Elementary differential equations 11th edition boyce solutions manual

Elementary Differential Equations 11th Edition Boyce Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/elementary-differential-equations-11th-edition-boyce-solutions-manual/

CHAPTER

6

The Laplace Transform

6.1 3.

The function f (t) is continuous.

6. Integration is a linear operation. It follows that Z Z Z A 1 A bt −st 1 A −bt −st cosh bt · e−st dt = e · e dt + e · e dt = 2 0 2 0 0 1 = 2

Z A e 0

(b−s)t

1 dt + 2

Z A

e−(b+s)t dt .

0

145

146

Chapter 6. The Laplace Transform Hence

Z A cosh bt · e 0

−st

1 1 − e−(b+s)A 1 1 − e(b−s)A + . dt = 2 s−b 2 s+b

Taking a limit, as A → ∞ ,

Z ∞ 1 1 1 1 s −st cosh bt · e dt = . + = 2 2 s − b 2 s + b s − b2 0 Note that the above is valid for s > |b| .

7. Proceeding as in Problem 6,

Z A 1 1 − e(b−s)A 1 1 − e−(b+s)A sinh bt · e−st dt = − .

2 s−b 2 s+b 0 Taking a limit, as A → ∞ ,

Z ∞ 1 1 1 1 b . sinh bt · e−st dt = − = 2 2 s−b 2 s+b s − b2 0 The limit exists as long as s > |b| .

8. Using the linearity of the Laplace transform,

1 ibt 1 L [ sin bt] = L e − L e−ibt .

2i 2i Since Z ∞ 1 , e(a+ib)t e−st dt = s − a − ib 0 we have

Z ∞ 0

Therefore L [sin bt] =

e± ibt e−st dt =

1 . s ∓ ib

1 1 1 b − = 2 .

2i s − ib s + ib s + b2

The formula holds for s > 0.

9. Using the linearity of the Laplace transform,

1 1 L [ cos bt] = L eibt + L e−ibt .

2 2 From Problem 8, we have Z ∞ 1 e± ibt e−st dt = . s ∓ ib 0 Therefore

1 1 1 s L [cos bt] = + = 2 .

2 s − ib s + ib s + b2

The formula holds for s > 0.

6.1

147 11. Using the linearity of the Laplace transform, i 1 h i

1 h L eat cos bt = L e(a+ib)t + L e(a−ib)t .

2 2 Based on the integration in Problem 8, Z ∞ e(a ± ib)t e−st dt = 0

1 . s − a ∓ ib

Therefore

at 1 1 1 s−a .

L e cos bt = + = 2 s − a − ib s − a + ib (s − a)2 + b2 The above is valid for s > a .

12. Integrating by parts, Z A Z A A 1 (a−s)t t e(a−s)t at −st + e dt = te · e dt = − s − a s − a 0 0 0 =

1 − eA(a−s) + A(a − s)eA(a−s) . (s − a)2

Taking a limit, as A → ∞ , Z ∞

teat · e−st dt =

0

1 . (s − a)2

Note that the limit exists as long as s > a .

14. Integrating by parts, Z A Z A A n n−1 (a−s)t tn e(a−s)t n at −st + t e · e dt = − t e dt = s − a s − a 0 0 0 An e−(s−a)A =− + s−a

Z A 0

n n−1 (a−s)t t e dt . s−a

Continuing to integrate by parts, it follows that Z A An e(a−s)A nAn−1 e(a−s)A tn eat · e−st dt = − − − s−a (s − a)2 0 −

n!Ae(a−s)A n!(e(a−s)A − 1) − · · · − . (n − 2)!(s − a)3 (s − a)n+1

That is, Z A

tn eat · e−st dt = pn (A) · e(a−s)A +

0

n! , (s − a)n+1

in which pn (ξ) is a polynomial of degree n . For any given polynomial, lim pn (A) · e−(s−a)A = 0 ,

A→∞

148

Chapter 6. The Laplace Transform as long as s > a . Therefore, Z ∞

tn eat · e−st dt =

0

n! . (s − a)n+1

15. First we observe that, by integrating by parts twice Z Z 1 a sin(at)e−st dt = − sin(at)e−st + cos(at)e−st dt s s

Z 1 a 1 a sin(at)e−st dt = − sin(at)e−st + − cos(at)e−st − s s s s 2 Z 1 a a = − sin(at)e−st − 2 cos(at)e−st − 2 sin(at)e−st dt s s s Solving for the original integral yields Z −e−st (s sin(at) + a cos(at)) sin(at)e−st dt = 2 s + a2 By a similar process, we find that Z −e−st cos(at)e−st dt = 2 (s cos(at) − a sin(at)) s + a2 We now find the Laplace transforms of t sin(at) and t cos(at). To find the Laplace transform of t sin(at), which is Z ∞ Z A L{t sin(at)} = t sin(a)e−st dt = lim t sin(a)e−st dt A→∞

0

0

we use integration parts with u = t and dv = sin(at)e−st dt. Then du = dt, v = −e−st s2 +a2 (s sin(at) + a cos(at)), and: !

Z A −st A e −te−st (s sin(at) + a cos(at)) + (s sin(at) + a cos(at)) dt L{t sin(at)} = lim 2 2 A→∞ s2 + a2 0 s +a 0 Z A Z A s a −st = 2 lim sin(at)e lim cos(at)e−st dt dt + 2 s + a2 A→∞ 0 s + a2 A→∞ 0 s a = 2 L{sin(at)} + 2 L{cos(at)} 2 s +a s + a2 s a a s = 2 · + 2 · s + a2 s2 + a2 s + a2 s2 + a2 2as = 2 (s + a2 )2 Proceeding in similar fashion, we find that the Laplace transform of t cos(at) is L{t cos(at)} =

s2 − a2 (s2 + a2 )2

.

Now, to find the Laplace transform of f (t) = t2 sin(at), which is Z ∞ Z A 2 2 −st F (s) = L{t sin(at)} = t sin(at)e dt = lim t2 sin(at)e−st dt 0

A→∞

0

6.1

149 we use integration by parts with u = t2 and dv = sin(at)e−st dt. Then du = 2tdt, −st v = −e s2 +a2 (s sin(at) + a cos(at)), and: !

Z A A te−st −t2 e−st (s sin(at) + a cos(at)) + 2 (s sin(at) + a cos(at)) dt F (s) = lim 2 2 A→∞ s2 + a2 0 s +a 0 Z A Z A 2s 2a −st lim t sin(at)e dt + lim t cos(at)e−st dt = 2 s + a2 A→∞ 0 s2 + a2 A→∞ 0 2s 2a = 2 L{t sin(at)} + 2 L{t cos(at)} s + a2 s + a2 s 2as a s2 − a2 · 2 + 2 · 2 = 2 2 2 2 2 s + a (s + a ) s + a (s + a2 )2 2a(3s2 − a2 ) = (s2 + a2 )3 This is valid for s > 0.

17. Using the definition of the Laplace transform, we get that Z ∞ Z 1 Z ∞ L [f (t)] = e−st f (t) dt = e−st t dt + e−st dt = 0

0

1

e−s e−s 1 e−s e−s 1 =− − 2 + 2+ = − 2 + 2. s s s s s s 20. Integrating by parts, Z A Z A A t e−t dt = −t e−t + e−t dt = 1 − e−A − Ae−A .

0

0

Taking a limit, as A → ∞ ,

Z ∞

0

t e−t dt = 1.

0

Hence the integral converges .

21. Based on a series expansion, note that for t > 0 , et > 1 + t + t2 /2 > t2 /2. It follows that for t > 0 , t−2 et > 1/2. Hence for any finite A > 1 , Z A A−1 . t−2 et dt > 2 1 It is evident that the limit as A → ∞ does not exist.

23.(a) Let p > 0 . Integrating by parts, Z A Z A Z A A e−x xp dx = −e−x xp + p e−x xp−1 dx = −Ap e−A + p e−x xp−1 dx.

0

0

0

Taking a limit, as A → ∞ , Z ∞ e 0

−x p

0

Z ∞

x dx = p 0

e−x xp−1 dx .

150

Chapter 6. The Laplace Transform That is, Γ(p + 1) = p Γ(p) . (b) Setting p = 0 ,

Z ∞ Γ(1) =

e−x dx = 1 .

0

(c) Let p = n . Using the result in part (a), Γ(n + 1) =n Γ(n) =n(n − 1)Γ(n − 1) .. . =n(n − 1)(n − 2) · · · 2 · 1 · Γ(1) .

Since Γ(1) = 1 , Γ(n + 1) = n! . (d) Using the result in part (a), Γ(p + n) =(p + n − 1) Γ(p + n − 1) =(p + n − 1)(p + n − 2)Γ(p + n − 2) .. . =(p + n − 1)(p + n − 2) · · · (p + 1)p Γ(p) .

Hence Γ(p + n) = p(p + 1)(p + 2) · · · (p + n − 1) . Γ(p) √ Given that Γ(1/2) = π , it follows that √ π 3 1 1 Γ( ) = Γ( ) = 2 2 2 2 and √ 11 9 7 5 3 3 945 π Γ( ) = · · · Γ( ) = .

2 2 2 2 2 2 32

6.2 1. Write the function as

3 3 2 = . s2 + 4 2 s2 + 4

Hence L−1 [Y (s)] = 3 sin 2t /2.

3. Using partial fractions,

2 2 1 1 = − . s2 + 3s − 4 5 s−1 s+4 Hence L−1 [Y (s)] = 2(et − e−4t )/5.

6.2

151

4. Note that the denominator s2 + 2s + 5 is irreducible over the reals. Completing the square, s2 + 2s + 5 = (s + 1)2 + 4 . Now convert the function to a rational function of the variable ξ = s + 1 . That is, 2(s + 1) 2s + 2 = . s2 + 2s + 5 (s + 1)2 + 4 We know that L−1

2ξ = 2 cos 2t . ξ2 + 4

Using the fact that L [eat f (t)] = L [f (t)]s→s−a ,

2s + 2 = 2e−t cos 2t .

L−1 2 s + 2s + 5 5. Using partial fractions,

2s − 3 1 7 1 = + . s2 − 4 4 s−2 s+2 Hence L−1 [Y (s)] = (e2t + 7e−2t )/4. Note that we can also write s 3 2 2s − 3 =2 2 − . s2 − 4 s − 4 2 s2 − 4 6. Using partial fractions, 8s2 − 4s + 12 1 s 2 =3 +5 2 −2 2 . s(s2 + 4) s s +4 s +4 Hence L−1 [Y (s)] = 3 + 5 cos 2t − 2 sin 2t .

7. The denominator s2 + 4s + 5 is irreducible over the reals. Completing the square, s2 + 4s + 5 = (s + 2)2 + 1 . Now convert the function to a rational function of the variable ξ = s + 2 . That is, 1 − 2s 5 − 2(s + 2) = . s2 + 4s + 5 (s + 2)2 + 1 We find that −1

L

5 2ξ − = 5 sin t − 2 cos t . ξ2 + 1 ξ2 + 1

Using the fact that L [eat f (t)] = L [f (t)]s→s−a ,

1 − 2s L−1 2 = e−2t (5 sin t − 2 cos t) . s + 4s + 5 9. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 3 [s Y (s) − y(0)] + 2 Y (s) = 0 .

152

Chapter 6. The Laplace Transform Applying the initial conditions, s2 Y (s) + 3s Y (s) + 2 Y (s) − s − 3 = 0 .

Solving for Y (s), the transform of the solution is Y (s) =

s+3 . s2 + 3s + 2

Using partial fractions, s+3 s2 + 3s + 2

=

2 1 − . s+1 s+2

Hence y(t) = L−1 [Y (s)] = 2 e−t − e−2t .

10. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) − 2 [s Y (s) − y(0)] + 2 Y (s) = 0 .

Applying the initial conditions, s2 Y (s) − 2s Y (s) + 2 Y (s) − 1 = 0 .

Solving for Y (s), the transform of the solution is Y (s) =

1 s2 − 2s + 2

.

Since the denominator is irreducible, write the transform as a function of ξ = s − 1 .

That is, 1 1 = .

2 s − 2s + 2 (s − 1)2 + 1 First note that −1

L

1 = sin t . ξ2 + 1

Using the fact that L [eat f (t)] = L [f (t)]s→s−a ,

1 −1 L = et sin t . s2 − 2s + 2 Hence y(t) = et sin t .

12. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 2 [s Y (s) − y(0)] + 5 Y (s) = 0 .

Applying the initial conditions, s2 Y (s) + 2s Y (s) + 5 Y (s) − 2s − 3 = 0 .

Solving for Y (s), the transform of the solution is Y (s) =

2s + 3 . s2 + 2s + 5

6.2

153 Since the denominator is irreducible, write the transform as a function of ξ = s + 1 .

That is, 2(s + 1) + 1 2s + 3 = .

2 s + 2s + 5 (s + 1)2 + 4 We know that L−1

1 1 2ξ = 2 cos 2t + sin 2t . + ξ2 + 4 ξ2 + 4 2

Using the fact that L [eat f (t)] = L [f (t)]s→s−a , the solution of the IVP is

2s + 3 1 y(t) = L−1 2 = e−t (2 cos 2t + sin 2t ) . s + 2s + 5 2 14. Taking the Laplace transform of the ODE, we obtain s4 Y (s) − s3 y(0) − s2 y 0 (0) − s y 00 (0) − y 000 (0) − Y (s) = 0 .

Applying the initial conditions, s4 Y (s) − Y (s) − s3 − s = 0 .

Solving for the transform of the solution, Y (s) =

s s2 − 1

.

By inspection, it follows that y(t) = L−1 [Y (s)] = cosh t .

18. Let f (t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 4 Y (s) = L [f (t)] .

Applying the initial conditions, s2 Y (s) + 4 Y (s) = L [f (t)] .

Based on the definition of the Laplace transform, Z ∞ Z 1 Z ∞ 1 e−s −st −st L [f (t)] = f (t) e dt = t e dt + e−st dt = 2 − 2 . s s 0 0 1 Solving for the transform, Y (s) =

1 1 − e−s 2 2 . s2 (s2 + 4) s (s + 4)

21.(a) Assuming that the conditions of Theorem 6.2.1 are satisfied, Z ∞ Z ∞

d ∂ −st F 0 (s) = e−st f (t)dt = e f (t) dt = ds 0 ∂s 0 Z ∞ Z ∞

−t e−st f (t) dt = e−st [−tf (t)] dt . = 0

0

154

Chapter 6. The Laplace Transform (b) Using mathematical induction, suppose that for some k ≥ 1 , Z ∞

(k) F (s) = e−st (−t)k f (t) dt .

0

Differentiating both sides, Z ∞ Z ∞

d ∂ −st (k+1) −st k e F (s) = (−t) f (t) dt = e (−t)k f (t) dt = ds 0 ∂s 0 Z ∞ Z ∞

e−st (−t)k+1 f (t) dt . = −t e−st (−t)k f (t) dt = 0

0

22. We know that L eat =

1 . s−a

Based on Problem 21,

d 1 at L −t e = . ds s − a Therefore, L t eat =

1 . (s − a)2

25. Using the translation property of the Laplace transform, L eat sin bt =

b . (s − a)2 + b2

Therefore,

at

L te

d b 2b(s − a) sin bt = − . = ds (s − a)2 + b2 ((s − a)2 + b2 )2

26.(a) Taking the Laplace transform of the given Bessel equation, L [ t y 00 ] + L [ y 0 ] + L [ t y ] = 0 .

Using the differentiation property of the transform, −

d d L [ y 00 ] + L [ y 0 ] − L[y] = 0. ds ds

That is, −

d 2 d s Y (s) − s y(0) − y 0 (0) + s Y (s) − y(0) − Y (s) = 0 . ds ds

It follows that (1 + s2 )Y 0 (s) + s Y (s) = 0 . (b) We obtain a first-order linear ODE in Y (s): s Y 0 (s) + 2 Y (s) = 0 , s +1

6.2

155 with integrating factor µ(s) = e

R

s ds s2 +1

=

p s2 + 1 .

The first-order ODE can be written as i d hp 2 s + 1 · Y (s) = 0 , ds with solution c Y (s) = √ . s2 + 1 (c) In order to obtain negative powers of s , first write

−1/2 1 1 1 √ = 1+ 2 . s s s2 + 1 Expanding (1 + 1/s2 )−1/2 in a binomial series, 1 p

1 + (1/s2 )

=1−

1 −2 1 · 3 −4 1 · 3 · 5 −6 s + s − s + ··· , 2 2·4 2·4·6

valid for s−2 < 1 . Hence, we can formally express Y (s) as

1 1 1 1·3 1 1·3·5 1 Y (s) = c − + − + ··· . s 2 s3 2 · 4 s5 2 · 4 · 6 s7 Assuming that term-by-term inversion is valid,

1 t2 1 · 3 t4 1 · 3 · 5 t6 y(t) = c 1 − + − + ··· 2 2! 2 · 4 4! 2 · 4 · 6 6!

4! t4 6! t6 2! t2 + − + ··· . =c 1− 2 2 2! 22 · 42 4! 22 · 42 · 62 6!

It follows that

∞ X 1 1 1 (−1)n 2n y(t) = c 1 − 2 t2 + 2 2 t4 − 2 2 2 t6 + · · · = c t .

2 2 ·4 2 ·4 ·6 22n (n!)2 n=0 The series is evidently the expansion, about x = 0 , of J0 (t).

28. By definition of the Laplace transform, given the appropriate conditions, Z t

Z ∞ Z ∞Z t L [ g(t)] = e−st f (τ )dτ dt = e−st f (τ )dτ dt .

0

0

0

0

Assuming that the order of integration can be exchanged, Z ∞

−sτ Z ∞ Z ∞ e −st f (τ ) e dt dτ = f (τ ) L [ g(t)] = dτ. s 0 τ 0 (Note the region of integration is the area between the lines τ (t) = t and τ (t) = 0.)

Hence Z 1 1 ∞ f (τ ) e−sτ dτ = L [ f (t)] .

L [ g(t)] = s 0 s

156

Chapter 6. The Laplace Transform

6.3 1.

2.

4.

6.3

157 6.(a)

(b) f (t) = 1 − 2u1 (t) + 2u2 (t) − 2u3 (t) + u4 (t).

7.(a)

(b) f (t) = 1 + (e−(t−2) − 1)u2 (t).

8.(a)

(b) f (t) = t + (2 − t)u2 (t) + (5 − t)u5 (t) + (t − 7)u7 (t).

9. Using the Heaviside function, we can write f (t) = (t − 2)2 u2 (t). The Laplace

158

Chapter 6. The Laplace Transform transform has the property that L [ uc (t)f (t − c)] = e−cs L [f (t)]. Hence L (t − 2)2 u2 (t) =

2 e−2s . s3

10. The function can be expressed as f (t) = (t − π) [uπ (t) − u2π (t)]. Before invoking the translation property of the transform, write the function as f (t) = (t − π) uπ (t) − (t − 2π) u2π (t) − π u2π (t) .

It follows that L [ f (t)] =

e−2πs πe−2πs e−πs − − . s2 s2 s

11. It follows directly from the translation property of the transform that L [ f (t)] =

e−3s e−4s e−s +2 −6 . s s s

12. Before invoking the translation property of the transform, write the function as f (t) = (t − 2) u2 (t) − u2 (t) − (t − 3) u3 (t) − u3 (t) .

It follows that L [ f (t)] =

e−2s e−3s e−3s e−2s − − − . s2 s s2 s

13. Using the fact that L [eat f (t)] = L [f (t)]s→s−a ,

3! −1 L = t3 e2t . (s − 2)4 16. Write the function as e−s e−2s e−3s e−4s + − − . s s s s It follows from the translation property of the transform, that −s

+ e−2s − e−3s − e−4s −1 e L = u1 (t) + u2 (t) − u3 (t) − u4 (t) . s F (s) =

17.(a) By definition of the Laplace transform, Z ∞ L [ f (ct)] = e−st f (ct)dt .

0

Making a change of variable, τ = ct , we have Z Z 1 ∞ −s(τ /c) 1 ∞ −(s/c)τ L [ f (ct)] = e f (τ )dτ = e f (τ )dτ . c 0 c 0 Hence L [ f (ct)] = (1/c) F (s/c) , where s/c > a .

6.3

159 (b) Using the result in part (a),

t L f = k F (ks). k Hence −1

L

1 [F (ks)] = f k

t . k

(c) From part (b), L−1 [F (as)] = (1/a)f (t/a) Note that as + b = a(s + b/a). Using the fact that L [ect f (t)] = L [f (t)]s→s−c ,

t −1 −bt/a 1 f .

L [F (as + b)] = e a a 18. First write F (s) =

n! . ( 2s )n+1

Let G(s) = n!/sn+1 . Based on the results in Problem 25, 1 −1 h s i L = g(2t), G 2 2 in which g(t) = tn . Hence L−1 [F (s)] = 2 (2t)n = 2n+1 tn .

21. By definition of the Laplace transform, Z ∞ L [ f (t)] = e−st u1 (t)dt .

0

That is, Z 1 L [ f (t)] = 0

e−st dt =

1 − e−s . s

22. First write the function as f (t) = u0 (t) − u1 (t) + u2 (t) − u3 (t) . It follows that Z 1 Z 3 L [ f (t)] = e−st dt + e−st dt .

0

2

That is, L [ f (t)] =

e−2s − e−3s 1 − e−s + e−2s − e−3s 1 − e−s + = . s s s

25. The given function is periodic, with T = 2 . Using the result of Problem 24, Z 2 Z 1 1 1 −st L [f (t)] = e f (t)dt = e−st dt .

1 − e−2s 0 1 − e−2s 0 That is, L [f (t)] =

1 − e−s 1 = . −2s s(1 − e ) s(1 + e−s )

160

Chapter 6. The Laplace Transform 27. The function is periodic, with T = 1 . Using the result of Problem 24, Z 1 1 L [f (t)] = t e−st dt .

1 − e−s 0 It follows that L [f (t)] =

1 − e−s (1 + s) . s2 (1 − e−s )

28. The function is periodic, with T = π . Using the result of Problem 24 Z π 1 sin t · e−st dt .

L [f (t)] = 1 − e−πs 0 We first calculate

Z π

sin t · e−st dt =

0

Hence L [f (t)] =

1 + e−πs .

1 + s2

1 + e−πs .

(1 − e−πs )(1 + s2 )

29.(a)

We get that L [f (t)] = L [1] − L [u1 (t)] =

(b)

1 e−s − . s s

6.3

161 Let F (s) = L [1 − u1 (t)]. Then

Z t 1 1 − e−s [1 − u1 (τ )] dτ = F (s) = .

L s s2 0 (c)

Let G(s) = L [g(t)]. Then L [h(t)] = G(s) − e−s G(s) =

1 − e−s (1 − e−s )2 1 − e−s − e−s = .

2 2 s s s2

30.(a)

(b) The given function is periodic, with T = 2 . Using the result of Problem 24, Z 2 1 L [f (t)] = e−st p(t)dt .

1 − e−2s 0 Based on the piecewise definition of p(t), Z 2 Z 1 Z 2 1 e−st p(t)dt = t e−st dt + (2 − t)e−st dt = 2 (1 − e−s )2 . s 0 0 1 Hence L [p(t)] =

(1 − e−s ) . s2 (1 + e−s )

162

Chapter 6. The Laplace Transform (c) Since p(t) satisfies the hypotheses of Theorem 6.2.1 , L [p 0 (t)] = s L [p(t)] − p(0).

Using the result of Problem 26 from the Student Solutions Manual, L [p 0 (t)] =

(1 − e−s ) . s(1 + e−s )

We note the p(0) = 0 , hence L [p(t)] =

1 (1 − e−s ) . s s(1 + e−s )

6.4 2. (a)

(b) Let h(t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 2 [s Y (s) − y(0)] + 2 Y (s) = L [h(t)] .

Applying the initial conditions, s2 Y (s) + 2s Y (s) + 2 Y (s) − 1 = L [h(t)] .

The forcing function can be written as h(t) = uπ (t) − u2π (t) . Its transform is e−πs − e−2πs . s Solving for Y (s), the transform of the solution is L [h(t)] =

Y (s) = First note that

1 e−πs − e−2πs + . s2 + 2s + 2 s(s2 + 2s + 2)

1 1 = . s2 + 2s + 2 (s + 1)2 + 1

Thus the inverse transform of the first term is

1 1 −1 −1 L =L = e−t sin t . s2 + 2s + 2 (s + 1)2 + 1

6.4

163 Now let G(s) =

1 s(s2 + 2s + 2)

.

Using partial fractions,

1 1 1 (s + 1) + 1 G(s) = − .

2 s 2 (s + 1)2 + 1 so L−1 [G(s)] =

1 −1 1 1 (s + 1) + 1 1 1 1 L − L−1 = − e−t cos t − e−t sin t .

2 2 s 2 (s + 1) + 1 2 2 2

Using Theorem 6.3.1,

1 1 L−1 e−cs G(s) = uc (t) − e−(t−c) [cos(t − c) + sin(t − c)] uc (t) .

2 2 Hence the solution of the IVP is 1 1 y(t) = e−t sin t + uπ (t) − e−(t−π) [cos(t − π) + sin(t − π)] uπ (t) 2 2 1 1 −(t−2π) − u2π (t) + e [cos(t − 2π) + sin(t − 2π)] u2π (t) .

2 2 That is, y(t) = e−t sin t +

1 1 [uπ (t) − u2π (t)] + e−(t−π) [cos t + sin t] uπ (t) 2 2

1 + e−(t−2π) [cos t + sin t] u2π (t) .

2 (c)

(d) The solution starts out as free oscillation, due to the initial conditions. The amplitude increases, as long as the forcing is present. Thereafter, the solution rapidly decays.

164

Chapter 6. The Laplace Transform 4.(a)

(b) Let f (t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 3 [s Y (s) − y(0)] + 2 Y (s) = L [f (t)] .

Applying the initial conditions, s2 Y (s) + 3s Y (s) + 2 Y (s) = L [f (t)] .

The transform of the forcing function is L [f (t)] =

1 e−10s − . s s

Solving for the transform, Y (s) =

1 s(s2 + 3s + 2)

−

e−10s s(s2 + 3s + 2)

.

Using partial fractions,

1 1 1 1 2 = + − . s(s2 + 3s + 2) 2 s s+2 s+1 Hence −1

L

1 1 e−2t = + − e−t . s(s2 + 3s + 2) 2 2

Based on Theorem 6.3.1,

i e−10s 1h −1 −2(t−10) −(t−10) L = 1 + e − 2e u10 (t) . s(s2 + 3s + 2) 2 Hence the solution of the IVP is y(t) =

i 1 e−2t 1 h −(2t−20) [1 − u10 (t)] + − e−t − e − 2e−(t−10) u10 (t) .

2 2 2

6.4

165 (c)

(d) The solution increases to a temporary steady value of y = 1/2. After the forcing ceases, the response decays exponentially to y = 0 .

7.(a)

(b) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 4 Y (s) =

e−πs e−3πs − . s s

Applying the initial conditions, s2 Y (s) + 4 Y (s) =

e−πs e−3πs − . s s

Solving for the transform, Y (s) =

e−πs e−3πs − . s(s2 + 4) s(s2 + 4)

Using partial fractions,

1 1 1 s = − . s(s2 + 4) 4 s s2 + 4 Taking the inverse transform yields

1 1 −1 L = (t − cos 2t) . s(s2 + 4) 4

166

Chapter 6. The Laplace Transform Based on Theorem 6.3.1, the solution of the IVP is 1 1 [1 − cos(2t − 2π)] uπ (t) − [1 − cos(2t − 6π)] u3π (t) 4 4 1 1 = [uπ (t) − u3π (t)] − cos 2t · [uπ (t) − u3π (t)] .

4 4

y(t) =

(c)

(d) Since there is no damping term, the solution responds immediately to the forcing input. There is a temporary oscillation about y = 1/4 .

8.(a)

(b) Taking the Laplace transform of the ODE, we obtain s4 Y (s) − s3 y(0) − s2 y 0 (0) − s y 00 (0) − y 000 (0)+

1 e−πs + 5 s2 Y (s) − s y(0) − y 0 (0) + 4 Y (s) = − . s s Applying the initial conditions, s4 Y (s) + 5s2 Y (s) + 4Y (s) =

1 e−πs − . s s

Solving for the transform of the solution, Y (s) =

1 e−πs − . s(s4 + 5s2 + 4) s(s4 + 5s2 + 4)

6.4

167 Using partial fractions,

1 3 s 4s 1 = + − . s(s4 + 5s2 + 4) 12 s s2 + 4 s2 + 1 It follows that −1

L

1 1 = [3 + cos 2t − 4 cos t] .

4 2 s(s + 5s + 4) 12

Based on Theorem 6.3.1, the solution of the IVP is y(t) =

1 1 1 [1 − uπ (t)] + [cos 2t − 4 cos t] − [cos 2(t − π) − 4 cos(t − π)] uπ (t) .

4 12 12

That is, y(t) =

1 1 1 [1 − uπ (t)] + [cos 2t − 4 cos t] − [cos 2t + 4 cos t] uπ (t) .

4 12 12

(c)

(d) After an initial transient, the solution oscillates about ym = 0 .

9. The function is defined by   0 ≤ t < t0 0, h f (t) = k (t − t0 ), t0 ≤ t < t0 + k   h, t ≥ t0 + k which can also be written as f (t) =

h h (t − t0 ) ut0 (t) − (t − t0 − k) ut0 +k (t) . k k

10. The function is defined by  0, 0 ≤ t < t0     h (t − t ), t0 ≤ t < t0 + k 0 g(t) = k h  − (t − t − 2k), t0 + k ≤ t < t0 + 2k 0    k 0, t ≥ t0 + 2k

168

Chapter 6. The Laplace Transform which can also be written as h 2h h g(t) = (t − t0 ) ut0 (t) − (t − t0 − k) ut0 +k (t) + (t − t0 − 2k) ut0 +2k (t). k k k 12. We consider the initial value problem y 00 + 4y =

1 [(t − 5) u5 (t) − (t − 5 − k) u5+k (t)] , k

with y(0) = y 0 (0) = 0 . (a) The specified function is defined by   0≤t<5 0, 1 f (t) = k (t − 5), 5 ≤ t < 5 + k   1, t≥5+k so k controls the point at which f (t) reaches 1. When k = 5, f (t) = g(t) in Ex.2.

(b) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 4 Y (s) =

e−5s e−(5+k)s − . ks2 ks2

Applying the initial conditions, s2 Y (s) + 4 Y (s) =

e−5s e−(5+k)s − .

2 ks ks2

Solving for the transform, Y (s) =

e−(5+k)s e−5s − . ks2 (s2 + 4) ks2 (s2 + 4)

Using partial fractions,

1 1 1 = − 2 . s2 (s2 + 4) 4 s2 s +4 1

It follows that L

−1

1 s2 (s2 + 4)

=

1 1 t − sin 2t .

4 8

6.4

169 Using Theorem 6.3.1, the solution of the IVP is y(t) =

1 [h(t − 5) u5 (t) − h(t − 5 − k) u5+k (t)] , k

in which h(t) = t/4 − sin 2t /8. (c) Note that for t > 5 + k , the solution is given by y(t) =

1 1 1 1 sin k − sin(2t − 10) + sin(2t − 10 − 2k) = − cos(2t − 10 − k) .

4 8k 8k 4 4k

So for t > 5 + k , the solution oscillates about ym = 1/4 , with an amplitude of A=

|sin(k)| .

4k

(a) k = 3/2

(b) k = 1

(c) k = 1/2

(a) n = 3

(b) n = 4

(c) n = 5

15.(a)

(b) Taking the Laplace transform of both sides of the ODE, we obtain n

s2 Y (s) − s y(0) − y 0 (0) + Y (s) =

1 X (−1)k e−kπs + . s s k=1

Applying the initial conditions, n

s2 Y (s) + Y (s) =

1 X (−1)k e−kπs + . s s k=1

170

Chapter 6. The Laplace Transform Solving for the transform, n

Y (s) =

X (−1)k e−kπs 1 + .

2 s(s + 1) s(s2 + 1) k=1

Using partial fractions, 1 s(s2 + 1)

=

1 s − 2 . s s +1

Let −1

h(t) = L

1 = 1 − cos t . s(s2 + 1)

Applying Theorem 6.3.1, term-by-term, the solution of the IVP is y(t) = h(t) +

n X

(−1)k h(t − kπ) ukπ (t) .

k=1

Note that h(t − kπ) = u0 (t − kπ) − cos(t − kπ) = ukπ (t) − (−1)k cos t .

Hence y(t) = 1 − cos t +

n X k=1

(−1)k ukπ (t) − (cos t)

n X

ukπ (t) .

k=1

(c)

The ODE has no damping term. Each interval of forcing adds to the energy of the system, so the amplitude will increase. For n = 15, g(t) = 0 when t > 15π. Therefore the oscillation will eventually become steady, with an amplitude depending on the values of y(15π) and y 0 (15π) . (d) As n increases, the interval of forcing also increases. Hence the amplitude of the transient will increase with n . Eventually, the forcing function will be constant.

In fact, for large values of t , ( 1 , n even g(t) = 0 , n odd

6.4

171 Further, for t > nπ , y(t) = 1 − cos t − n cos t −

1 − (−1)n .

2

Hence the steady state solution will oscillate about 0 or 1 , depending on n , with an amplitude of A = n + 1 . In the limit, as n → ∞ , the forcing function will be a periodic function, with period 2π . From Problem 33, in Section 6.3 , L [g(t)] =

1 . s(1 + e−s )

As n increases, the duration and magnitude of the transient will increase without bound.

16.(a) Taking the initial conditions into consideration, the transform of the ODE is n

s2 Y (s) + 0.1 s Y (s) + Y (s) =

1 X (−1)k e−kπs + . s s k=1

Solving for the transform, n

Y (s) =

X (−1)k e−kπs 1 + . s(s2 + 0.1s + 1) s(s2 + 0.1s + 1) k=1

Using partial fractions, 1 s(s2 + 0.1s + 1)

=

1 s + 0.1 − 2 . s s + 0.1s + 1

Since the denominator in the second term is irreducible, write s + 0.1 (s + 0.05) + 0.05 = . s2 + 0.1s + 1 (s + 0.05)2 + (399/400) Let

1 (s + 0.05) 0.05 − − 2 s (s + 0.05) + (399/400) (s + 0.05)2 + (399/400) " # √ √ 399 1 399 −t/20 sin( =1−e cos( t) + √ t) .

20 20 399

h(t) = L−1

Applying Theorem 6.3.1, term-by-term, the solution of the IVP is y(t) = h(t) +

n X

(−1)k h(t − kπ) ukπ (t) .

k=1

For odd values of n, the solution approaches y = 0 . (On the next figure, n = 5.)

172

Chapter 6. The Laplace Transform

For even values of n, the solution approaches y = 1 . (On the next figure, n = 6.)

√ (b) The solution is a sum of damped sinusoids, each of frequency ω = 399 /20 ≈ 1 .

Each term has an initial amplitude of approximately 1 . For any given n , the solution contains n + 1 such terms. Although the amplitude will increase with n, the amplitude will also be bounded by n + 1 . (c) Suppose that the forcing function is replaced by g(t) = sin t . Based on the methods in Chapter 3, the general solution of the differential equation is √

# √ 399 399 c1 cos( t) + c2 sin( t) + yp (t) .

20 20

" −t/20

y(t) = e

Note that yp (t) = A cos t + B sin t . Using the method of undetermined coefficients, A = −10 and B = 0 . Based on the initial conditions, the solution of the IVP is " y(t) = 10 e

−t/20

# √ 399 1 399 t) + √ sin( t) − 10 cos t . cos( 20 20 399 √

Observe that both solutions have the same frequency, ω =

√

399/20 ≈ 1 .

6.4

173

17.(a) Taking the initial conditions into consideration, the transform of the ODE is n

s2 Y (s) + Y (s) =

X (−1)k e−(11k/4)s 1 +2 . s s k=1

Solving for the transform, n

Y (s) =

X (−1)k e−(11k/4)s 1 + 2 . s(s2 + 1) s(s2 + 1) k=1

Using partial fractions, 1 s(s2 + 1)

=

1 s − 2 . s s +1

Let h(t) = L−1

1 = 1 − cos t . s(s2 + 1)

Applying Theorem 6.3.1, term-by-term, the solution of the IVP is

y(t) = h(t) + 2

n X

(−1)k h(t −

k=1

11k ) u11k/4 (t) .

4

That is,

y(t) = 1 − cos t + 2

n X

11k (−1)k 1 − cos(t − ) u11k/4 (t) .

4

k=1

(b) On the figure we see the solution for n = 35.

174

Chapter 6. The Laplace Transform

(c) Based on the plot, the slow period appears to be 88 . The fast period appears to be about 6 . These values correspond to a slow frequency of ωs = 0.0714 and a fast frequency ωf = 1.0472 . (d) The natural frequency of the system is ω0 = 1 . The forcing function is initially periodic, with period T = 11/2 = 5.5 . Hence the corresponding forcing frequency is w = 1.1424 . Using the results in Section 3.8, the slow frequency is given by ωs =

|ω − ω0 | = 0.0712 2

and the fast frequency is given by |ω + ω0 | = 1.0712 .

2 Based on theses values, the slow period is predicted as 88.247 and the fast period is given as 5.8656 . ωf =

6.5 2.(a) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 4Y (s) = e−πs − e−2πs .

Applying the initial conditions, s2 Y (s) + 4Y (s) = e−πs − e−2πs .

Solving for the transform, e−πs − e−2πs e−πs e−2πs = − . s2 + 4 s2 + 4 s2 + 4 Applying Theorem 6.3.1, the solution of the IVP is Y (s) =

y(t) =

1 1 1 sin(2t − 2π)uπ (t) − sin(2t − 4π)u2π (t) = sin(2t) [ uπ (t) − u2π (t) ] .

2 2 2

6.5

175 (b)

6.(a) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y(0) − y 0 (0) + 4Y (s) = 2 e−(π/4)s .

Applying the initial conditions, s2 Y (s) + 4Y (s) = 2 e−(π/4)s .

Solving for the transform, Y (s) =

2 e−(π/4)s . s2 + 4

Applying Theorem 6.3.1, the solution of the IVP is π y(t) = sin(2t − )uπ/4 (t) = − cos(2t) uπ/4 (t) .

2 (b)

7.(a) Taking the initial conditions into consideration, the transform of the ODE is s s2 Y (s) + 2s Y (s) + 2 Y (s) = 2 + e−(π/2)s . s +1 Solving for the transform, Y (s) =

s (s2 + 1)(s2 + 2s + 2)

+

e−(π/2)s . s2 + 2s + 2

176

Chapter 6. The Laplace Transform Using partial fractions, s (s2 + 1)(s2 + 2s + 2) We can also write

=

1 s 2 s+4 + − .

5 s2 + 1 s2 + 1 s2 + 2s + 2

s+4 s2 + 2s + 2

=

Let Y1 (s) = Then L−1 [Y1 (s)] =

(s + 1) + 3 . (s + 1)2 + 1 s

(s2 + 1)(s2 + 2s + 2)

.

1 2 1 cos t + sin t − e−t [cos t + 3 sin t] .

5 5 5

Applying Theorem 6.3.1, −(π/2)s

π π e = e−(t− 2 ) sin t − uπ/2 (t) .

L−1 2 s + 2s + 2 2 Hence the solution of the IVP is π 1 2 1 y(t) = cos t + sin t − e−t [cos t + 3 sin t] − e−(t− 2 ) cos(t) uπ/2 (t) .

5 5 5 (b)

8.(a) Taking the initial conditions into consideration, the transform of the ODE is s4 Y (s) − Y (s) = e−s .

Solving for the transform, Y (s) =

e−s s4 − 1

.

Using partial fractions,

1 1 1 1 = − . s4 − 1 2 s2 − 1 s2 + 1 It follows that −1

L

1 1 1 = sinh t − sin t . s4 − 1 2 2

6.5

177 Applying Theorem 6.3.1, the solution of the IVP is y(t) =

1 [sinh(t − 1) − sin(t − 1)] u1 (t) .

2

(b)

10.(a) The Laplace transform of the ODE is 1 s2 Y (s) + s Y (s) + Y (s) = e−s .

2 Solving for the transform of the solution, Y (s) =

e−s . s2 + s/2 + 1

First write 1 1 = .

1 2 s2 + s/2 + 1 (s + 4 ) + 15 16 Taking the inverse transform and applying both shifting theorems, √ 4 15 y(t) = √ e−(t−1)/4 sin (t − 1) u1 (t) .

4 15

178

Chapter 6. The Laplace Transform

(b) As shown on the graph, the maximum is attained at some t1 > 2. Note that for t > 2, √ 15 4 −(t−1)/4 (t − 1) . y(t) = √ e sin 4 15 Setting y 0 (t) = 0 , we find that t1 ≈ 2.3613 . The maximum value is calculated as y(2.3613) ≈ 0.71153 . (c) Setting γ = 1/4 , the transform of the solution is Y (s) =

e−s s2 + s/4 + 1

.

Following the same steps, it follows that √ 8 3 7 (t − 1) u1 (t) . y(t) = √ e−(t−1)/8 sin 8 3 7

Once again, the maximum is attained at some t1 > 2 . Setting y 0 (t) = 0 , we find that t1 ≈ 2.4569 , with y(t1 ) ≈ 0.8335 . (d) Now suppose that 0 < γ < 1 . Then the transform of the solution is Y (s) = First write

e−s s2 + γs + 1

.

1 1 = . s2 + γs + 1 (s + γ/2)2 + (1 − γ 2 /4)

It follows that −1

h(t) = L

p 1 2 =p e−γt/2 sin( 1 − γ 2 /4 · t).

2 2 s + γs + 1 4−γ

Hence the solution is y(t) = h(t − 1) u1 (t) .

The solution is nonzero only if t > 1 , in which case y(t) = h(t − 1) . Setting y 0 (t) = 0 , we obtain hp i 1p tan 1 − γ 2 /4 · (t − 1) = 4 − γ2 , γ

6.5

179 that is, tan

hp i 1 − γ 2 /4 · (t − 1) 2 p = .

2 γ 1 − γ /4

As γ → 0, we obtain the formal equation tan(t − 1) = ∞. Hence t1 → 1 + π2 . Setting t = π/2 in h(t), and letting γ → 0 , we find that y1 → 1 . These conclusions agree with the case γ = 0 , for which it is easy to show that the solution is y(t) = sin(t − 1) u1 (t) .

11.(a) See Problem 10. It follows that the solution of the IVP is √ 15 4k −(t−1)/4 sin (t − 1) u1 (t) . y(t) = √ e 4 15 This function is a multiple of the answer in Problem 10(a). Hence the peak value occurs at t1 ≈ 2.3613. The maximum value is calculated as y(2.3613) ≈ 0.71153 k.

We find that the appropriate value of k is k1 = 2/0.71153 ≈ 2.8108. (b) Based on Problem 10(c), the solution is √ 8k 3 7 y(t) = √ e−(t−1)/8 sin (t − 1) u1 (t) .

8 3 7 Since this function is a multiple of the solution in Problem 14(c), we have that t1 ≈ 2.4569, with y(t1 ) ≈ 0.8335 k. The solution attains a value of y = 2 when k1 = 2/0.8335, that is, k1 ≈ 2.3995. (c) Similar to Problem 10(d), for 0 < γ < 1 , the solution is y(t) = h(t − 1) u1 (t) , in which

p 2k h(t) = p e−γt/2 sin( 1 − γ 2 /4 · t).

4 − γ2

It follows that t1 − 1 → π/2 . Setting t = π/2 in h(t), and letting γ → 0 , we find that y1 → k . Requiring that the peak value remains at y = 2 , the limiting value of k is k1 = 2 . These conclusions agree with the case γ = 0 , for which it is easy to show that the solution is y(t) = k sin(t − 1) u1 (t) .

12.(a) Taking the initial conditions into consideration, the transformation of the ODE is

e−(4+k)s 1 e−(4−k)s 2 − . s Y (s) + Y (s) = 2k s s Solving for the transform of the solution,

1 e−(4−k)s e−(4+k)s Y (s) = − .

2k s(s2 + 1) s(s2 + 1)

180

Chapter 6. The Laplace Transform Using partial fractions, 1 1 s = − 2 . s(s2 + 1) s s +1 Now let h(t) = L−1

1 = 1 − cos t . s(s2 + 1)

Applying Theorem 6.3.1, the solution is φ(t , k) =

1 [h(t − 4 + k) u4−k (t) − h(t − 4 − k) u4+k (t)] .

2k

That is, 1 [ u4−k (t) − u4+k (t)] − 2k 1 − [cos(t − 4 + k) u4−k (t) − cos(t − 4 − k) u4+k (t)] .

2k

φ(t , k) =

(b) Consider various values of t. For any fixed t < 4, φ(t, k) = 0, as long as 4 − k > t.

If t ≥ 4 , then for 4 + k < t , φ(t , k) = −

1 [cos(t − 4 + k) − cos(t − 4 − k) ] .

2k

It follows that lim φ(t , k) = lim −

k→0

k→0

cos(t − 4 + k) − cos(t − 4 − k) = sin(t − 4) .

2k

Hence lim φ(t , k) = sin(t − 4) u4 (t) .

k→0

(c) The Laplace transform of the differential equation y 00 + y = δ(t − 4), with y(0) = y 0 (0) = 0 , is s2 Y (s) + Y (s) = e−4s .

Solving for the transform of the solution, Y (s) =

e−4s . s2 + 1

It follows that the solution is φ0 (t) = sin(t − 4) u4 (t) , and this means that lim φ(t, k) = φ0 (t).

k→0

(d) We can see the convergence on the graphs.

6.5

181

14.(b) The transform of the ODE (given the specified initial conditions) is 20 X

s2 Y (s) + Y (s) =

(−1)k+1 e−kπs .

k=1

Solving for the transform of the solution,

Y (s) =

20 X 1 (−1)k+1 e−kπs . s2 + 1 k=1

Applying Theorem 6.3.1, term-by-term,

y(t) =

20 X

(−1)k+1 sin(t − kπ) ukπ (t) = − sin(t) ·

20 X

ukπ (t) .

k=1

k=1

(c) For t > 20π, y(t) = −20 sin t.

16.(b) Taking the initial conditions into consideration, the transform of the ODE is s2 Y (s) + Y (s) =

40 X k=1

(−1)k+1 e−(11k/4)s .

182

Chapter 6. The Laplace Transform Solving for the transform of the solution, 40 X

Y (s) =

(−1)k+1

k=1

e−(11k/4)s . s2 + 1

Applying Theorem 6.3.1, term-by-term, y(t) =

40 X

(−1)k+1 sin(t −

k=1

11k ) u11k/4 (t) .

4

(c) For t > 110, y(t) =

40 X

(−1)

k=1

k+1

11k sin t − 4

≈ −5.13887 cos(56.375 − t).

17.(b) The transform of the ODE (given the specified initial conditions) is 2

s Y (s) + 0.1s Y (s) + Y (s) =

20 X

(−1)k+1 e−kπs .

k=1

Solving for the transform of the solution, Y (s) =

20 X k=1

e−kπs . s2 + 0.1s + 1

First write 1 s2 + 0.1s + 1 It follows that −1

L

=

1

.

1 2 ) + 399 (s + 20 400

√

1 20 −t/20 399 √ = e sin( t). s2 + 0.1s + 1 20 399

Applying Theorem 6.3.1, term-by-term, y(t) =

20 X k=1

(−1)k+1 h(t − kπ) ukπ (t) ,

6.5

183 in which √ 399 20 −t/20 h(t) = √ e sin( t).

20 399

(c) For t > 20π, each term in the sum for y(t) will contain a decaying exponential term multiplied by a bounded term. Thus y(t) → 0.

18.(b) Taking the initial conditions into consideration, the transform of the ODE is s2 Y (s) + 0.1s Y (s) + Y (s) =

15 X

e−(2k−1)πs .

k=1

Solving for the transform of the solution, 15 X e−(2k−1)πs Y (s) = . s2 + 0.1s + 1 k=1

As shown in Problem 23, −1

L

√

1 20 −t/20 399 =√ e sin( t). s2 + 0.1s + 1 20 399

Applying Theorem 6.3.1, term-by-term,

y(t) =

15 X

h [t − (2k − 1)π] u(2k−1)π (t) ,

k=1

in which √ 20 −t/20 399 h(t) = √ e sin( t).

20 399

184

Chapter 6. The Laplace Transform

(c) For t > 29π, each term in the sum for y(t) will contain a decaying exponential term multiplied by a bounded term. Thus y(t) → 0.

6.6 2. Let f (t) = et . Then Z t (f ∗ 1) (t) =

e

t−τ

· 1 dτ = e

t

Z t

0

e−τ dτ = et − 1 .

0

3. It follows directly that Z Z t 1 1 t [cos(t − 2τ ) − cos(t)]dτ = (sin t − t cos t). (f ∗ f ) (t) = sin(t − τ ) sin(τ ) dτ = 2 0 2 0 The range of the resulting function is R .

5. We have L [e−t ] = 1/(s + 1) and L [sin t] = 1/(s2 + 1). Based on Theorem 6.6.1, Z t

1 1 1 L e−(t−τ ) sin(τ ) dτ = · 2 = .

2 + 1) s + 1 s + 1 (s + 1)(s 0 6. We have f (t) = (g ∗ h) (t) , in which g(t) = sin t and h(t) = cos t . The transform of the convolution integral is Z t

1 s s L g(t − τ )h(τ ) dτ = 2 · 2 = 2 . s +1 s +1 (s + 1)2 0 8. It is easy to see that

1 L−1 = e−t s+1 Applying Theorem 6.6.1,

L−1

and

L−1

s = cos 2t . s2 + 4

Z t s = e−(t−τ ) cos 2τ dτ . (s + 1)(s2 + 4) 0

6.6

185 9. We first note that

−1

L

1 = t e−t (s + 1)2

−1

and

L

1 1 = sin 2t .

2 s +4 2

Based on the convolution theorem,

Z 1 1 t (t − τ )e−(t−τ ) sin 2τ dτ = L−1 (s + 1)2 (s2 + 4) 2 0 Z 1 t −τ = τ e sin(2t − 2τ ) dτ .

2 0 10.(a) By definition, Z t f ∗g =

(t − τ )m τ n dτ .

0

Set τ = t − t u , dτ = −t du, so that Z t Z 0 Z 1 m n m n m+n+1 (t − τ ) τ dτ = − (tu) (t − tu) t du = t um (1 − u)n du .

0

1

0

(b) The Convolution Theorem states that L [f ∗ g] = L [f ] · L [g]. Noting that L tk =

k! , sk+1

it follows that m! n! (m + n + 1) ! = sm+1 sm+1 sm+n+2 Therefore

Z 1

um (1 − u)n du =

0

Z 1

um (1 − u)n du .

0

m!n! . (m + n + 1) !

(c) If k is not an integer, we can write L[tk ] =

Γ(k + 1) sk+1

and Γ(m + 1) Γ(n + 1) Γ(m + 1) · = m+n+2 sm+1 sn+1 s so

Z 1

um (1 − u)n du =

0

Z 1

um (1 − u)n du,

0

Γ(m + 1)Γ(n + 1) . Γ(m + n + 2)

11. Taking the initial conditions into consideration, the transform of the ODE is s2 Y (s) − 1 + ω 2 Y (s) = G(s).

Solving for the transform of the solution, Y (s) =

G(s) 1 + 2 . s2 + ω 2 s + ω2

186

Chapter 6. The Laplace Transform As shown in a related situation, Problem 11,

Z 1 t G(s) −1 = sin(ω(t − τ )) g(τ ) dτ .

L s2 + ω 2 ω 0 Hence the solution of the IVP is y(t) =

1 1 sin(ωt) + ω ω

Z t sin(ω(t − τ )) g(τ ) dτ .

0

12. The transform of the ODE (given the specified initial conditions) is 4s2 Y (s) + 4s Y (s) + 17 Y (s) = G(s).

Solving for the transform of the solution, Y (s) = First write

G(s) 4s2 + 4s + 17

.

1 1 4 . = 4s2 + 4s + 17 (s + 12 )2 + 4

Based on the elementary properties of the Laplace transform,

1 1 L−1 = e−t/2 sin 2t .

4s2 + 4s + 17 8 Applying the convolution theorem, the solution of the IVP is Z 1 t −(t−τ )/2 y(t) = e sin 2(t − τ ) g(τ ) dτ .

8 0 15. Taking the initial conditions into consideration, the transform of the ODE is s4 Y (s) − s3 + 5s2 Y (s) − 5s + 4 Y (s) = G(s).

Solving for the transform of the solution, Y (s) =

G(s) s3 + 5s + . (s2 + 1)(s2 + 4) (s2 + 1)(s2 + 4)

Using partial fractions, we find that

s3 + 5s 1 4s s = − , (s2 + 1)(s2 + 4) 3 s2 + 1 s2 + 4 and

1 1 1 1 = − . (s2 + 1)(s2 + 4) 3 s2 + 1 s2 + 4

It follows that L

−1

and −1

L

s(s2 + 5) 4 1 = cos t − cos 2t, (s2 + 1)(s2 + 4) 3 3

1 1 1 = sin t − sin 2t . (s2 + 1)(s2 + 4) 3 6

6.6

187 Based on the convolution theorem, the solution of the IVP is Z 1 1 t 4 [2 sin(t − τ ) − sin 2(t − τ )] g(τ ) dτ . y(t) = cos t − cos 2t + 3 3 6 0 17.(a) Taking the Laplace transform of the integral equation, with Φ(s) = L [φ(t)] , Φ(s) +

1 2 · Φ(s) = 2 . s2 s +4

Note that the convolution theorem was applied. Solving for the transform Φ(s), Φ(s) =

2s2 (s2 + 1)(s2 + 4)

.

Using partial fractions, we can write

2s2 2 4 1 = − . (s2 + 1)(s2 + 4) 3 s2 + 4 s2 + 1 Therefore the solution of the integral equation is φ(t) =

4 2 sin 2t − sin t .

3 3

(b) Differentiate both sides of the equation, we get Z t φ0 (t) + (t − t)φ(t) + φ(ξ)dξ = 2 cos 2t.

0

Clearly, t − t = 0, so differentiating this equation again we obtain φ00 (t) + φ(t) = −4 sin 2t.

Plugging t = 0 into the original equation gives us φ(0) = 0. Also, t = 0 in the first equation here in part (b) gives φ0 (0) = 2. (c) Taking the Laplace transform of the ODE, with Φ(s) = L [φ(t)] , 8 s2 Φ(s) − 2 + Φ(s) = − 2 . s +4 Solving for the transform of the solution, Φ(s) =

2s2 (s2 + 1)(s2 + 4)

.

This is identical to the Laplace transform we obtained in part (a), so the solution will be the same.

18.(a) Taking the Laplace transform of both sides of the integral equation (using the Convolution Theorem) Φ(s) +

1 1 Φ(s) = .

2 s s

It follows that Φ(s) =

s s2 + 1

188

Chapter 6. The Laplace Transform and −1

φ(t) = L

s = cos t . s2 + 1

(b) Differentiating both sides of the equation twice, we get Z t φ0 (t) + φ(ξ) dξ = 0, 0

and then φ00 (t) + φ(t) = 0. Plugging t = 0 into the original equation and the first equation above gives φ(0) = 1 and φ0 (0) = 0. (c) The function φ(t) = cos t clearly solves the initial value problem in part (b).

19.(a) The Laplace transform of both sides of the integral equation (using the Convolution Theorem) is Φ(s) +

1 2s Φ(s) = . s2 + 1 s+1

Solving for Φ(s): Φ(s) =

s2 + 1 . (s + 1)3

Rewriting, Φ(s) =

2 (s + 1)2 − 2(s + 1) + 2 1 2 − = + .

3 2 (s + 1) s + 1 (s + 1) (s + 1)3

The solution of the integral equation is

1 2 2 −1 φ(t) = L − + = e−t − 2 t e−t + t2 e−t . s + 1 (s + 1)2 (s + 1)3 (b) Differentiating both sides of the equation twice, we get Z t φ0 (t) + 2φ(t) − 2 sin(t − ξ)φ(ξ) dξ = −e−t , 0

and then φ00 (t) + 2φ0 (t) − 2

Z t

cos(t − ξ)φ(ξ) dξ = e−t .

0

Using the original equation, we can convert the second equation to φ00 (t) + 2φ0 (t) + φ(t) = 2e−t .

Plugging t = 0 into the original equation and the first equation above gives φ(0) = 1 and φ0 (0) = −3. (c) It is easily confirmed that the function φ(t) = e−t − 2te−t + t2 e−t solves the initial value problem in part (b).