A ordinary differential equations integrating factor

Differential Equations INTEGRATING FACTOR METHOD Graham S McDonald A Tutorial Module for learning to solve 1st order linear differential equations

● Table of contents ● Begin Tutorial

c 2004 g.s.mcdonald@salford.ac.uk

Table of contents 1.

2.

3.

4.

5.

6.

Theory Exercises Answers Standard integrals Tips on using solutions Alternative notation Full worked solutions

Section 1: Theory

3

1. Theory Consider an ordinary differential equation (o.d.e.) that we wish to solve to find out how the variable y depends on the variable x.

If the equation is first order then the highest derivative involved is a first derivative.

If it is also a linear equation then this means that each term can dy involve y either as the derivative dx OR through a single factor of y .

Any such linear first order o.d.e. can be re-arranged to give the following standard form: dy + P (x)y = Q(x) dx where P (x) and Q(x) are functions of x, and in some cases may be constants.

Toc

JJ

II

J

I

Back

Section 1: Theory

4

A linear first order o.d.e. can be solved using the integrating factor method.

After writing the equation in standard form, P (x) can be identified.

One then multiplies the equation by the following “integrating factor�: R

IF= e

P (x)dx

This factor is defined so that the equation becomes equivalent to: d dx (IF y)

= IF Q(x),

whereby integrating both sides with respect to x, gives: IF y =

R

IF Q(x) dx

Finally, division by the integrating factor (IF) gives y explicitly in terms of x, i.e. gives the solution to the equation.

Toc

JJ

II

J

I

Back

Section 2: Exercises

5

2. Exercises In each case, derive the general solution. When a boundary condition is also given, derive the particular solution.

Click on Exercise links for full worked solutions (there are 10 exercises in total) Exercise 1. dy + y = x ; y(0) = 2 dx Exercise 2. dy + y = e−x ; y(0) = 1 dx Exercise 3. dy x + 2y = 10x2 ; y(1) = 3 dx ● Theory ● Answers ● Integrals ● Tips ● Notation Toc

JJ

II

J

I

Back

Section 2: Exercises

Exercise 4. dy x − y = x2 ; y(1) = 3 dx Exercise 5. dy x − 2y = x4 sin x dx Exercise 6. dy x − 2y = x2 dx Exercise 7. dy + y cot x = cosec x dx

● Theory ● Answers ● Integrals ● Tips ● Notation Toc JJ II J I Back

6

Section 2: Exercises

7

Exercise 8. dy + y · cot x = cos x dx Exercise 9. dy (x2 − 1) + 2xy = x dx Exercise 10. dy = y tan x − sec x ; y(0) = 1 dx

● Theory ● Answers ● Integrals ● Tips ● Notation Toc

JJ

II

J

I

Back

Section 3: Answers

8

3. Answers 1. General solution is y = (x − 1) + Ce−x , and particular solution is y = (x − 1) + 3e−x , 2. General solution is y = e−x (x + C) , and particular solution is y = e−x (x + 1) , 3. General solution is y = 25 x2 + y = 21 (5x2 + x12 ) ,

C x2

, and particular solution is

4. General solution is y = x2 + Cx , and particular solution is y = x2 + 2x , 5. General solution is y = −x3 cos x + x2 sin x + Cx2 , 6. General solution is y = x2 ln x + C x2 ,

Toc

JJ

II

J

I

Back

Section 3: Answers

9

7. General solution is y sin x = x + C , 8. General solution is 4y sin x + cos 2x = C , 9. General solution is (x2 − 1)y =

x2 2

+C ,

10. General solution is y cos x = C − x , and particular solution is y cos x = 1 − x .

Toc

JJ

II

J

I

Back

Section 4: Standard integrals

10

4. Standard integrals f (x) n

x

1 x x

e sin x cos x tan x cosec x sec x sec2 x cot x sin2 x cos2 x

R

f (x)dx

xn+1 n+1

(n 6= −1)

ln |x| ex − cos x sin x − ln

|cos x| ln tan x2

ln |sec x + tan x| tan x ln |sin x| x sin 2x 2 − 4 x sin 2x 2 + 4

Toc

JJ

R

f (x) n

0

[g (x)] g (x) g 0 (x) g(x) x

a sinh x cosh x tanh x cosech x sech x sech2 x coth x sinh2 x cosh2 x

II

J

f (x)dx

[g(x)]n+1 n+1

(n 6= −1)

ln |g (x)| ax (a > 0) ln a cosh x sinh x ln cosh x

ln tanh x2

2 tan−1 ex tanh x ln |sinh x| sinh 2x − x2 4 sinh 2x + x2 4

I

Back

Section 4: Standard integrals

f (x) 1 a2 +x2

√ 1 a2 −x2

√

a2 − x2

11

R

f (x) dx

f (x)

1 a

tan−1

1 a2 −x2

R

(a > 0)

1 x2 −a2

f (x) dx

a+x

1 2a ln a−x (0 < |x| < a)

x−a

1 ln

2a x+a (|x| > a > 0)

sin−1

x a

√ 1 a2 +x2

√ 2 2

ln x+ aa +x (a > 0)

(−a < x < a)

√ 1 x2 −a2

√

2 2

ln x+ xa −a (x > a > 0)

a2 2

−1 sin √

+x

Toc

x a

x a

a2 −x2 a2

JJ

√ i √

a2 +x2

a2 2

x2 −a2

II

a2 2

J

h h

sinh−1

− cosh−1

I

x a

i √ x a2 +x2 2 a i √ 2 2 + x xa2−a

+

x a

Back

Section 5: Tips on using solutions

12

5. Tips on using solutions ● When looking at the THEORY, ANSWERS, INTEGRALS, TIPS or NOTATION pages, use the Back button (at the bottom of the page) to return to the exercises. ● Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct. ● Try to make less use of the full solutions as you work your way through the Tutorial.

Toc

JJ

II

J

I

Back

Section 6: Alternative notation

13

6. Alternative notation The linear first order differential equation: dy + P (x) y = Q(x) dx R

has the integrating factor IF=e

P (x) dx

.

The integrating factor method is sometimes explained in terms of simpler forms of differential equation. For example, when constant coefficients a and b are involved, the equation may be written as: dy + b y = Q(x) a dx In our standard form this is: dy b Q(x) + y= dx a a with an integrating factor of: R

IF = e Toc

JJ

II

b a

dx

bx

=ea J

I

Back

Solutions to exercises

14

Full worked solutions Exercise 1.

Compare with form:

dy + P (x)y = Q(x) (P, Q are functions of x) dx

Integrating factor:

P (x) = 1.

Integrating factor,

R

IF = e P (x)dx R = e dx = ex

Multiply equation by IF: dy + ex y = ex x dx d x i.e. [e y] = ex x dx ex

Toc

JJ

II

J

I

Back

Solutions to exercises

15

Integrate both sides with respect to x: ex y Z { Note:

dv u dx dx u

i.e. y

=

ex (x − 1) + C Z

=

uv −

v

du dx i.e. integration by parts with dx

dv ≡ ex dxZ → xex − ex dx ≡

x,

→ xex − ex = ex (x − 1) } = (x − 1) + Ce−x .

Particular solution with y(0) = 2: 2

= (0 − 1) + Ce0 = −1 + C i.e. C = 3 and y = (x − 1) + 3e−x .

Return to Exercise 1 Toc

JJ

II

J

I

Back

Solutions to exercises

16

Exercise 2.

Integrating Factor:

P (x) = 1 ,

R

IF = e

P dx

R

=e

dx

= ex

Multiply equation: ex i.e.

dy + ex y = ex e−x dx d x [e y] = 1 dx

Integrate: ex y y

i.e.

= x+C = e−x (x + C) .

Particular solution: y=1 x=0

gives

1

and y

= e0 (0 + C) = 1.C i.e. C = 1 −x = e (x + 1) .

Return to Exercise 2

Toc

JJ

II

J

I

Back

Solutions to exercises

17

Exercise 3.

Equation is linear, 1st order i.e.

dy dx

+ x2 y = 10x,

Integrating factor : IF =

dy + P (x)y = Q(x) dx P (x) = x2 , Q(x) = 10x

i.e. where R

e

P (x)dx

= e2

2

= e2 ln x = eln x = x2 .

dy + 2xy = dx d 2 x ·y = dx

i.e.

Integrate: i.e.

JJ

dx x

x2

Multiply equation:

Toc

R

II

J

x2 y

=

y

=

I

10x3 10x3 5 4 x +C 2 C 5 2 x + 2 2 x

Back

Solutions to exercises

18

Particular solution i.e.

3 =

5 2

·1+

i.e.

6 2

=

5 2

+C

i.e.

C=

1 2

∴

y = 52 x2 +

y(1) = 3

i.e. y(x) = 3

when x = 1.

C 1

1 2x2

=

1 2

5x2 +

1 x2

.

Return to Exercise 3

Toc

JJ

II

J

I

Back

Solutions to exercises

19

Exercise 4. dy − dx

Standard form: Compare with

1 y=x x

dy + P (x)y = Q(x) , giving dx R

Integrating Factor: IF = e

P (x)dx

= e−

R

dx x

P (x) = −

1 x −1

= e− ln x = eln(x

Multiply equation: 1 1 dy − 2 y = x dx x d 1 i.e. y = dx x

1 1

Integrate:

i.e.

Toc

JJ

1 y x y II

= x+C = x2 + Cx .

J

I

Back

)

=

1 . x

Solutions to exercises

20

Particular solution with y(1) = 3: i.e.

Particular solution is

3 = 1+C C = 2 y

= x2 + 2x .

Return to Exercise 4

Toc

JJ

II

J

I

Back

Solutions to exercises

Exercise 5.

Linear in y :

21

dy dx

− x2 y = x3 sin x

Integrating factor: IF = e−2

R

dx x

−2

= e−2 ln x = eln x

=

1 x2

1 dy 2 − y = x sin x x2 dx x3 d 1 y = x sin x dx x2 Z y = −x cos x− 1 · (− cos x)dx+C 0 x2

Multiply equation: i.e.

Integrate:

[Note: integration by parts, R dv R u dx dx = uv − v du dx dx, u = x, i.e. i.e.

y x2 y

dv dx

= sin x]

= −x cos x + sin x + C = −x3 cos x + x2 sin x + Cx2 .

Return to Exercise 5

Toc

JJ

II

J

I

Back

Solutions to exercises

22

Exercise 6.

2 y=x x

Standard form:

dy − dx

Integrating Factor:

P (x) = −

R

IF = e

P dx

Multiply equation:

Toc

= e−2

R

dx x

2 x −2

= e−2 ln x = eln x

=

1 x2

1 dy 2 1 − y= x2 dx x3 x

d 1 1 i.e. y = dx x2 x

JJ

II

J

I

Back

Solutions to exercises

23

1 y= x2

Integrate:

Z

dx x

i.e.

1 y x2

= ln x + C

i.e.

y

= x2 ln x + Cx2 .

Return to Exercise 6

Toc

JJ

II

J

I

Back

Solutions to exercises

24

Exercise 7.

Of the form: where

dy + P (x)y = Q(x) (i.e. linear, 1st order o.d.e.) dx

P (x) = cot x.

R

Integrating factor :

IF = e

P (x)dx

cos x sin x dx

R

=e

R â&#x2030;Ąe

f 0 (x) dx f (x)

= eln(sin x) = sin x

sin x ¡

dy dx

+ sin x

i.e.

sin x ¡

dy dx

+ cos x ¡ y = 1

i.e.

d dx

Multiply equation :

Integrate:

cos x sin x

y=

[sin x ¡ y] = 1

(sin x)y = x + C .

Toc

JJ

sin x sin x

II

Return to Exercise 7 J

I

Back

Solutions to exercises

25

Exercise 8.

Integrating factor:

P (x) = cot x = R

IF = e

Note:

P dx

cos x f 0 (x) â&#x2030;Ą sin x f (x)

R

=e

cos x sin x dx

cos x sin x = eln(sin x) = sin x

Multiply equation: sin x ¡

Integrate:

Toc

cos x dy + sin x ¡ y ¡ = sin x ¡ cos x dx sin x d i.e. [sin x ¡ y] = sin x ¡ cos x dx Z y sin x = sin x ¡ cos x dx

JJ

II

J

I

Back

Solutions to exercises

26

Z { Note:

Z sin x cos x dx ≡

f (x)f (x)dx ≡ Z

≡ i.e.

i.e.

Z

0

f df =

y sin x =

1 2

sin2 x + C

=

1 2

· 12 (1 − cos 2x) + C

f

df · dx dx

1 2 f +C } 2

4y sin x + cos 2x = C 0

where C 0 = 4C + 1 = constant

.

Return to Exercise 8

Toc

JJ

II

J

I

Back

Solutions to exercises

27

Exercise 9. dy + dx

Standard form:

2x x2 − 1

P (x) =

Integrating factor:

R

IF = e

Multiply equation:

P dx

y=

x x2 − 1

2x x2 − 1 R

=e

(x2 − 1)

2x dx x2 −1

2

= eln(x −1) = x2 − 1

dy + 2x y = x dx

(the original form of the equation was half-way there!)

d 2 i.e. (x − 1)y = x dx Integrate:

(x2 − 1)y =

Toc

JJ

1 2 x +C .

2 II

Return to Exercise 9 J

I

Back

Solutions to exercises

28

Exercise 10.

P (x) = − tan x Q(x) = − sec x

IF = e−

R

tan x dx

= e−

R

sin x cos x dx

= e+

R

− sin x cos x dx

= eln(cos x) = cos x Multiply by IF: i.e. y(0) = 1

d dx

dy cos x dx − cos x ·

[cos x · y] = −1

sin x cos x y

= − cos x · sec x

i.e. y cos x = −x + C .

i.e. y = 1 when x = 0 gives

cos(0) = 0 + C ∴ C = 1 i.e. y cos x = −x + 1 .

Return to Exercise 10

Toc

JJ

II

J

I

Back