Dierential Equations

INTEGRATING FACTOR METHOD

Graham S McDonald

A Tutorial Module for learning to solve 1storder linear dierential equations

Table of contents Begin Tutorial

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

Table of contents1. Theory2. Exercises3. Answers4. Standard integrals5. Tips on using solutions6. Alternative notation

Full worked solutions

Section 1: Theory 3

1. Theory

Consider an ordinary dierential equation (o.d.e.) that we wish tosolve to find out how the variable y depends on the variable x.

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

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

Any such linear first order o.d.e. can be re-arranged to give the fol-lowing standard form:

dy

dx+ P (x)y = Q(x)

where P (x) and Q(x) are functions of x, and in some cases may beconstants.

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Section 1: Theory 4

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

After writing the equation in standard form, P (x) can be identified.One then multiplies the equation by the following integrating factor:

IF= eRP (x)dx

This factor is defined so that the equation becomes equivalent to:

ddx (IF y) = IFQ(x),

whereby integrating both sides with respect to x, gives:

IF y =RIFQ(x) dx

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

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Section 2: Exercises 5

2. Exercises

In each case, derive the general solution. When a boundarycondition is also given, derive the particular solution.Click on Exercise links for full worked solutions (there are 10exercises in total)

Exercise 1.

dy

dx+ y = x ; y(0) = 2

Exercise 2.

dy

dx+ y = ex ; y(0) = 1

Exercise 3.

xdy

dx+ 2y = 10x2 ; y(1) = 3

Theory Answers Integrals Tips Notation

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Section 2: Exercises 6

Exercise 4.

xdy

dx y = x2 ; y(1) = 3

Exercise 5.

xdy

dx 2y = x4 sinx

Exercise 6.

xdy

dx 2y = x2

Exercise 7.

dy

dx+ y cotx = cosecx

Theory Answers Integrals Tips Notation

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Section 2: Exercises 7

Exercise 8.

dy

dx+ y cotx = cosx

Exercise 9.

(x2 1)dydx

+ 2xy = x

Exercise 10.

dy

dx= y tanx secx ; y(0) = 1

Theory Answers Integrals Tips Notation

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Section 3: Answers 8

3. Answers

1. General solution is y = (x 1) + Cex , and particularsolution is y = (x 1) + 3ex ,

2. General solution is y = ex(x+ C) , and particular solution isy = ex(x+ 1) ,

3. General solution is y = 52x2 + Cx2 , and particular solution is

y = 12 (5x2 + 1x2 ) ,

4. General solution is y = x2 + Cx , and particular solution isy = x2 + 2x ,

5. General solution is y = x3 cosx+ x2 sinx+ Cx2 ,

6. General solution is y = x2 ln x+ C x2 ,

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Section 3: Answers 9

7. General solution is y sinx = x+ C ,

8. General solution is 4y sinx+ cos 2x = C ,

9. General solution is (x2 1)y = x22 + C ,

10. General solution is y cosx = C x , and particular solution isy cosx = 1 x .

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Section 4: Standard integrals 10

4. Standard integrals

f (x)Rf(x)dx f (x)

Rf(x)dx

xn xn+1

n+1 (n 6= 1) [g (x)]n g0 (x) [g(x)]n+1

n+1 (n 6= 1)1x ln |x| g

0(x)g(x) ln |g (x)|

ex ex ax ax

ln a (a > 0)sinx cosx sinhx coshxcosx sinx coshx sinhxtanx ln |cosx| tanhx ln coshxcosecx ln

tan x2 cosechx ln tanh x2 secx ln |secx+ tanx| sechx 2 tan1 exsec2 x tanx sech2 x tanhxcotx ln |sinx| cothx ln |sinhx|sin2 x x2 sin 2x4 sinh2 x sinh 2x4 x2cos2 x x2 +

sin 2x4 cosh

2 x sinh 2x4 +x2

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Section 4: Standard integrals 11

f (x)Rf (x) dx f (x)

Rf (x) dx

1a2+x2

1a tan

1 xa

1a2x2

12a ln

a+xax (0< |x| 0) 1x2a2

12a ln

xax+a (|x| > a>0)1p

a2x2 sin1 x

a1p

a2+x2lnx+pa2+x2a (a > 0)

(a < x < a) 1px2a2 ln

x+px2a2a (x>a>0)pa2 x2 a22

sin1

xa

pa2+x2 a

2

2

hsinh1

xa

+ x

pa2+x2

a2

i+x

pa2x2a2

i px2a2 a22

h cosh1 xa+ xpx2a2a2 i

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Section 5: Tips on using solutions 12

5. Tips on using solutions

When looking at the THEORY, ANSWERS, INTEGRALS, TIPSor NOTATION pages, use the Back button (at the bottom of thepage) to return to the exercises.

Use the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correct.

Try to make less use of the full solutions as you work your waythrough the Tutorial.

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Section 6: Alternative notation 13

6. Alternative notation

The linear first order dierential equation:dy

dx+ P (x) y = Q(x)

has the integrating factor IF=eRP (x) dx.

The integrating factor method is sometimes explained in terms ofsimpler forms of dierential equation. For example, when constantcoecients a and b are involved, the equation may be written as:

ady

dx+ b y = Q(x)

In our standard form this is:dy

dx+

b

ay =

Q(x)a

with an integrating factor of:

IF = eR

ba dx = e

bxa

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Solutions to exercises 14

Full worked solutions

Exercise 1.

Compare with form:dy

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

Integrating factor: P (x) = 1.

Integrating factor, IF = eRP (x)dx

= eRdx

= ex

Multiply equation by IF:

exdy

dx+ exy = exx

i.e.d

dx[exy] = exx

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Solutions to exercises 15

Integrate both sides with respect to x:

exy = ex(x 1) + C

{ Note:Z

udv

dxdx = uv

Zvdu

dxdx i.e. integration by parts with

u x, dvdx ex

! xex Z

exdx

! xex ex = ex(x 1) }i.e. y = (x 1) + Cex .

Particular solution with y(0) = 2:

2 = (0 1) + Ce0= 1 + C i.e. C = 3 and y = (x 1) + 3ex .

Return to Exercise 1

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Solutions to exercises 16

Exercise 2.Integrating Factor: P (x) = 1 , IF = e

RPdx = e

Rdx = ex

Multiply equation:

exdy

dx+ exy = exex

i.e.d

dx[exy] = 1

Integrate:exy = x+ C

i.e. y = ex(x+ C) .

Particular solution:

y = 1x = 0 gives 1 = e

0(0 + C)

= 1.C i.e. C = 1and y = ex(x+ 1) .

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Solutions to exercises 17

Exercise 3.

Equation is linear, 1st order i.e.dy

dx+ P (x)y = Q(x)

i.e. dydx +2xy = 10x, where P (x) =

2x , Q(x) = 10x

Integrating factor : IF = eRP (x)dx = e2

Rdxx = e2 ln x = eln x

2= x2 .

Multiply equation: x2dy

dx+ 2xy = 10x3

i.e.d

dx

x2 y = 10x3

Integrate: x2y =52x4 + C

i.e. y =52x2 +

C

x2

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Solutions to exercises 18

Particular solution y(1) = 3 i.e. y(x) = 3 when x = 1.

i.e. 3 = 52 1 + C1i.e. 62 =

52 + C

i.e. C = 12

) y = 52x2 + 12x2 = 125x2 + 1x2

.

Return to Exercise 3

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Solutions to exercises 19

Exercise 4.

Standard form:dy

dx1x

y = x

Compare withdy

dx+ P (x)y = Q(x) , giving P (x) = 1

x

Integrating Factor: IF = eRP (x)dx = e

Rdxx = e ln x = eln(x

1) =1x.

Multiply equation:1x

dy

dx 1

x2y = 1

i.e.d

dx

1xy

= 1

Integrate:1xy = x+ C

i.e. y = x2 + Cx .

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Solutions to exercises 20

Particular solution with y(1) = 3:

3 = 1 + Ci.e. C = 2

Particular solution is y = x2 + 2x .

Return to Exercise 4

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Solutions to exercises 21

Exercise 5.Linear in y : dydx 2xy = x3 sinxIntegrating factor: IF = e2

Rdxx = e2 ln x = eln x

2= 1x2

Multiply equation:1x2

dy

dx 2

x3y = x sinx

i.e.d

dx

1x2

y

= x sinx

Integrate:y

x2= x cosx

Z1 ( cosx)dx+C 0

[Note: integration by parts,Ru dvdxdx = uv

Rv dudxdx, u = x,

dvdx = sinx]

i.e.y

x2= x cosx+ sinx+ C

i.e. y = x3 cosx+ x2 sinx+ Cx2.Return to Exercise 5

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Solutions to exercises 22

Exercise 6.

Standard form:dy

dx2x

y = x

Integrating Factor: P (x) = 2x

IF = eRPdx = e2

Rdxx = e2 ln x = eln x

2=

1x2

Multiply equation:1x2

dy

dx 2

x3y =

1x

i.e.d

dx

1x2

y

=

1x

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Solutions to exercises 23

Integrate:1x2

y =Z

dx

x

i.e.1x2

y = ln x+ C

i.e. y = x2 ln x+ Cx2 .Return to Exercise 6

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Solutions to exercises 24

Exercise 7.

Of the form:dy

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

where P (x) = cotx.

Integrating factor : IF = eRP (x)dx = e

R cos xsin x dx

e

R f0(x)f(x) dx

= eln(sin x) = sinx

Multiply equation : sinx dydx + sinxcos xsin x

y = sin xsin x

i.e. sinx dydx + cosx y = 1

i.e. ddx [sinx y] = 1

Integrate: (sinx)y = x+ C . Return to Exercise 7

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Solutions to exercises 25

Exercise 8.Integrating factor: P (x) = cotx =

cosxsinx

IF = eRPdx = e

R cos xsin x dx = eln(sin x) = sinx

Note:

cosxsinx

f0(x)f(x)

Multiply equation:

sinx dydx

+ sinx y cosxsinx

= sinx cosx

i.e.d

dx[sinx y] = sinx cosx

Integrate: y sinx =Z

sinx cosx dx

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Solutions to exercises 26

{ Note:Z

sinx cosx dx Z

f(x)f 0(x)dx Z

fdf

dx dx

Z

fdf =12f2 + C }

i.e. y sinx = 12 sin2 x+ C

= 12 12 (1 cos 2x) + Ci.e. 4y sinx +cos 2x = C 0where C 0 = 4C + 1

= constant

.

Return to Exercise 8

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Solutions to exercises 27

Exercise 9.

Standard form:dy

dx+

2xx2 1

y =

x

x2 1

Integrating factor: P (x) =2x

x2 1

IF = eRPdx = e

R 2xx21dx = eln(x

21)

= x2 1Multiply equation: (x2 1)dy

dx+ 2x y = x

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

i.e.d

dx

(x2 1)y = x

Integrate: (x2 1)y = 12x2 + C .

Return to Exercise 9

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Solutions to exercises 28

Exercise 10.P (x) = tanxQ(x) = secx IF = e

R tan x dx = e R sin xcos xdx = e+ R sin xcos x dx= eln(cos x) = cosx

Multiply by IF: cosx dydx cosx sin xcos xy = cosx secx

i.e. ddx [cosx y] = 1 i.e. y cosx = x+ C .

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

cos(0) = 0 + C ) C = 1

i.e. y cosx = x+ 1 .Return to Exercise 10

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Table of contents1 Theory2 Exercises3 Answers4 Standard integrals5 Tips on using solutions6 Alternative notation Full worked solutions