boundary conditions. (a) Chebyshev’s equation

(1− x ) 2

d2y

−x

dy

dx2

dx

d2 y x + 2bx dx

dy

+2y = 0

(b) Euler’s equation 2

+ cy = 0

dx

(c) Hermite’s equation d 2 y 2x dy + 2 y = 0 − dx2 dx (d) Laguerre’s equation d2 y x

dx2

+ (1− x)

dy

+ y = 0

dx

Solution (a) Chebyshev’s equation

(

1− x2

)

d2y

−x

dx2

dy

+2y = 0

dx

Upon comparing this equation with Eq. (6.33), we see that a0(x) = 1 − x2, a1(x) = −x, a2(x) = 0, a3(x) = 1, and  = 2. Then using Eqs. (6.9) and (6.15), we find that   x p(x) = exp  dx = exp ln x2 − 1 = x2 − 1  x2 − 1    q(x) = 0

(

r(x) =

)

x2 − 1 = − 1 2 1− x x2 − 1

Then, the Sturm-Liouville form is dy  d  2 x −1 −  dx  dx 



y=0

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Since r(x) is less than zero, the functions will not be orthogonal. (b) Euler’s equation x2

d2y dy + bx + cy = 0 2 dx dx

Upon comparing this equation with Eq. (6.33), we see that a0(x) = x2, a1(x) = bx, a2(x) = 0, a3(x) = 1, and  = c. Then using Eqs. (6.9) and (6.15), we find that  x  p(x) = exp b dx = exp ln xb = xb   2  x

(

)

 

q(x) = 0 r(x) = Then, the Sturm-Liouville form is

xb = xb−2 x2 d  b dy  x + xb−2cy = 0 dx  dx 

Since r(x) is greater than zero, the functions will be orthogonal with respect to this weight function. (c) Hermite’s equation d 2 y 2x dy + 2 y = 0 − dx2 dx Upon comparing this equation with Eq. (6.33), we see that a0(x) = 1, a1(x) = −2x, a2(x) = 0, a3(x) = 1, and  = 2. Then using Eqs. (6.9) and (6.15), we find that 2 p(x) = exp−2 x dx  = e−x   q(x) = 0 r(x) = e−x

2

Then, the Sturm-Liouville form is 2 d  − x2 dy  e + 2e− x  y = 0 dx  dx 

Since r(x) is greater than zero, the functions will be orthogonal with respect to this weight function. (d) Laguerre’s equation d2 y x

dy + (1− x) + y = 0 dx dx2

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Upon comparing this equation with Eq. (6.33), we see that a0(x) = x, a1(x) = 1 − x, a2(x) = 0, a3(x) = 1, and  = . Then using Eqs. (6.9) and (6.15), we find that  1− x  p(x) = exp dx = exp[ln x − x] = xe−x  x    q(x) = 0 r(x) =

xe− x = e− x x

Then, the Sturm-Liouville form is

d 

xe− x

dx 

dy  dx 

+ e− x y = 0

Since r(x) is greater than zero, the functions will be orthogonal with respect to this weight function.

Section 6.1.2 6.3 Consider the Cauchy-Euler equation x2

d2y dy + x + ( + 2) y = 0 2 dx dx

where it is assumed that  + 2 > 0. The boundary conditions are dy(1)/dx = 0 and dy(2)/dx = 0.

Determine the eigenvalues and eigenfunctions.

Solution We start with Eq. (4.54) with a0 =  + 2, a1 = 1, and a2 = 1. Then, from Eq. (4.57) r1,2 =  j  + 2 This is case three, whose solution is given by Eq. (4.60) with b = 0, and d =  + 2. Then, y(x) = D1 cos

(  + 2 ln x)+ D sin(  + 2 ln x) 2

To satisfy the boundary conditions, we note that dy(x) dx

+2

= −D 1

x

sin

(  + 2 ln x )+ D  + 2 cos(  + 2 ln x) 2

x

Then, from the boundary conditions dy(1)/dx = 0, we find D2  + 2 cos

(  + 2)= 0 → D = 0 2

and from the boundary condition dy(2)/dx = 0, we find

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

−D 1

+2 2

sin

(  + 2 ln 2)= 0 → sin(  + 2 ln 2)= 0

Therefore,  n  2  + 2 ln 2 = n →  = n =  −2  ln 2  where n is the eigenvalue and the corresponding eigenfunction, is  n  y (x) = cos ln x n  ln 2 

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 7 Sections 7.2.2 7.1 Solve the Laplace equation 2  r 2

+

 r r

+

2  r 2  2

=0

with the boundary conditions on the semicircular annulus that are given by (r,0) = (r, ) = 0 (a, ) = sin (b, ) = 0

b a

Solution: From Eq. (T7.3c) of Table 7.3, the solution is given by

(

)

(r, ) = Crm + Dr −m ( Acos m + B sin m ) Using the first set of boundary conditions,

(

)

(r,0) = A Crm + Dr −m = 0

(

)

(r, ) = Crm + Dr −m ( Acos m ) = 0 Therefore, A = 0 and the solution reduces to

(

)

(r, ) = Cr m + Dr −m sin m

m0

where the constant B has been subsumed in the other constants.

The boundary condition at r = a and r = b give

( ) (b, ) = (Cb + Db )sin m = 0

(a, ) = Cam + Da−m sin m = sin m

−m

Therefore, to satisfy the second equation, D = −Cb2m To satisfy the first equation, Ca + Da−1 = 1 Cam + Da−m = 0

m=1 m = 2,3,...

Hence, a solution only exists for m = 1. Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

C= D=

a a − b2 −ab2 2

a2 − b2

and the solution is  (r, ) =

ar 2 − ab2

(

r a 2 − b2

)

sin

It is seen that the boundary conditions are satisfied.

7.2 Consider the rectangular region 0  x  a and 0  y  b that is governed by the Laplace equation

2 u x2

+

2 u  y2

=0

with the following boundary conditions u(0, y) = 0, x u(x,0) = 0, y

u(a, y) =0 x u(x,b) =  + x

Determine an expression for u.

Solution: From Eq. (T7.3c) of Table 7.3, the solution is given by u(x, y) = (a0 + b0 x)(c0 + d0 y)+ ( Acos( x) + B sin(x))(C cosh( y) + D sinh( y)) We note that u

= b (c

+ d y ) +  (− Asin(x) + B cos( x))(C cosh( y) + D sinh( y))

0 0 0 x u = d0 (a0 + b0 x)+  ( Acos( x) + B sin(x))(C sinh( y) + D cosh( y)) y

Then the boundary conditions yield u(0, y)

= b (c

+ d y)+ B(C cosh( y) + D sinh( y)) = 0

0 0 0 x u(a, y) + d y)+  (− Asin(a) + B cos(a))(C cosh( y) + D sinh( y)) = 0 = b0 (c 0 0 x u(x,0) = d 0 (a0 + b0 x)+ D( Acos( x) + B sin(x)) = 0 y

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(a)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Thus, to satisfy these for arbitrary y and x, respectively, we see that b0 = B = d0 = D = 0 and n a

sin(a) = 0 →  → n =

n = 1,2,...

Then Eq. (a) reduces to un (x, y) = a0 + Acos(n x)cosh(n y) Therefore, the total solution is 

u(x, y) = a0 +  An cos(n x)cosh(n y)

(b)

n=1

To determine An, we use the last boundary condition to obtain 

u(x,b) =  + x = a0 +  An cos(n x)cosh(nb)

(c)

n=1

We first integrate Eq. (c) with respect to x over the interval {0,a} to obtain a

a

0

0



a

( + x)dx =  a dx +  A cosh( b) cos( x) dx 0

n

n

n=1

n

0

1 a + a 2 = aa 0 2 1 a =  + a 0 2

Next, we multiply Eq. (c) by cos(mx) and integrate over the region to obtain a



a

a

( + x)cos( x) dx = a  cos( x) dx +  A cosh( b) cos( x)cos( x) dx m

0

a2

n2 2

(

0

n − 1+ (−1) =

m

n

a 0

n

n=1

n

m

0

) 2  A cosh( b) mn

n

n

n=1

An =

−4a n  cosh(nb) 2

n = 1,3,5,...

2

The complete solution is 1

4a



u(x, y) =  +  a − 2  2  n=1,3,5

cos( x)cosh( y) n

n

n cosh(n b) 2

7.3 Referring to Figure 7.5, obtain a solution to 2 u 

r 2

+

u r r

+

2 u = 0 r 2  2

0ar b

0     2

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(a)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

over the annular sector indicated and shown in Figure 7.5 with the following boundary conditions u(r,0) = 0 u(r, ) = uo

(b)

u(a, ) = 0 u(b, ) = 0 Solution From Eq. (T7.3c) of Table 7.3, we assume a solution of the form

(

u(r, ) = a0 + b0 + Crm + Dr −m

)( Acos m + B sin m )

(c)

Upon substituting Eq. (c) into the first boundary conditions given in Eq. (c), we arrive at

(

)

a0 + A Crm + Dr −m = 0

Therefore, for arbitrary r, ao = A = 0. Then Eq. (c) reduces to

(

)

u(r, ) = b  + Crm + Dr −m sin m 0

(d)

Upon substituting Eq. (d) into the second boundary conditions given in Eq. (c), we arrive at b  + Crm + Dr −m sin m = u

(

0

)

o

Therefore, for arbitrary r, bo = uo/ and sin m = 0 → m = n / 

n = 1,2,3,...

and the corresponding orthogonal eigenfunction is sin(n/). Then, Eq. (d) becomes

(

)

u(r, ) = u  /  +  C r n / + D r −n / sin(n /  ) 

o

n

n

(e)

n=1

We now substitute Eq. (e) into the last two boundary conditions given by Eq. (b) and obtain

(

)

u  /  +  C an / + D a−n / sin(n /  ) = 0 

o

n

n

n=1 

(

)

u  /  +  C bn / + D b−n / sin(n /  ) = 0 o

n

n

(f)

n=1

To determine Cn and Dn, we multiply each equation in Eq. (f) by sin(k/) and integrate from 0 to  and obtain C an / + D a− n / = − n

n

n /

Cb n

+Db

− n /

n

=−

4uo  n  4uo  n 

where we have used the facts that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(f)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab



2

 sin(n /  )d = n 

0

 sin2 (n /  )d = 0

n = 1,3,5,...

 2

Solving for Cn and Dn, we obtain 4uo  1 n / n  a + bn / n / D = − 4uo  (ab) n n  an / + bn /

Cn = −

(g)

Substituting Eq. (g) into Eq. (e) gives u  4  (r / a)n / + (r / b)−n / u(r, ) =



o

1−   n=1,3,5 

(

n 1+ (b / a) n /

7.4 Obtain a solution to the following system 1   u  2 u r + =0 r r  r  z2

0r a

)



sin(n /  ) 

0 z b

(e)

(a)

with the boundary conditions u(0, z) = finite, u(r,0) = f (r),

du(a, z)

=0 dr u(r,b) = g(r)

(b)

Solution From Eq. (T7.3d) of Table 7.3, the solution to Eq. (a) is u(r, z) = (a0 + b0 z)+ (C0 J0 ( r) + D0Y0 ( r))(E cosh( z) + F sinh( z))

(c)

To satisfy the first boundary condition of Eq. (b), we must set D0 = 0. Then, Eq. (c) simplifies to u(r, z) = a0 + b0 z + (E cosh( z) + F sinh( z)) J0 ( r)

(d)

Upon substituting Eq. (d) into the second boundary condition of Eq. (b), we arrive at −(E cosh( z) + F sinh( z)) J1( a) = 0 For Eq. (e) to be satisfied for arbitrary z, n = a, n = 1, 2, … must be solutions to J1(n ) = 0 Thus,  = n/a is the eigenvalue and the corresponding orthogonal eigenvector is J1(nr / a) Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(e)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

which has a weighting factor equal to r.

Then, Eq. (d) becomes 

u(r, z) = a0 + b0 z +  ( E n cosh(nz / a) + Fn sinh(nz / a)) J0 (nr / a)

(f)

n=1

Upon substituting Eq. (f) in to the second set of boundary conditions given in Eq. (b), we obtain 

 E J ( r / a) = f (r) − a n 0

n

0

n=1



(g)

 ( E cosh( b / a) + F sinh( b / a)) J ( r / a) = g(r) − a − b b n

n

n

n

0

n

0

0

n=1

To obtain the constants En and Fn, we multiply each equation in Eq. (g) by J1(mr/a)rdr and integrate from 0 to a to obtain 

a

 E  J ( r / a)J ( r / a)r dr = p n

n=1

0

n

0

m

n

0

En =  

a

pn Nn

  (E cosh( b / a) + F sinh( b / a))  J ( r / a)J ( r / a)r dr = s n=1



n

n

n

n

0

n

0

m

(h)

n

0

E cosh( b / a) + F sinh( b / a) = n

n

n

n

sn Nn

where a

pn =  ( f (r) − a0 ) J0 (mr / a)r dr 0

a

sn =  ( g(r) − a0 − b0b) J0 (mr / a)r dr n

 0 a

0

n

N = J 2 ( r / a)r dr 0

Therefore, F = n

1 N n sinh(nb / a)

(s − p cosh( b / a)) n

n

n

and Eq. (f) becomes

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(i)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

u(r, z) = a0 +b0z +  p   n cosh(n z / a) + n=1  N n 1 Nn sinh(nb / a)

(s − p cosh( b / a))sinh( z / a) J ( r / a) n

n

n



pn sinh(n (b − z) / a ) + sn sinh(nz / a)

n=1

Nn sinh(nb / a)

= a0 + b0 z + 



n

0

(j)

n

J0 (n r / a)

It is seen from Eq. (i) that a0 and b0 are determined from the form of f(r) and g(r). If neither of these functions has a constant term, then a0 = b0 = 0; otherwise, a0 and b0 are determined from these constants as indicated in the arguments of the integrals in Eq. (i). For example, let f(r) = h(r) + co, where co is known. Then in the argument of the integral for pn becomes h(r) + co − a0.

Therefore, we set a0 = co.

Section 7.2.3 7.5 Consider the equation 2 u 2 u + 2 + 2u = f o sin( x / a)sin( x / b) 2 x y

0 xa

0 yb

subject to the boundary conditions u(0, y) = u(a, y) = 0 u(x,0) = u(x,b) = 0 Using the results of Example 7.8, determine the solution when: (a)  and  are not integers; (b)  = k, an integer and  = l, an integer.

Solution: (a) From Example 7.7, the solution to the homogeneous equation and these boundary conditions is 



u(x, y) =  Anm sin(n x / a)sin(m y / b) m n=1

Substituting this result into the differential equation, we arrive at 2    A  2 −  2  n2 + m  sin(n x / a)sin(m y / b) = f sin( x / a)sin( y / b)    nm 2  a 2 o   m n=1 b   We multiply this equation by sin(kx/a)sin(ly/b)dxdy and integrate over the region to obtain

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

−1  A = fo ab g h  2 −  2 n2 + m2   nm  a2 4 n m b2  

where gn = hm =

(−1)n+1nsin( )

(

 2 n2 −  2

)

(−1)m+1msin( )

(

 2 m2 −  2

)

(b) When  = k, an integer and  = l, an integer, we find that a 2 h = b ml m 2 gn =  nk

that is, we have a one term solution.

Section 7.2.4 7.6 Consider the transient symmetrical distribution of non-dimensional temperature  in a solid sphere of non-dimensional radius 1 that is subject to a constant non-dimensional heat rate g. The governing equation is  

=

1   2   +g 2    

(a)

The boundary conditions are (1,) = 1 and (0,) is finite and the initial condition is (,0) = 0.

Determine the temperature distribution in the sphere.

Solution: From Eq. (T7.7h) of Table 7.7, the solution to Eq. (a) is (, ) = a0 + b0 /  + ( Aj0 () + By0 ())T ( )

(b)

For Eq. (a) to remain finite at  = 0, we set b0 = 0 and B = 0. Then, Eq. (b) reduces to (, ) = a0 + j0 ()T ( ) where the constant A is subsumed in T(t). However, from Eq. (5.113) j0 () =

sin() 

Then, Eq. (c) becomes

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(c)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(, ) = a0 +

sin() T ( ) 

(d)

Upon substituting Eq. (d) into the boundary condition at  = 1, we get sin() a + T ( ) = 1 0 

(e)

For Eq. (e) to be satisfied for arbitrary , a0 = 1 and sin() = 0 →  = n = n

(f)

where n are the eigenvalues and sin(n)/(n) is the eigenfunction. Then, using superposition, the solution so far is 

sin(n) Tn( ) n n=1

(, ) = 1+ 

(g)

Upon substituting Eq. (g) into Eq. (a) and using the fact that 1   2   sin(n) sin(n)  = −n2 2 n 2     n

we obtain

 sin(n)  dTn ( ) n2 2T ( ) = g +  n  n  d 

(h)

n=1

Noting from Eq. (6.99A) that the weight factor is 2, we multiply Eq. (h) by sin(m)/(m)2d to arrive at    dTn ( ) + n2 2T ( )  1 1 sin(n)sin(m) d = g 1  sin(m) d  d n m   nm 2 0 n=1  0  dTn ( ) 2 2  + n  T ( ) 1 g(−1)n+1 = n  d  2n2 2 n2 2 or

dTn ( )

+ n2 2T ( ) = 2(−1)n+1 g

d

(i)

n

Using Eq. (4.25), the solution to Eq. (i) is n+1 2 2 2(−1) g Tn ( ) = + Cn e− n   2 2 n

Then, Eq. (g) yields 

 2(−1)n+1 g

(, ) = 1+  n=1 

n 2

2

− 2 2  sin(n)

+ Cn e n  



n

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(j)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Using Eq. (j) in the initial condition, we find that   2(−1)n+1 g  sin(n)  n2 2 + Cn  n = −1 n=1

(k)

We multiply Eq. (k) by sin(m)/(m)2d to arrive at   2(−1)n+1 g  1 1 1 1   n2 2 + Cn  nm 2  sin(n)sin(m) d = − m   sin(m) d n=1  0 0  2(−1)n+1 g  1 (−1)n+1 + Cn  =− 2 2  n2 2 n  2n2 2 or

 g  C = −2(−1)n+1 1+ n  n2 2 

(l)

Placing the results of Eq.(l) into Eq. (j), we obtain   g  g  −n2 2  sin(n) (, ) = 1+ 2 (−1)n+1 − 1+ e   n  n2 2  n2 2  n=1

7.7 Consider the transient symmetrical distribution of non-dimensional temperature  in a solid infinitely long cylindrical rod of non-dimensional radius 1 that is subject to a constant nondimensional heat rate H. The governing equation is      =  +H 0 1   0 (a)         where  is a known constant. The steady-state temperature at the center of the rod is c and is an unknown quantity at this point. The boundary condition at  = 1 is (1, ) = −h(1, ) where the prime indicates the derivative with respect to  and h is a constant. The boundary condition at  = 0 is that (0,) is finite. The initial condition is (,0) = 0. Determine the temperature distribution in the rod. Hint: Assume that (,) = s() + d(,), where s() is the solution to the nonhomogeneous equation containing H. Then, s(0) = c and c is subsequently determined from s (1, ) = −hs (1, ) Solution: To solve this equation with the boundary conditions and initial conditions as stated, we assume that (, ) = s () + d (, )

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(a)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

where the differential equations and boundary and initial conditions that s and d are specified subsequently. Then the governing equation becomes d 

   d   d  ds  =  +  +H      d       d 

and the boundary condition at  = 1 is s (1, ) + d (1, ) = −hs (1, ) − hd (1, ) The initial condition is s () + d (,0) = 0 We take s() to be the solution to 1 d 

d  H  s =−  d  d  

(a)

and the boundary conditions s(0) = c, a finite value, and s (1) = −hs (1) The solution to Eq. (a) is obtained by integration to give H2  =− + c ln + c2 s 1 4 Since the result must remain finite at  = 0, we set c1 = 0. Then H2 s = c2 − 4 However, at the center of the rod, s (0) = c

 c2 = c

and we obtain s = c −

H2 4

Upon substituting this result into Eq. (b), we obtain  H − = −h  H −  c 2  4  H  1 1  = + c 2  h 2  Therefore,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(b)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 = s

(

)

 H 1 1 + 1− 2  2  h 2

(c)

We take d(,) to be the solution to d 

=

   d        

(d)

with the boundary condition d (0, ) = finite

(e)

d (1, ) = −hd (1, ) The initial condition is 

d (,0) = −s () =

H2 4

− c

(f)

From Eq. (T7.7f) of Table 7.7, the solution to Eq. (d) is d (, ) = a0 + b0 ln + ( AJ0 () + BY0 ())e−  2

(g)

For Eq. (g) to remain finite at  = 0, we set b0 = 0 and B = 0. Then, Eq. (g) reduces to d (, ) = a0 + AJ0 ()e−  2

(h)

Upon substituting Eq. (h) into the second boundary condition given by Eq. (e), we obtain J0( ) + hJ0( ) = 0 −J1( ) + hJ0( ) = 0 This equation is satisfied when  = n, where n are the solutions to n J1(n ) − hJ0 (n ) = 0

n = 1,2,3,...

Then, the solution becomes 2

 d (, ) = An J0 ( n)e− n where, because of the boundary conditions, J0(n) is an orthogonal function with a weighting factor . Using superposition, this portion of the solution is 

 (, ) =  A J ( )e− n d

2

n 0

(i)

n

n=1

Upon placing Eqs. (c) and (i) in Eq. (a), we obtain (, ) = c −

H2  −   An J0 (n)e n 4 +  n=1 2

To determine An, we use the initial condition given by Eq. (f) in Eq. (j) and obtain

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(j)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 − c

H2 4

+



 A J ( ) = 0 n

n 0

n=1

or 

 An J0 (n) = n=1

H2 − c 4

(l)

We multiply Eq. (l) by J0(m)d and integrate over the region to obtain 

1

1

n=1

0

0

 H2   An  J0 (n)J0 (m) d =   4 − c  J0 (m) d An = where n

h =

2  H 4

1

  1 n H2  2 n n 3 n   0 n  4  −  J ( ) d = 2J ( ) −  J ( ) − c J ( ) c

0 1

hn Nn

1

2

Nn =  J0 (n) d = 0



n

n 2

2



J ( ) + J1 (n ) 2 0 n

The solution, therefore, is  ( , ) = c −

H2 4



+ n=1

h n

Nn

−  2

J 0 (n)e

n

(j)

 0

(a)

7.8 Obtain a solution to the following non-dimensional system  

=

1         

0  1

with the boundary conditions ( , ) = −1 (1, ) + k(1, ) = 0

(b)

(,0) = 

(c)

and the initial condition The quantities k and  are constants and the prime () denotes the derivative with respect to .

Solution From Eq. (T7.7f) of Table 7.7, we assume a solution to Eq. (a) to be (, ) = F0 + G0 ln + (FJ0 () + GY0 ())e−  2

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(d)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Upon substituting Eq. (d) into Eq. (b), we obtain G0 2 − (FJ ( ) + GY ( ))e−  = −1 1 1 2  G + kF +  F (kJ ( ) − J ( ) ) + G (kY ( ) − Y ( )) e−  = 0 0 0 0 1 0 1  

(e)

To satisfy Eq. (e) for arbitrary , G0

= −1→ G = −



0

G0 + kF0 = 0 → F0 =

(f)

 k

and FJ1( ) + GY1( ) = 0

F (kJ0 ( ) − J1( ))+ G (kY0 ( ) − Y1( )) = 0

(g)

To obtain a non-trivial solution, we require that the determinant of the coefficients of Eq. (g) equal zero. Thus, if we denote those values of  as n, then n are the eigenvalues that satisfy Y1(n )(kJ0 (n ) − J1(n ))− J1(n )(kY0 (n ) − Y1(n )) = 0

(h)

From the first equation of Eq. (g) and the appropriate part of Eq. (d), the corresponding eigenfunction is n () = Cn J1(n) + Y1(n)

(i)

where Y (  ) Cn = − J1 (n  ) 1

n

The function n() is an orthogonal eigenfunction whose weight factor is .

Then the solution that satisfies Eqs. (a) and (b) is

(

)



(, ) =  k − ln +  An n ()e −1

−  n2

(j)

n=1

To satisfy the initial condition, Eq. (j) in substituted into Eq. (c)

(

)



 k −1 − ln +  An n () = 

(k)

n=1

Since m() is an orthogonal function, we multiply Eq. (k) by m()d and integrate from  to 1 to arrive at

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab



1

 A   () () d = g n

n

m

n



n=1

(l)

g An = n Nn where 1

(

)) () d

(

gn =   −  k −1 − ln 

and we have used the fact that n

 1

m

n

N =  2 () d 

The complete solution is

(

)



(, ) =  k −1 − ln  + 

gn

(C J ( ) + Y ( ))e−

n=1 Nn



2

n 1

n

1

(m)

n

n

7.9 Obtain a solution to the following non-dimensional system  1    2       =  + 0 0 z 1 0       z2

(a)

with the boundary conditions (0, z, ) = finite, (,0, ) = 1,

d( , z, ) d d(,1, )

+ k( , z, ) = 0 (b)

=0

dz and the initial condition (, z,0) = f (, z)

(c)

Solution From Eq. (T7.7e) of Table 7.7, we assume a solution to Eq. (a) to be (, z, ) = (F0 + G0 ln)(D0 + E0 z ) + (FJ0 () + GY0())(D cos( z) + E sin( z))e−  2

(d)

where  2 = 2 −  2

(e)

To satisfy the first boundary condition of Eq. (b), we set G = G0 = 0 and Eq. (d) simplifies to (, z, ) = D0 + E0 z + (D cos( z) + E sin( z)) J0 ()e−  2

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(f)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

where F0 and F have been incorporated into the remaining constants. Upon substituting Eq. (f) into the remaining boundary conditions given by Eq. (b), we arrive at

(kJ ( ) − J ( ))( D cos( z) + E sin( z))e− = 0 2

0



1

D0 + DJ0 ()e−

2

=1

(g)

E0 + (− D sin( ) + E cos( )) J0 ()e

−  2

=0

Thus, for arbitrary , , and z, we find that kJ0( ) − J1( ) = 0 D0 = 1,

D=0

(h)

E0 = 0 cos( ) = 0 From the first equation of Eq. (h),  = n, where n are the eigenvalues that satisfy kJ0 (n ) − J1(n ) = 0 → n = n / 

(i)

and the corresponding eigenfunction is J0 (n /  )

(j)

which is an orthogonal eigenfunction with a weight factor . From the last equation of Eq. (h),  = m, where m are the eigenvalues that satisfy cos( m ) = 0 →  m =

2m − 1  2

m = 1,2,3,...

(k)

and the corresponding eigenfunction is sin( mz)

(l)

which is an orthogonal eigenfunction.

Then, using Eqs. (h) through (l) in Eq. (f), the solution that satisfies Eqs. (a) and (b) becomes  ( , z, ) = 1+





 A  (, z)e

(

)

−  2 + 2  n

m

(m)

nm nm

m=1 n=1

where nm (, z) = J0 (n /  )sin( mz)

(n)

and we have used Eq. (e).

We now determine An from the initial condition given by Eq. (c). Thus, upon substituting Eq. (m) into Eq. (c), we arrive at 



 A J (  /  )sin( z) = f (, z) − 1 nm 0

n

m

m=1 n=1

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(o)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Since Jn(n/) and sin(nz) are orthogonal functions, we multiply Eq. (o) by Jp(p/)d and integrate from 0 to  and additionally multiply Eq. (o) by sin(sz) integrate from 0 to 1 to arrive at 





1

 A  J (  /  )J (  /  ) d  sin( z)sin( z) dz = f nm

m=1 n=1

0

n

0

p

m

0

s

ps

0

An =

fnm

(p)

Nn Mm

where  1

(

)

f ps =   f (, z) − 1 J0 (p /  )sin( sz) d dz 0 0

and N =  J 2 (  /  ) d = n



0



n

(J ( )z + J ( )) 2

0

2

0 1

2

2

n

1

n

1

Mm =  sin ( m z) dz = 2

0

2

The complete solution is   f  (, z) −( + )  e n m ( , z, ) = 1+ 2 nm nm Nn m=1 n=1 2

2

7.10 Obtain a solution to the following non-dimensional system u 

=

1   u   +q      o

0 1

 0

(a)

with the boundary conditions u(0, ) = finite u(1, ) = 0

(b)

u(,0) = 0

(c)

and the initial condition

Solution From Eq. (T7.3f) of Table 7.3 and because of the nature of the differential equation, we assume a solution of the form u(r, ) = F0 + G0 ln + (FJ0 () + GY0 ())T ( )

(d)

It is seen from the first boundary condition of Eq. (b) that we must set G0 = G = 0. Then Eq. (d) reduces to

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

u(, ) = F0 + FJ0 ()T ( )

(e)

Upon substituting Eq. (e) into the second boundary condition of Eq. (b), we find that F0 + FJ0( )T ( ) = 0 Therefore, for arbitrary , F0 = 0 and  = n, where n, n = 1, 2, … are the solutions to J0 (n ) = 0

(f)

Thus, n is the eigenvalue and the corresponding eigenfunction is J0 (n) This eigenfunction is orthogonal with respect to the weight function . Equation (d) can now be written as 

u(, ) = Tn ( )J 0 (n)

(g)

n=1

We now substitute Eq. (g) into Eq. (a) and obtain    dT ( ) 1 d  dJ ( )  n ( ) − T ( ) 0 n J  =q   n o 0 n  d d   n=1   d 

(h)

However, from Table 7.7, 1   J 0 (n)   = − 2 J ( ) n 0 n      

Then, Eq. (h) becomes 



 dTn( )

 n=1 

d

 +  2T ( ) J ( ) = q n n o  0 n

(i)

We multiply Eq. (i) by J0(m)d and integrate over the region to1 obtain 1    dTn( ) +  2T ( ) J ( )J ( ) d = q J ( ) d n n 0 m  0 n o 0 m  d 0 n=1  0 dTn ( )

+  2T ( ) = n n

d

2qo n J1(n )

where we have used the relations 1

1

N n =  J n ( n) d = 2

0 1

1

( J ( ) + J ( )) = 2 J ( ) 2 2

0

2

n

1

2

n

1

n

1

gn =  J0 (n) d =  J1(n ) 0

n

and Eq. (f).

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(i)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

The solution to Eq. (i) is 2qo



2

T ( ) = A e− n + n

n

(j)

 3 J ( ) n

1

n

The initial condition for Eq. (j) is obtained from Eq. (c) as Tn (0) = 0

(j)

Upon substituting Eq. (i) into Eq. (j), we find that An = −

2qo

 J ( ) 3

n 1

n

Therefore, Eq. (i) can be written as T ( ) = n

2qo

 J ( ) 3

n

1

(1− e ) −  2

(k)

n

n

The complete solution is obtained by substituting Eq. (k) into Eq. (g), which gives

(

 J03 (n ) 1− e− n  u(, ) = 2q o  J ( ) n=1

n

1

2

)

n

Section 7.2.5 7.11 Given the following partial differential equation 2 u u 2 u + 2 = 0  x 1 t x2 t2

0  1

with the boundary conditions u(0,1) = u(1,t) = 0 and the initial conditions u(x,0) = 0 u (x,0) = x sin 2 x Obtain the solution for u.

Solution: From Eq. (T7.9b) of Table 7.9 and because of the nature of the differential equation, we assume a solution of the form u(x,t) = ( Acos x + B sin x)T (t) where T(t) is to be determined Using the boundary conditions, we find that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

u(0,t) = AT (t) = 0 → A = 0 u(1,t) = ( Acos  + B sin  )T (t) = B sin( )T (t) = 0 →  = n = n Then un (x,t) = sin(nx)Tn (t) and using superposition, the solution becomes 

u(x,t) = Tn (t)sin(nx) n=1

Substituting this expression into the differential equation yields   2T  T n n 2 sin  x + 2 +  T  n  2  =0 n n t  t  n=1 Thus, this equation is satisfied when 2T n t2

+ 2

Tn t

+  2T = 0 n

n

The solution is this equation is given by Eq. (d) of Example 4.10 as

(

)

(

T (t) = C e−t cos t  2 −  2 + D e−t sin t  2 −  2 n

n

n

Thus,

n

(

u(x,t) = sin( x)e−t C cos t  2 −  2 n n  n n=1

n

)

)+ D sin(t  −  ) 2

n

n

2



Substituting this result into the first initial conditions gives 

u(x,0) = Cn sin(nx) = 0 → Cn = 0 n=1

Hence,

(

u(x,t) = e−t sin( x)D sin t  2 −  2 

n

n

n

)

n=1

Before using the second initial condition, we note that u (x,t) =  sin( x)D e−t − sin t  2 −  2 +



n

(



n

n=1

n

)

(

)

 2 −  2 cos t  2 −  2  n n 

Therefore, 

u!( x,0) =   2 −  2 sin(n x)D = x sin 2 x n

n

n=1

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

From this relation, we determine Dn by multiplying both sides by sin(mx) and integrating from 0 to 1. Thus,

  2 −  2 D  sin(n x)sin(m x) dx = f 

1

n

n

n=1

m

0

Dn =

2 fn  n2 −  2

where 1

fn =  x sin 2 x sin(n x) dx 0

=

1

n=2

4 = 0 n = 4,6,... −8n = n = 1,3,5,...

2  2 n2 − 4

(

)

Then, the solution is 

fn sin(n x)

n=1

(n )2 −  2

u(x,t) = 2e−t 

(

sin t (n )2 −  2

7.12 Find the solution to the non-dimensional equation 2 u 2 u = −h u 0    2

2

o

)

 1

0

for the boundary conditions u(0,) = u(1,) = 0 and for the initial conditions 0   1/ 2 2uo  = u( , 0)  2uo (− + 1) 1 / 2    1 u (, 0) = 0 Solution: Assume the separable solution u(,) = X()T(). Then the partial differential equation becomes 2T 2 X X = 2 T − ho XT  2  2T 2 X + h = = −k2 o T  2 X  2 Therefore,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2T + (ho + k 2 )T = 0 → T ( ) = C sin + D cos  2 = ho + k 2 2  2 X + 2 k X = 0 → X() = A sin k + B cos k 2 Substituting these results into the boundary conditions yields u(0, ) = T ( ) ( A sin k0 + B cos k0 ) = 0 → B = 0 u(1, ) = T ( )sin k = 0 → sin k = 0 → k = n Thus, the solution that satisfies the boundary conditions is 

u(, ) = sin n (Cn sinn + Dn cosn ) n=1

where n = ho + (n  )2 Using this result in the initial conditions, we obtain 

u(, 0) =  Dn sin n n=1 

u!(, 0) = n sin n (Cn cos(n 0) − Dn sin(n 0)) = 0 n=1 

= nCn sin n = 0 → Cn = 0 n=1

Therefore, 

 D sin n = 2u  (u() − u( − 1 / 2)) + (1 − )(u( − 1 / 2) − u( − 1)) n

o

n=1

Since sinn is an orthogonal function over the interval 0    1, 

1

1/2

1

 D  sin n sin md = 2u   sin md + 2u  (1 − )sin md n

n=1

o

o

0

0

1/2

 8 cos(n / 4)sin3(n / 4)  = 2u o   n2 2 2   32uo D = cos(n / 4)sin3(n / 4) m 2 2 n

Dm

Then the solution is u(, ) =

sin n cos( h + (n )2 )  3 32u cos(n / 4)sin (n / 4) o



2

n=1

n2

o

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

7.13 Consider the following circular region that is governed by the wave equation and subjected to an external excitation f(,). The governing equation is

2 u

+

2

u  

−

2 u  2



= f (, )

0

 1 0

subject to the initial conditions u(r,0) = 0

u(r,0)

and



=0

and the boundary conditions u(0, ) = finite u(1, ) = 0 Obtain an expression for u that satisfies these equations.

Solution: From Eq. (T7.9d) of Table 7.9 and because of the nature of the differential equation, we assume a solution of the form u(, ) = (CJ0 (k) + DY0 (k))T ( ) where T() is to be determined. For the solution to be finite at  = 0, we must set D = 0. Then, u(, ) = J0 (k)T ( ) Substituting this solution in to the remaining boundary condition, we find u(1, ) = J0 (k )T ( ) = 0 → J0 (k ) = 0 Therefore, k → kn, where kn are those values that satisfy J0(kn ) = 0

n = 1,2,3,...

Then, using superposition, 

u(, ) =  J0 (kn)Tn ( ) n=1

where J0(kn) is an orthogonal function with weighting factor .

Substituting this equation in the governing equation, we obtain 2    (k )T ( ) −   2 J (k )  Tn( ) = f (, ) + J 2    0 n n 0 n  2 

n=1

n=1

However,  2  2

+

 

2

J (k ) = −kn J0 (kn)    0 n

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(a)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Therefore, Eq. (a) can be rewritten as  2T ( ) + k T ( ) J (k ) = − f (, )   n 2 n2 n  0 n n=1 

We now multiply this equation by J0(km)d and integrate over the region to obtain 2  1 (k )J (k ) d = − 1 f (, )J (k ) d    T ( )   n 2 + kn2Tn ( )  J0 n 0 m 0 0 m n=1 0 2   N   Tn( ) + k 2T ( ) = − g ( )  nm n n n n    2 n=1

2Tn( )  2

g ( ) + k 2T ( ) = − n n n Nn

where 1

1 2 2  2  Nn =  J0 (kn) d =  J0 (kn ) + J1 (kn ) = J1 (kn ) 2 2 2

1

0 1

gn ( ) =  f (, )J0 (km) d 0

We use Table 4.3 and when the initial conditions are zero and  = 0 and find that T ( ) = − n



1

kn N n 0

(

)

sin kn ( −  ) g n ( ) d

The complete solution is 1   u(, ) = −  J 0 (k n)  sin(k ( −  )) f (, )J (k ) d d n=1

kn Nn

n 0



0

0

m



Section 7.2.7 7.14 (a) Find the general solution to    2 2 −  =0 t  

where 2 =

2 r

+ 2

1  1 2 + r r r 2  2

when

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 (r, ,t) = (r)sinejt Hint: Let  = 1 + 2 and choose the appropriate equations that each will satisfy. (b) Select that portion of the solution in (a) that remains finite as r → .

Solution (a) Substitution of the assumed solution into the original differential equation gives 2 2 − j = 2 − j 2 = 0 r

(

) (

r

)

r

r

where 2 2 = d + 1 d − 1 r dr 2 r dr r 2

Therefore, we seek solution to the following equations d 21 1 d 1 1 2   = 0→ + − =0 r 1 2 dr 2 r dr r and   − j = 0 → r 2 r

2

2

d 2 2

2

dr

+r

2

d 2

(

(a)

)

−  2r 2 + 1  = 0

dr

(b)

2

where =

j

Equation (b) is the modified Bessel equation of order 1. Its solution is 2 = C3 I1(r) + C4 K1(r) The solution to Eq. (a) is an Euler equation [Eq. (4.67)] with a2 = 1, a1 = 1, and a1 = −1. The solution is given by Case 1 [i.e., Eq. (4.73)]; thus, [r1 = 1 and r2 = −1] 1 = C1r + C2 / r Therefore,  = 1 + 2 = C1r + C2 / r + C3 I1(r) + C4 K1(r) (b) For the solution to remain finite as r → , we set C1 = C3 = 0. [Recall Eq. (5.131)].

7.15 Consider the following equation describing a square plate subject to a non-dimensional uniform load P 4u = P

0  1

0   1

where

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2 =

2 2

+

2  2

On the edges  = 0 and  = 1, the boundary conditions are u(0, ) = u(1, ) = 0 2 2 u(0, ) = u(1, ) = 0 2 2 and those on the edges  = 0 and  = 1 are u(,0) = u(,1) = 0  u(,0) = u(,1) = 0   

Determine the solution to this system.

Solution We start with Eqs. (7.88) and (7.92); that is, w(, ) = (A cos + B sin)Y ( ) Upon substituting this equation into the boundary conditions at  = 0 and  = 1, we find that w(0, ) = AY ( ) = 0 w(1, ) = (A cos + B sin )Y ( ) = 0 2 w(0, ) = − A 2Y ( ) = 0 2 2 w(1, ) = − 2 (A cos + B sin )Y ( ) = 0 2 Therefore, A = 0 and  = n and the solution becomes wn (, ) = sin nYn ( ) Then, using superposition, 

w(, ) = Yn ( )sin n n=1

Substituting this result into the governing differential equation. we arrive at 2   4  d Yn 2 2 d Yn − 2n  + n4 4Y sin n = P   n  dy4 dy2  n=1

Multiplying this equation by sin(m)d and integrating over the interval, we obtain d 4Yn d 2Yn P − 2n2 2 + n4 4Y = n dy4 dy2 2

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

By inspection, we see that the particular solution is Yp,n =

P 2n4 4

Then, from Eq. (7.98), which is the homogeneous solution, Yn ( ) = C1 cosh(n ) + C2 sinh(n ) +  (C3 cosh(n ) + C4 sinh(n )) +

P 2n 4 4

Before proceeding, we note that  Yn ( ) = C1n sinh(n ) + C2n cosh(n ) + C3 (cosh(n ) + n sinh(n ))  + C4 (sinh(n ) + n cosh(n )) Substituting this result into the boundary conditions at  = 0 and  = 1, we obtain   1 0 0 0  C1   cosh(n ) sinh(n ) cosh(n ) sinh(n )   C   2    0 n 1 0  = C    3  cosh(n ) + sinh(n ) +  n sinh(n ) n cosh(n )  C  n sinh(n ) n cosh(n )   4    −P / (2n4 4 )     −P / (2n4 4 )    0     0   Using Mathematica, the solution is C1 = − C2 =

P 2 4n4 2 P sinh (n / 2)

 4n4 (sinh(n) + n) P sinh2 (n / 2) C3 = − 3 3  n (sinh(n) + n) P sinh(n) C4 = 2 3n3(sinh(n) + n)

Then, doing the appropriate substitutions, we get for the final result P  1 1 1 sinh2 (n / 2)sinh(n ) w(, ) = 4  4  − cosh(n ) +  n=1 n  2 2 (sinh(n) + n) 2 n sinh (n / 2) cosh(n ) n sinh(n) sinh(n )  − + sin n  (sinh(n) + n) 2(sinh(n) + n)  Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solutions to Exercises in Chapter 8 Section 8.1.2 8.1 Obtain the Laplace transform of cosh t sin t using Eq. (8.16).

Solution: We express this product as g(t) =

1

(e + e )sin t = 1 e sin t + 1 e sin t −t

t

−t

t

2

2

2

Then, using Eq. (8.16) 1 L et sin t  + L e−t sin t  2 2 1 1 1 1 = + 2 2 (s − 1) + 1 2 (s + 1)2 + 1 s2 + 2 = (s + 1)2 + 1 (s − 1)2 + 1

L[g(t)] = G(s) =

(

1

)(

)

Mathematica verification LaplaceTransform[Cosh[t] Sin[t],t,s] 8.2 Obtain the Laplace transform of e−tt cost Solution: From Eq. (8.16), f (t)ebt  F (s − b) and from Eq. (8.7) L tn f (t)  (−1) n

dn F (s) dsn

Then, from Eq. (8.13)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 d  L e−tt cost  L[cost] = − d  2 s    ds  s + 1  = −     ds  s→s+1  s→s+1    2  2 1 2s s + 1− 2s2 = − −   = − 2 2   s2 + 1 2   s2 + 1 s + 1 s→s+1 s→s+1    2  s − 12 (s + 1)2 − 1 2  =  2 = s +1 (s + 1)2 + 1

(

(

 =

)

)

(

s→s+1 s(s + 2)

(

)

)

(s + 2s + 2)

2

2

Mathematica verification LaplaceTransform[Exp[-t] t Cos[t],t,s] 8.3 Use Eqs. (8.6) and (8.7) to obtain the solution to the following equation d2 y dy x 2 − 2(x + 1) + (x + 2) y = 0 dx dx Solution: From Eq. (8.6), we note that L[ y] = Y (s)  dy  L dx = sY (s) − y(0)      d2y  2 L  = s Y (s) − sy(0) − y(0) 2 dx  

(a)

and from Eq. (8.7) that L[xy] = −

d

Y (s) ) ( ds

 dy  d L  x  = − (sY (s) − y(0)) dx ds   d 2 y  d

(

s Y (s) − sy(0) − y(0) L x 2  = − ds  dx  2

(b)

)

Then the Laplace transform of the differential equation is  d2 y   dy   dy  L  x 2  − 2L  x  − 2L   + L[xy] + 2L[ y] = 0 dx  dx   dx   

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

which upon using Eqs. (a) and (b) gives d 2 d − s Y (s) − sy(0) − y(0) + 2 (sY (s) − y(0)) ds ds d − 2(sY (s) − y(0))− (Y (s))+ 2Y (s) = 0 ds

(

)

Performing the indicated operations and collecting terms, we arrive at    dY (s) − 2sY (s) + s2 − y(0) + dY (s)  + s  2Y (s)  ds ds  dY (s) − 2 (sY (s) − y(0) )− + 2Y (s) = 0 ds which reduces to

(s − 2s + 1) dYds(s) + 4(s − 1)Y (s) = 3y(0) 2

dY (s) 4 3y(0) + Y (s) = ds s −1 (s − 1)2

(c)

To obtain a solution to Eq. (c), we use Eq. (4.24) with 4

p(s) = q(s) =

s −1 3y(0)

(s − 1)

2

Then, from Eq. (4.22) the integrating factor is  ds  = exp4 ln(s − 1) = (s − 1)4 (s) = exp4   s − 1   From Eq. (4.24), we obtain  3y(0) (s − 1)4 C 1  2 ds + C  = Y (s) = (s − 1) ds + 4 3y(0) 4  (s − 1)  (s − 1)2 (s − 1)4  (s − 1) 3y(0)  s3  C 2 = + s − s + 4 4 (s − 1)  3  (s − 1)

(d)

The inverse of the first term of Eq. (d) requires an extensive table of Laplace transform pairs [see Sanders and Kaufman in the Bibliography] or the use of Mathematica. The second term is obtained from pair 20 of Table 8.2. Then, using Mathematica, the inverse Laplace transform of Eq. (d) is y(0) x C y(x) = e 6 + x3 + x3ex 6 6

(

)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Mathematica verification InverseLaplaceTransform[3 y0/(s-1)^4 (s-s^2+s^3/3)+ c/(s-1)^4,s,x] Section 8.1.3 8.4 Use partial fractions to determine the inverse Laplace transform of a3  −1  L  3  s(s + a)  Solution: We would like to expand this into the form B B B a3 A 1 2 3 = + + + 3 2 s s + a s(s + a) (s + a) (s + a)3 From Eq. (8.20), we find that 

 3  =1  s→0 s(s + a)  

A = lim From Eq. (8.22), we find that 1 d2 

sa3

2 a3 d  1 B1 = (s + a) s(s + a)3  = 2! ds2  s  2! ds2  3

a3

a3  2 



= s=− a

2  s3  s=− a

= −1

d  1  −1 = a3  2  = −a    s  s=− a ds  s  s=− a  1 B = a3 = −a2  s 3 B2 = a3

s=− a

Then,  a  − L−1  a2   1  1  −1 −1 −1    L  = L  −L  −L   3 2 3 s   s + a   (s + a)   s(s + a)   (s + a)  − 

a3

1



(

)

= 1− 1+ at + a2t2 e−at where we have used Eqs. (8.11) and (8.15).

Mathematica verification ap=Apart[a^3/(s (s+a)^3)] InverseLaplaceTransform[ap,s,t]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

8.5 Use partial fractions to determine the inverse Laplace transform of 6 + (s + 1)(s2 − 5s + 11) F (s) = s(s − 1)(s − 2)(s + 1) Solution: Since there are no repeating roots, we would like to expand this into the form A A2 A3 A4 F (s) = 1 + + + s (s − 1) (s − 2) (s + 1) From Eq. (8.20), we find that  6 + (s + 1)(s2 − 5s + 11)  6 + 11 17 A1 = lim  = 2 = 2 s→0  (s − 1)(s − 2)(s + 1)   6 + (s + 1)(s2 − 5s + 11)  6 + 14 A2 = lim   = −2 = −10 s→1 s(s − 2)(s + 1)    6 + (s + 1)(s2 − 5s + 11)  6 + 15 7 = A3 = lim  = s→2 6 s(s − 1)(s + 1) 2   2  6 + (s + 1)(s − 5s + 11)  6 A4 = lim   = −6 = −1 s→−1 s(s − 1)(s − 2)   Then, −  6 + (s + 1)(s

− 5s + 11)  17 1  1  7  1  − L−1  1  −1 −1 −1 − 10L      L  L  = + L  s s − 1 2 s − 2 s + 1 s(s − 1)(s − 2)(s + 1) 2           17 7 = − 10et + e2t − e−t 2 2 2

1



Mathematica verification ap=Apart[(6+(s+1) (s^2-5 s+11))/(s (s-1) (s-2) (s+1))] InverseLaplaceTransform[ap,s,t] 8.6 Determine the inverse Laplace transform of  − s  L−1  e 2   1+ s  Solution: Using pair 6 of Table 8.2,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

−1 −s − = − −  e−s    − L−1  = L G(s)e g(t )u(t ) sin(t )u(t ) 2    1+ s 

since L−1

 1  = sin t 1+ s2 

Mathematica verification InverseLaplaceTransform[Exp[-Pi s]/(1+s^2),s,t] 8.7 Determine the Laplace transform of t

e

−t

 e− cos[ (t −  )]d

0

Solution: We write this equation as e(t) = e−t f (t) where t

f (t) =  e− cos[ (t −  )]d

(a)

0

If we use the convolution relation given by Eq. (8.26), we see that t

L [G(s)H (s)] =  g( )h(t −  ) d

(b)

t  G(s)H (s) = L   g( )h(t −  ) d  0 

(c)

−1

0

or

Upon comparing Eq. (a) with Eq. (b), we identify L−1  H (s) = e− L−1 G(s) = cos( ) Therefore, H (s) = L e−  = 

1



(s + 1)2 s G(s) = L cos( ) =   s2 +  2

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

where we have used Eqs. (8.18) and (8.13). Hence, t  − F (s) = L[ f (t)] = L   e  cos[ (t −  )]d  = H (s)G(s) 0  s =

(s +  )(s + 1) 2

2

2

and from Eq. (8.16), L[e(t)] = E(s) = L e−t f (t) = F (s + 1) =

s +1

((s + 1) +  )(s + 2) 2

2

2

Mathematica verification Simplify[LaplaceTransform[Exp[-t]Integrate[Exp[-x] x * Cos[w (t-x)],{x,0,t}],t,s]] 8.8 Determine the inverse Laplace transform of s

(s + k ) 2

2

2

Solution: We write this equation as s

(s + k ) 2

2

2

=

s

1

s + k s + k2 2

2

2

Therefore, we let F (s) = G(s) =

s s2 + k 2 1 s2 + k 2

and from Eqs. (8.12) and (8.13), we have L−1  F (s) = cos kt 1 L−1 G(s) = sin kt k Using Eq. (8.26),

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

t

L−1[F (s)H (s)] =

 h( ) f (t −  ) d 0

t s 1  = 1 cos(k )sin (k(t −  )) d −1   L  s2 + k 2 s2 + k 2  k 0  t = sin(kt) 2k

Mathematica verification InverseLaplaceTransform[s/(s^2+k^2)^2,s,t] Section 8.1.6 8.9 Obtain the Laplace transform of the following periodic waveform (sawtooth wave) of period T f (t) = t / T

0t T

Solution: From Eq. (8.37) F (s) =

(

1− e−sT

1T

) T e −1

− s

d = 2

Ts 1− e

0

= =

2

Ts 1

Ts2

(

1

1− e−sT

−

)

−sT

)

1− e−sT (1+ sT )

−sT −sT 1− e − sTe )

e−sT

(

(

1

=

s 1− e−sT

1

) Ts

2

1

−

(

)

s esT − 1

Mathematica verification Simplify[1/T Integrate[x Exp[-s x],{x,0,T}]/(1-Exp[-s T])] 8.10 Obtain the Laplace transform of the following periodic waveform (triangular wave) of period T f (t) = 2t / T = 2(1− t / T )

0t T /2 T /2t T

Solution: From Eq. (8.37)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

F (s) =

(

1− e−sT

)

−1

T 2 T /2  s − −s d e d + 2 (1−  / T )e     T 0  T /2

2−e )(

=

1 s T 1− e−sT

=

2e− sT esT /2 − 1 s2T 1− e−sT

2

(

( (

−sT /2 (2 + sT )

) + e (2 + e −sT

)

−sT /2(−2 + sT )

2(1− e )= ) ) s T (1− e )(1+ e ) 2

− sT /2

2

− sT /2

2

− sT /2

 sT  ( ) 2 = = tanh   4  s T (1+ e ) s T 2 1− esT /2

2

− sT /2

2

where we have used the results of Example 8.7.

Mathematica verification FullSimplify[(2/T Integrate[x Exp[-s x],{x,0,T/2}]+ 2 Integrate[(1-x/T) Exp[-s x],{x,T/2,T}])/(1-Exp[-s T])] Section 8.1.7 8.11 Use the Bromwich inversion integral to determine the inverse Laplace transform of F (s) =

1 (s + b)cosh(as)

Solution: The poles of this expression are at s = −b and cosh(as) = cosh( j ) = cos  = 0 →  = n = (2n − 1) / 2 → s =  j(2n − 1) / (2a) Therefore,  ezt zt + lim   (z + b)e (z + b)d cosh(az) / dz f (t) = (z + b)cosh(az) z=− b n=1 z→zn   − bt   zt lim  e  e = cosh(ab) +  z→ j n /( 2 a ) a(z + b)sinh(az) n=1   j t /( 2 a ) − bt  e e 1 = + 2 cosh(ab) ( j + 2ab) ( j sin( / 2)) n

n=1

n

n

e− j t /( 2 a ) 1 + 2 (− j + 2ab) (− j sin( / 2)) n=1 n n   e− bt (−1)n+1 e j t /( 2 a ) (−1)n e− j t /( 2 a ) = + 2 + 2 cosh(ab) n=1 j(− j n + 2ab) k =1 j( j n + 2ab) 

n

n

n

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

where n = (2n − 1). Upon further manipulation, we find that   (−1)n e j t /( 2 a ) (−1)n e− j t /( 2 a ) e− bt + 2 j − 2 j cosh(ab) (− j + 2ab) ( j + 2ab) n=1 n=1 n n  n  (−1) ( j + 2ab)e j nt /( 2 a ) (−1)n (− j + 2ab)e− j nt /( 2 a ) e− bt n n = + 2 j − 2 j cosh(ab)  2n + 4a2b2  n2 + 4a2b2 n=1 n=1 j nt /( 2 a )  − (− j + 2ab)e− j nt /( 2 a ) ( j + 2ab)e e− bt n n = + 2 j (−1)n n

f (t) =

n



cosh(ab) = =

 n2 + 4a2b2

n=1

 2 j n cos( nt / (2a) ) + 4 jabsin( nt / (2a)) e− bt + 2 j (−1)n cosh(ab)  n2 + 4a2b2 n=1 

 4 cos( nt / (2a) ) + 8absin( nt / (2a)) e− bt −  (−1)n n cosh(ab) n=1  n2 + 4a2b2 

Mathematica verification of one term in series r1=Residue[Exp[z t]/((z+b) Cosh[a z]),{z,-b}]; q=I (2 n-1) Pi/(2 a); r2=FullSimplify[Exp[z t]/(z+b)/D[Cosh[a z],z]/.z->q, Assumptions->nIntegers]; r3=FullSimplify[Exp[z t]/(z+b)/D[Cosh[a z],z]/.z->-q, Assumptions->nIntegers]; r1+Simplify[ComplexExpand[r2+r3],Assumptions->nIntegers] 8.12 Use the Bromwich inversion integral to determine the inverse Laplace transform of cosh(x s / a )

F (s) =

scosh(k s / a ) Solution: The poles of this expression are at s = 0 and

(

)

cosh k s / a = cosh( j ) = cos  = 0 Therefore,  = n = (2n − 1) / 2

n = 1,2,3,...

and s = −a n2 / k 2 Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

 )    + lim  d f (t) =  cosh k z / a  ( )  z dz  2a z / a cosh x z / a e  ( )  = 1+ lim    kz sinh(k z / a )    



n=1

(

cosh x z / a ezt

z→zn

z=0

zt



n=1

z→zn



= 1+ 

2ajn cosh( jnx / k )

n=1 

k

2

z=zn

2

e−an t / k

(−a / k )sinh( j ) 2

2

n

n

2

cos(nx / k )e−ant / k 2



= 1− 2

n

(−1) − a t / k cos(  n x / k )e n n=1 n

= 1+ 2

2

n=1

2

n sin(n )

2

Mathematica verification of one term in series q=-a b^2/k^2 ; r2=FullSimplify[(Exp[z t]Cosh[x Sqrt[z/a]]/ D[Cosh[k Sqrt[z/a]],z])/z/.z->q,Assumptions->nIntegers] 8.13 Use the Bromwich inversion integral to determine the inverse Laplace transform of F (s) =

sinh(a s ) ssinh( s )

Solution: The poles of this expression are at s = 0 and

( s ) = sinh( j ) = j sin  = 0

sinh Therefore,

 = n = n

n = 1,2,3,...

and s = sn = −n2 2 Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

  zt   sinh(a z )e +  lim   d n=1 z=− n    sinh( z ) z=0 z sinh( z )  dz    2 z sinh(a z )ezt  a z 1+ a2z / 6 + ... ezt +  =  lim2 2  z=− n   z cosh( z )  z (1+ z / 6 + ...) z=0 2 2 n=1 2 2  2 jn sinh( jan )e−n  t 2  sin(an )e−n  t 

zt f (t) = sinh(a z )e

2

(

2

)

= a+

= a+   n=1 −n2 2 cosh( jn )  n 2 2 2 (−1) == a +  sin(an )e−n  t  n=1 n n=1

ncos(n )

Mathematica verification of one term in series q=-n^2Pi^2 ; a+FullSimplify[(Exp[z t]Sinh[a Sqrt[z]]/D[Sinh[Sqrt[z]],z])/z /.z->q,Assumptions->nIntegers] 8.14 Use the Bromwich inversion integral to determine the inverse Laplace transform of

( )

sinh a s

F (s) =

(a)

Solution: Using Mathematica, a series expansion of F(s) about s = 0 is F (s) 

a s(a + mo )

−

a3m

o

3(a + mo )

2

+

(

s a 6 mo + 6a5m2o

)

45(a + mo )

(b)

3

Thus, we have a simple pole at s = 0 whose residue is a/(a + mo). The other poles are at the roots of

( )

tanh a s = −a s o

(c)

where o = a/mo. If we let jn = a s in Eq. (c), Eq. (c) becomes tan(n ) = − on

(d)

Therefore, for a given o, n, n = 1, 2, …, are those value that satisfy Eq. (d). The corresponding values of s → sn = − 2n / a2.

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Then

 ( )    z(a + m )   z d (m z cosh(a z ) + sinh(a z ))  dz    2 z sinh(a z )e a  = +  lim   a+m z ((a + m )cosh(a z ) + am z sinh(a z )) 

 f (t) = lim 

 +    lim 

zaezt

n=1 z=−  n / a 2

z=0

sinh a z ezt

o

2

o



zt

n=1

o

=

(2 j / a)sinh( j )e − 2 / a2 ((a + m )cosh( j ) + j m sinh( j ))  sin( )e



a+m a

n

+

a

=

z=−  2 / a2



o

o

n

n



−  n2t / a2

n=1 o

a + mo

o

n

n

+ 2 n=1

n

n

o

n

− t/a 2

2

n

( / a)((a + m )cos( ) −  m sin( )) n

o

n

n

o

n

Equation (e) can be further simplified by using Eq. (d).

Mathematica verification of one term in series q=-b^2/a^2 ; a/(a+mo)+FullSimplify[(Exp[z t]Sinh[a Sqrt[z]]/ D[(mo Sqrt[z]Cosh[a Sqrt[z]]+Sinh[a Sqrt[z]]),z]/z) /.z->q,Assumptions->nIntegers] Section 8.2 8.15 Use Laplace transforms to determine the general solution to

d2y

dy 2 + k y = f (t) − 2k 2 dt dt

Solution: Taking the Laplace transform of each term, we find that s2Y (s) − y(0) − sy(0) − 2k (sY (s) − y(0)) + k 2Y (s) = F (s)

(s2 − 2ks + k 2 )Y (s) = y(0) + (s + 2k ) y(0) + F (s) (s − k )2Y (s) = C + sC + F (s) 1

Y (s) =

2

C1 sC2 + F (s) 2 + (s − k ) 2 (s − k ) (s − k )2

Taking the inverse Laplace transform using Eqs. (14) and (24) and noting that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(e)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

A + B 2 A(s − k) + B s = (s − k) = (s − k)2 (s − k)2 s − k and, therefore, As = s → A = 1 −kA + B = 0 → B = k we obtain

 y(t) = C 1L−1 

 s  −1  F (s)  1  + C2L−1  +L  2 (s − k ) (s − k )2  (s − k )2         k  t  1  k(t− ) + = C1tekt + C2 L−1 + C2 L−1  f (t)dt 2   (t − )e (s − k )     0  s −k  t = C1 tekt + C2 (ekt + ktekt ) + (t − )ek(t− ) f (t)dt  0

t

= C2ekt + C3tekt +  (t − )ek(t− ) f (t)dt 0

Mathematica verification with f(t) = 0 lp=LaplaceTransform[y''[t]-2 k y'[t]+k^2 y[t]==0,t,s]/.LaplaceTransform[y[t],t,s]->Ys; yss=Ys/.Solve[lp,Ys][[1]] InverseLaplaceTransform[yss,s,t] InverseLaplaceTransform[1/Denominator[yss],s,t] 8.16 The non-dimensional form of the one-dimensional heat flow equation in a solid is given by

d

=

d

2 (a)

2

It is assumed that the boundary conditions are  (0, ) 

= 0,

 (1, ) =  1

(b)

and the initial condition are  (,0) = 1

(c)

Use the Laplace transform with respect to  to obtain a solution to Eq. (a) subject to the conditions given by Eqs. (b) and (c).

Solution

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

To obtain a solution, we take the Laplace transform with respect to time of these equations. Then Eq. (a) becomes 2 = s − (, 0) = s −1 2

(d)

where we have used Eq. (c). The boundary conditions given by Eq. (b) become

  (1, s) = 1

 (0, s) = 0, 

(e)

s

The solution to the homogeneous portion of Eq. (d); that is, 2h  2

− sh = 0

is

h (, s) = Acosh s + B sinh s and the solution to inhomogeneous portion of Eq. (d); that is, is 2 p − s p = −1 2  is, by inspection,

 = p

1 s

Therefore, +

 (, s) =  h (, s) +p = Acosh s B sinh s +

1 s

(f)

Since  (, s) 

= A s sinh

+ B s cosh

we find upon substituting Eq. (f) into Eq. (e), the boundary conditions, that  (0, s) =0→B=0    (1, s) = 1 → A cosh s

1 1 1 −1 s + − → A= s s s cosh s

The Eq. (f) becomes

1 cosh s  (, s) = + ( −1 ) 1 s s cosh s

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(g)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

To determine the inverse transform, we use Eq. (66) and Eq. (g) to arrive at (1 −1)  + j cosh z  (, ) = u( ) + ezt dz  2 j  − j z cosh z We have a simple pole at z = 0 and at the poles 2  2n −1   2n −1  cosh z = 0 → zn = j   → zn = −  2   2   2   

(h)

Then 

  zezt cosh z   ezkt cosh z    k  (, ) = u( ) + (1 − 1) Lim  + d   z→0  cosh z  z cosh z  k =1 z  k   dz z= zk     zkt     znt    k e cosh z = u( ) + ( − 1) 1+ 2  e cosh z n  = u( ) + ( − 1) 1+       1 1 1 sinh  sinh z n=1  z  k =1 z   n   n  zk k   2 zk     eznt cosh z  n  =  + 2( − 1 )   1 1 sinh z  n=1  n   zn However, from Eq. (h),   2n −1   sinh = j  2n −1   sinh   zn zn  j 2   2       2n −1   2n −1   2n −1  = −   sin    = −   (−1)n−1    2 2 2        2n −1   = (−1)n   2    Hence, 

((

)

)

(−1)n e−(2 n−1)2 2t /4 cos  2n − 1  / 2 

 (, ) = 1 + 2(1 − 1)  n=1 

(2n − 1) / 2

  

Mathematica verification of one term in series zn=-((2 n-1) Pi/2)^2; r2=FullSimplify[Exp[z t]Cosh[x Sqrt[z]]/z/D[Cosh[Sqrt[z]],z] /.z->zn,Assumptions->nIntegers]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

8.17 Consider a hanging chain of length L that is free at x = 0 and fixed at x = L. The governing equation of motion for this system is   u  1 2u (a) x = x  x  g t2 where g is the gravity constant. The boundary conditions are u(L, t) = Asin t

(b)

u(0, t) = finite

that is, one end is oscillating harmonically at a frequency . The initial conditions are assumed to be zero. Use the Laplace transform with respect to  to obtain a solution to Eq. (a) subject to the conditions given by Eqs. (b) and (c).

Solution: To solve this problem, we take the Laplace transform of Eqs. (a) and (b) to arrive at   u (x, s)  s2 x − u (x, s) = 0 x  x  g x

2u (x, s) x

2

+

(c)

u (x, s) s u (x, s) = 0 − x g 2

and A

u (L, s) =

s2 +  2 u (0, s) = finite

(d)

We have shown in Case 2 of Table 5.2 that 2 dy 2d y 2 2 2b 2 2 x + x(1− 2ab) + b k x − b p y = 0 dx2 dx

(

)

(e)

y(x) = xab C1 J p (kxb ) + C2Y p (kxb )

(f)

has as a solution of the form

Comparing Eq. (c) with Eq. (e) we see that a = 0, p = 0, b = 1/2, and b2k2 = k2/4 = −s2/g. In addition, we note that Y does not remain finite at x = 0 and, therefore, we must set C2 = 0. Then Eq. (f) is used to obtain a solution to Eq. (c) when its form is

(

)

(

u (x, s) = C1J0 2 js x / g = C1I0 2s x / g

)

where we have used the relation

I0 (x) = J0 ( jx) To determine C1, we substitute Eq. (g) into Eq. (d) and find that

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(g)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(

C = 1

A

A

) s +

u (L, s) = C1I0 2s L / g =

2

2

( s 2 + 2 ) I (2s L / g ) 0

Therefore, the solution in the Laplace transform domain is

(

AI0 2s x / g

u (x, s) =

)

s +  I 0 2s L / g 2

2

=

(

AI0 s x / L

)

(h)

( s 2 +  ) I (s ) 2

0

where  2 = 4L / g. Equation (h) has simple poles at s = j and at the simple poles when

I0 (s ) = 0 → I0 ( jn ) = J0 (n ) = 0 → sn =  j

n 

To find the inverse transform, we substitute Eq. (h) into Eq. (66) and use residue theory to obtain  + j

u(x, t) = A 

ezt I 0

(z x / L )

( z +  ) I ( z )

 − j

2

2

(

dz

0

)

(

)

   ezt I 0 z x / L   ezt I 0 z x / L  = A z→ lim  + lim   j  ( z + j ) I 2z L / g  z→− j  ( z − j ) I 2z L / g   0 0      z  x / L     ezt I 1 0   2 2 lim +  lim ( z +  )  z→s z→sn   n  n=1 dI0 ( z ) dz       ezt I z x / L      0 1  lim  +  lim   z→− sn  n=1 ( z2 + 2 )  z→− sn  dI0 ( z ) dz     

(

)

(

)

(

)

(

)

We note that d I0 ( z ) =  I1 ( z ) dz and I1 (sn ) = I1 (  jn ) =  jJ1 (  n )

(

) (

I0 sn x / L = I0  jn

)

(

x / L = J0  n x / L

)

Then Eq. (i) becomes

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(i)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(

)

(

− jt jt u(x,t) = A  e I0 j x / L e I 0 − j x / L −  2 j I 2 j I0 (− j ) j ( ) 0  j t /



e n I 0 j n x / L

n=1

  2 −  n2 /  2 I1 j n

+

(

)

+

(

e

− j nt /

e n J0 n x / L

  2 −  n2 /  2 I1 − j n

j t /

+ n=1

(

j  2 −  n2 /  2 J1  n

)

    

I 0 − j n x / L

jt e− jt J 0  x / L = A  e J0  x / L  2 j J ( ) − 2 j J0 ( ) 0  

)

−

e

− j nt /

)

J0 n x / L

j  2 −  n2 /  2 J1  n

(

)

    

 J  x / L  J0 n x / L 1  0 1 = A  e jt − e− jt +  e jnt / − e− jnt /    J0 ( ) 2 j 2 2 2  J 1( n) j n=1   −  n /     J  x / L J0 n x / L =A 0 sint + 2 A  sin nt /  2 J ( )   − 2 / 2 J 

(

(

)

)

0

(

n=1

(

( n

)

)

(

)

) ( ) 1

n

Mathematica verification of one term in series arg[z_,s_]:=Exp[z t] BesselI[0,z g Sqrt[x/L]]/(z+I w)/(z-I w)/ If[s==0,BesselI[0,z g],D[BesselI[0,z g],z]] r1=Expand[(z-I w) arg[z,0]]/.z->I w; r2=Expand[(z+I w) arg[z,0]]/.z->-I w; r3=arg[z,1]/.z->I b/g; r4=arg[z,1]/.z->-I b/g; FullSimplify[w(r1+r2)] FullSimplify[w (r3+r4)] 8.18 Consider the non-dimensional equation governing the static displacement of beam having an in-span elastic support at s, 0 < s < 1, and subjected to an externally applied load q() = qo( − o), 0 < o < 1 d 4w + K   − =   −   ( ) y q ( ) 0 1 (d) s s o o d4 The boundary condition at  = 0 are

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

w(0) = −KLw(0),

w(0) = KtLw(0)

(e)

w(1) = −KRw(1),

w(1) = KtRw(1)

(f)

and those at  = 1 are where the prime indicates the derivative with respect to  and the K’s are constants that can vary between 0 and .

Using the Laplace transform method, obtain the general solution for w() in terms of w(s), K, and qo. Judiciously define groups of terms at appropriate solution stages in the solution process; this will greatly simplify the algebra.

Solution: We shall solve Eqs. (d) to (f) by using the Laplace transform with respect to . Then the Laplace Transform of Eq. (d) is s4W (s) = q oe−os − K se−ssw( s ) + s3w(0) + s2w(0) + sw(0) + w(0) Using Eq. (e) in Eq. (g), we obtain 1 W (s) = q e−os − K e−ssw( ) + s3w(0) + s2w(0) + sK w(0) − K w(0) s s tL L  s14  o s s 3 2 − − = q e o − K e s w( ) + s − K w(0) + s + sK w(0) tL s s L  s4  o −os −ss 1 3 1 2 s − K w(0) + s + sK w(0) = q e − K w( ) e + o s s L tL s4 s4 s4 s4

(

)

(

)

)

(

(

(g)

(h)

)

Using Eqs. (8.9) and (8.32), the inverse Laplace transform of Eq. (h) is q K w() = o ( −  )3u( −  ) − s w( )( −  )3u( −  ) o o s s s 6 6     K K +  1− L  3  w(0) +   + tL 2  w(0) 6 2     where u() is the unit step function.

To determine w(0) and w(0), we must substitute Eq. (i) into Eq. (f). Before doing this, we note that K K 1 w() = q ( −  )2 u( −  ) − s w( )( −  )2 u( −  ) − L 2w(0) + (1+ K )w(0) o o o s s s tL 2 2 2 w() = qo ( − o )u( − o ) − Ksw(s )( − s )u( − s ) − KLw(0) + KtLw(0) w() = qou( − o ) − Ksw(s )u( − s ) − KLw(0) where we have used the fact that du( − )/d = ( − ) = 0 at  = 1. Then

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(i)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

  K K  K  1 w(1) = (1−  )3 − s w( )(1−  )3 + 1− L w(0) + 1+ tL w(0) s o s  6 6  6  2  K K 1 w(1) = q (1−  )2 − s w( )(1−  )2 − L w(0) + (1+ K )w(0) o o s s tL 2 2 2 w(1) = qo (1− o ) − Ksw(s )(1− s ) − KLw(0) + KtLw(0)

(j)

w(1) = qo − Ksw(s ) − KLw(0) Then, substituting Eq. (j) into Eq. (f), we find that qo − Ksw(s ) − KLw(0)  qo    K K  K  +K (1−  )3 − s w( )(1−  )3 + 1− L w(0) + 1+ tL w(0) = 0   R s o s  2  6  6  6  qo K  q − K w( ) + K (1−  )3 − s w( )(1−  )3 o s s R o s s  6 6      KL  KtL  + K 1−  R  − K L  w(0) + K R 1+ 2  w(0) = 0 6  

(k)

qo (1− o ) − Ksw(s )(1− s ) − KLw(0) + KtLw(0)  qo K K  −K (1−  )2 − s w( )(1−  )2 − L w(0) + (1+ K )w(0) = 0  tL tR  s s o 2 2 2   qo Ks 2 2 (1−  ) − w( )(1−  ) q (1−  ) − K w( )(1−  ) − K  o  o o s s s tR s s 2 2    K K + tR L − K w(0) +  K − K (1+ K ) w(0) = 0  L tR tL   tL  2 

(l)

and

We choose to write Eq. (l) in matrix form as  a a12   w(0)   − A + K w( ) A  11 1 s s 2   =   a a22  w(0)   − B1 + Ksw(s )B2   21    where

 KL  1− −K , 11 R  L 6   KtR KL a = −K , 21 L 2 a =K

a = K 1 KtL  12 R  + 2    a = K − K (1+ K ) 22

tL

tR

tL

and

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(m)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

qo

A =q +K 1

o

R

A = 1+ 2

KR

(1−  )3 o

6

(1−  )3 s

6

B = q (1−  ) − 1

o

KtRqo

o

B = (1−  ) −

KtR

s

2

2

(1−  )2 o

2 (1−  )2 s

The solution to Eq. (m) is w(0) =

1 C + K w( )C   11

s s 12 D 1 w(0) = C21 + Ksw(s )C22  D

(n)

where D = a11a22 − a12a21 C11 = a12 B1 − a22 A1 C12 = a22 A2 − a12 B2 C21 = a21 A1 − a11B1 C22 = a11B2 − a21 A2 Substituting Eq. (n) into Eq. (i), we arrive at q K w() = o ( −  )3u( −  ) − s w( )( −  )3u( −  ) o o s s s 6  6 KL 3 1   K 1  + C + K w( )C   + tL 2 + C + K w( )C  1− s s 12   s s 22  D  11   D  21   6 2 q K = o ( −  )3u( −  ) − s w( )( −  )3u( −  ) o o s s s 6 6   K   K  + 1 C 1− L 3 + C  + tL 2  21    D  11  6  2   KL 3 KtL 2     K w(s )  1−  +C + + s  C12  22   2  D     6 qo K w() = ( −  )3u( −  ) − s w( )( −  )3u( −  ) + K w( )E () + E () o o s s s s s 1 2 6 6 where

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(p)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

1

   1− KL  3 + C  + KtL  2  E1 () = 22    C 6   2 D  12   1   KL  3  + C  + KtL  2  − E2 () =  21   C11 1 6   2  D 8.19 Take the Laplace transform with respect to  to obtain a solution to the following nondimensional equation u 2 u = 0  1   0 2   with zero initial conditions and with the boundary conditions u(0, ) = −1 

u(1, ) =0 

Solution: Taking the Laplace transform of the governing equation and the boundary conditions and using the initial condition gives 2U(, s) − sU( , s) = 0 2 and U(0, s) 1 =−  s

U(1, s) =0 

The solution to the transformed equation is U(, s) = A cosh

( s)+ B sinh( s)

Substituting this result into the boundary conditions yields U(0, s) 1 = − = A s sinh s 0 + B s cosh s 0  s 1 →B=− s s U(1, s) = A s sinh s + B s cosh s = 0 

( )

( ) cosh( s ) →A= s s sinh( s )

( )

( )

and, therefore,

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

( s )cosh( s)− sinh( s )sinh( s) s s sinh( s ) cosh( s[ − 1]) = s s sinh( s )

U(, s) =

1 cosh

It is seen that we have an second order pole at s = 0 and an infinite set of simple poles at

( s ) = 0 → − j sin j s = 0 → − j s = n → s = −n  2

sinh

2

Using Bromwich’s integral, 2 j d  z cosh u(, ) =  2 j dz  2

( z[ − 1])e  z

2

1

(

z[ −  +  d  z=0 2 j n=1 z z dz

( z[ − 1])e 

d  z cosh =  dz 

2 j  cosh

z

z=zk =− n2 2

cosh( jn [ − 1])e−n   +  −n2 2 cosh ( jn ) z=0 n=1 2 2 2   2  cosh( jn [ − 1])e−n 

2 2

− + + − 2  n2 cosh( jn )  n=1 3 2 2   1 2  cosh( jn )cosh( jn) − sinh( jn )sinh( jn) e−n = − + + − 2  n2 cosh( jn )  n=1 2 3 2 2   −n 2  1 2  cos ( n ) e = − + + − 2  2  n=1 2 3 n =

2 2

Mathematica verification of one term in series Limit[D[Sqrt[z] Cosh[(x-1) Sqrt[z]] Exp[z t]/Sinh[Sqrt[z]],z], z->0] FullSimplify[ Cosh[(x-1) Sqrt[z]] Exp[z t]/z^(3/2)/ D[Sinh[Sqrt[z]],z]/.z->-n^2 Pi^2,Assumptions->nIntegers] 8.20 Take the Laplace transform with respect to  to obtain a solution to the following partial differential equation u 

=

2 u 

2

+

u  

0  1   0

with the boundary conditions

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

lim | u(, ) |  →0

u(1, ) = sin and with the initial condition u(,) = 0.

Solution: Taking the Laplace transform of the partial differential equation and the boundary conditions, respectively, we obtain 2

2U(, s) 2

+

U(, s) 

− s 2U( , s) = 0

and lim | U(, s) |  →0

 U(1, s) = s2 + 2 The solution to the transformed equation is

( s)+ BK ( s)

U(, s) = AI0

0

For this solution to remain finite at  → 0, B = 0. Then, U(, s) = AI0

( s)

Substituting this into the second boundary condition yields s →A= U(1, s) =  = AI

( )

s2 + 2



(s +  )I ( s )

0

2

2

0

Thus, I0 U(, s) =

( s)

s 2 + 2 I 0

s

There are simple poles at s = j and at the roots of I0

( s)= 0

Using Eq. (5.126), I0

( s ) = J ( j s ) → I (− jy) = J (y) = 0 → J ( ) = 0 0

0

0

0

n

where y = jx. Therefore, since y = n, x = s → x = − jy → s = − jn → sn = − n2 The inverse is obtained by using the Bromwich integral, which gives

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

u(, ) =

(z + j)I 0

+

(z − j)I 0 z= j

z=− j



I 0

n=1

z 2 + 2

+

= =

I0

( − j)e

− j2I0 jI0

d I dz 0

− j

+

−

( − j)e 2I − j)

− j

−

z z=− n2

I 0

jI0

( j)e

( j)e

j

+ n=1

j



− 2 = n 1

0

I0



( − )e 2 n

 n4 +  2 I

− n2

− 2 / 2 − 2

j I ( j )e−  2

( +  ) I ( j ) n

0

4

n

n

2

n

1

n

Again using Eq. (5.126), Im (x) = j − m Jm ( jx) → Im ( jx) = j − m Jm (− x) = (− j)−m J m (x) Therefore, u(, ) =

j  I0  2 

=

j  I0 2 

( − j)e ( − j)e

− j

j

−

2

( j ) I ( j)e  −  J ( )e − 2 I0

− j

 j J ( )e − n  n 0 n  − 2 4 2  n=1 ( n +  ) J 1 ( n ) / (− j)

0



j

 

= ( 4 + 2 )J ( ) n

0

n 1

n

n

1

− n2

n

Mathematica verification of one term in series arg[z_,s_]:=Exp[z t] Om BesselI[0,x Sqrt[z]]/(z+I Om)/(z-I Om)/ If[s==0,BesselI[0,Sqrt[z]],D[BesselI[0,Sqrt[z]],z]] r1=Expand[(z-I Om) arg[z,0]]/.z->I Om; r2=Expand[(z+I Om) arg[z,0]]/.z->-I Om; Simplify[(r1+r2)] FullSimplify[PowerExpand[arg[z,1]/.z->-an^2]] 8.21 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation u 2 u = a2 2 t x

0xL

t0

with the boundary conditions

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

u(0,t ) =0 x u(L,t )  =  (1 − t ) x and with zero initial condition.

Solution: Taking the Laplace transform of the governing equation, we obtain 2U s − U=0 2 a2 x

(a)

and the boundary conditions become U(0, s) =0 x U(L, s)   1 1 =  − x  s s2 

(b)

where U = U(x,s) is the Laplace transform of u(x,t).

The solution to Eq. (a) is U(x, s) = A cosh(qx) + B sinh(qx)

(c)

where q = s a. Substituting Eq. (b) into the first boundary condition of Eq. (c) gives dU(x, s)

= Aq sinh(qx) + Bq cosh(qx) →

dx

dU(0, s)

= B=0

dx

Then Eq. (c) becomes U(x, s) = A cosh(qx) Substituting Eq. (d) into the second boundary condition of Eq. (b) yields U(L, s)   1 1 =  − x  s s2   1 1 Aq sinh(qL) = −  s2   s  1 1 A= −  s2  q sinh(qL) s

(d)

(e)

Then, Eq. (d) becomes U(x, s) =

 1 1  cosh(qx)  −  2   s s qsinh(qL)

(f)

To determine the inverse Laplace transform of Eq. (f), we use the Bromwich inversion integral, which for this case becomes

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

1  cosh(qx) 1 zt U(x,t ) =   z −  z2  q sinh(qL) e dz 2 j   −j 1 

 +j

(g)

where, now, q = z a. It is straightforward to show that zqsinh(qL) has a second order pole at z = 0 and that z2qsinh(qL) has a third order pole at z = 0. The other simple poles are those where sinh(qL) = 0. Thus, if qL = j, then sinh(j) = sin() = 0 and  = n. Therefore, qL = jn zL = jn a z → zn = −

n2 2a2 L2

Then, using residue theorem,   d  z2 cosh(qx)ezt    cosh(qx)ezt 1 U(x,t ) = lim   + lim   d  z→0 dz  zq sinh(qL)   k=1 z→zn  zq sinh(qL)   dz    zt  d2    3 zt cosh(qx)e 1 − lim  z cosh(qx)e   −  lim  z 2q d sinh(qL)  2  z→0 2!dz  z 2 q sinh(qL)   k=1 z→zn  dz   

(h)

We use the following Mathematica program to determine the limits of the first and third terms. q=Sqrt[z]/a; der1=Limit[D[z Cosh[q x]Exp[t z]/(q Sinh[q l]),z],z->0] der2=Limit[D[z Cosh[q x]Exp[t z]/(q Sinh[q l]),z,z],z->0] The results are d  z2 cosh(qx)ezt  = a2  t + x2 L2   2   − 2 z→0 dz 6a   zq sinh(qL)  L  2a 2 2 2 3 zt 2  7L4  d  z cosh(qx)e  = a t + tx 4 2 2 tL2 x L x  2 4 4 4  lim 2  2 − 2+ − + z→0 2!dz 6a 24a 12a L  2 2a 360a   z q sinh(qL)  lim











The evaluation of the second and fourth terms yields

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(i)

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

2 2 2 2    cosh(qx)ezt   cosh(qx)ezt  = − 2L cosh( jn x / L)e−n  a t /L 1 2  2 2 lim   = lim  d z→zn z→zn zL cosh(qL) / (2a )  zq n  cosh( jn ) sinh(qL)    dz   2 2 2 2    cosh(qx)ezt   = 2L3 cosh( jn x / L)e−n  a t /L  cosh(qx)ezt 1 2  lim  2 4 4  = lim  2 2 d z→zn z→zn z L cosh(qL) / (2a )  z q sinh(qL) a n  cosh( jn )    dz  

where we have used the fact that q = jn/L. Noting that cosh(jn) = cos(n) = (−1)n and using Eq. (i) , Eq. (h) becomes U(x,t ) =

a 2  x2 L2  2L  (−1)n 2 2 2 2 −n  a t /L t + − cos(n x / L)e   2 2 2 −  L  2a 6a   2 k=1 n a2  t 2 tx2 tL2 x4 L2 x2 7L4  − + − + − +  L  2 2a2 6a2 24a4 12a4 360a4  2 2 2 2 2L3  (−1)n −n  a t /L − 4 cos(n x / L)e 4   k=1 n

8.22 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation u 2 u = a2 2 t x

0x

t0

with the boundary conditions that u is finite as x →  and u(0,t) = f(t) and that the initial condition is zero.

Solution: Taking the Laplace transform of the governing equation with respect to t and using the initial condition, we obtain d 2u (x, s) s − 2 u (x, s) = 0 dx2 a The Laplace transform of the boundary condition at x = 0 is u (0, s) = F(s) where F(s) is the Laplace transform of f(t). The solution is u (x, s) = C e−qx + C eqx 1

2

where q = s / a. Since the solution must remain finite as x → , we set C2 = 0. Then u (x, s) = C1e−qx

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Substituting this result into the remaining boundary condition, we find that u (0, s) = F(s) = C1 Then the solution is u (x, s) = F(s)e−qx Taking the inverse Laplace transform with the aid of convolution, we have first that L−1 F(s) = f (t ) x e− x /( 4 ta ) − sx/a   L e  = 2a  t 3/2 2

2

−1

Then, from Eq. (8.25) u(x,t) =

e− x /( 4a ) 2

t

x

 f (t − )

2a  0

2

3/2

d

Mathematica verification InverseLaplaceTransform[Exp[-p Sqrt[s]],s,t]

8.23 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation u t

=

2 u x

2

− a 2u

−    1 x 1 t 0

with the boundary conditions u(−1,t ) = 1 u(1,t ) = 0 and zero initial condition; that is, u(x,0) = 0.

Solution: Taking the Laplace transform of the governing equation with respect to t and using the initial condition, we obtain 2 u (x, s) (s + a2 )u (x, s) = 0 − x2 and the Laplace transform of the boundary conditions are 1 u (−1, s) = s u (1, s) = 0 The solution to the transformed differential equation is

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

( s + a x ) + C sinh( s + a x)

u (x, s) = C1 cosh

2

2

2

Substituting this result into the boundary conditions gives

(

) ( ) 1s C cosh( s + a ) + C sinh( s + a ) = 0

C1 cosh − s + a2 + C2 sinh − s + a2 = 2

2

1

2

Placing in matrix form gives  cosh    cosh 

( s + a ) −sinh( s + a )   C   1  2

2

( s + a ) sinh( s + a ) 2

2

Solving for C1 and C2 yields

( s+a )

1 −sinh C1 =

1 s D 0

=  s  C   2   0   

1

2

=

1

sinh

( s+a ) 2

( s + a ) sD 1 s+a cosh ) s = − 1 cosh s + a ( 1 C = ( ) D sD cosh( s + a ) 0 2

sinh

2

2

2

2

where

( s + a )sinh( s + a ) = sinh(2 s + a )

D = 2 cosh

2

2

2

Therefore, u (x, s) =

sinh( s + a )cosh( s + a x ) − cosh( s + a )sinh( s + a x) s sinh(2 s + a ) 1

2

(

sinh (1 − x) s + a2 =

2

2

2

2

(

)

s sinh 2 s + a 2

To obtain the inverse transform, we note that s is a simple pole and we have an infinite number of simple poles that are obtained from n2 2 − a2 = − sinh 2 s + a2 = 0 → jn = 2 s + a2 → s = sn = − n 4

(

)

Then, the inverse transform is given by

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

(

sinh (1 − x) z + a 1 u(x,t ) =  z sinh 2 z + a 2 2 j  −j  +j

2

(

) e dz zt

Then, the residue at z = 0 is

(

)

  z sinh (1 − x) z + a2 sinh(a(1 − x)) zt   Res[z = 0] = lim e = z→0   sinh(2a)   The residues at z = −n are

)

  sinh (1 − x) zn + a 2 eznt  1 Res[z = −n ] =    d zn sinh 2 z + a2 n=1   dz z=zn     −t  sinh (1 − x) − n + a 2 e  1 =   d −n sinh 2 z + a2 n=1   dz  z=zn   2   sinh (1 − x) − + a e− nt −n + a 2   =   −n n=1 cosh 2 −n + a 2  

(



)

(

(

)

n

(

)

(

But, −n + a 2 = −a2 − n2 2 / 4 + a2 = jn / 2 Therefore,

 j sin(n (1 − x) / 2)e−(a2 +n2 2 /4 )t jn / 2  Res[z = −n ] =   cos(n ) − (a 2 + n2 2 / 4) n=1  2 2 2   (−1)n n sin(n (1 − x) / 2)e−(a +n  /4 )t =  2 n=1 (a2 + n2 2 / 4) 

and the solution is u(x,t ) =

sinh(a(1 − x)) sinh(2a)

  (−1)n n sin(n (1 − x) / 2)e−(a +n  /4 )t 2

+

2 n=1

2 2

(a + n2 2 / 4) 2

Mathematica verification of one term in series arg[z_,s_]:=Exp[z t]Sinh[(1-x) Sqrt[z+a^2]]/z/ If[s==0,Sinh[2 Sqrt[z+a^2]],D[Sinh[2 Sqrt[z+a^2]],z]] PowerExpand[Expand[z arg[z,0]]/.z->0]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Simplify[PowerExpand[arg[z,1]/.z->(-n^2 Pi^2/4-a^2)], Assumptions->nIntegers]

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab