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# Ordinary Differential Equations Direct

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Differential Equations

DIRECT INTEGRATION

Graham S McDonald

A Tutorial Module introducing ordinarydifferential equations and the method of

direct integration

c 2004 [email protected]

http://www.cse.salford.ac.uk/mailto:[email protected]:[email protected]://www.cse.salford.ac.uk/
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2. Theory

3. Exercises

5. Standard integrals

6. Tips on using solutions

Full worked solutions

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Section 1: Introduction 3

1. Introduction

d2ydx 2

+dydx

3

= x 7 is an example of an ordinary differential equa-

tion (o.d.e.) since it contains only ordinary derivatives such as dydxand not partial derivatives such as yx .

The dependent variable is y while the independent variable is x (ano.d.e. has only one independent variable while a partial differentialequation has more than one independent variable).

The above example is a second order equation since the highest or-der of derivative involved is two (note the presence of the d

2 ydx 2 term).

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Section 1: Introduction 4

An o.d.e. is linear when each term has y and its derivatives onlyappearing to the power one. The appearance of a term involving the

product of y anddy

dx would also make an o.d.e. nonlinear .

In the above example, the term dydx3

makes the equation nonlin-ear .

The general solution of an n th order o.d.e. has n arbitrary con-stants that can take any values.

In an initial value problem , one solves an n th order o.d.e. to ndthe general solution and then applies n boundary conditions (ini-tial values/conditions) to nd a particular solution that does nothave any arbitrary constants.

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Section 2: Theory 5

2. TheoryAn ordinary differential equation of the following form:

dydx

= f (x )

can be solved by integrating both sides with respect to x :

y = f (x ) dx .This technique, called DIRECT INTEGRATION , can also be ap-plied when the left hand side is a higher order derivative.

In this case, one integrates the equation a sufficient number of timesuntil y is found.

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Section 3: Exercises 6

3. Exercises

Click on Exercise links for full worked solutions (there are 8exercises in total)

Exercise 1.

Show that y = 2 e2x is a particular solution of the ordinary

differential equation: d2y

dx 2 dydx 2y = 0Exercise 2.

Show that y = 7 cos 3 x

2sin2x is a particular solution of

d2ydx 2

+ 2 y = 49cos3x + 4 sin 2 x

q Theory q Answers q Integrals q Tips

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Section 3: Exercises 7

Exercise 3.

Show that y = A sin x + B cos x , where A and B are arbitrary

constants, is the general solution of d2ydx 2

+ y = 0

Exercise 4.

Derive the general solution of dy

dx= 2 x + 3

Exercise 5.

Derive the general solution of d2ydx 2

= sin x

Exercise 6.

Derive the general solution of d2ydt 2

= a , where a = constant

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Section 3: Exercises 8

Exercise 7.

Derive the general solution of d3y

dx 3= 3 x 2

Exercise 8.

Derive the general solution of exd2ydx 2

= 3

q Theory q Answers q Integrals q Tips

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1. HINT: Work out d2 y

dx2 and dy

along with the given form of y, into the differential equation ,

2. HINT: Show that d2 y

dx 2 = 63cos3x + 8 sin 2 x and substitutethis, along with the given form of y, into the differentialequation ,

3. HINT: Show that d2 y

dx 2 = A sin x B cos x ,4. y = x 2 + 3 x + C ,

5. y = sin x + Ax + B ,

6. y = 12 at2 + Ct + D ,

7. y = 120 x5 + C x 2 + Dx + E ,

8. y = 3 ex + Cx + D .

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Section 5: Standard integrals 10

5. Standard integrals

f (x ) f (x )dx f (x ) f (x )dxx n x n +1n +1 (n = 1) [g (x )]n g (x ) [g (x )] n +1n +1 (n = 1)1x ln |x |

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

ex ex a x ax

ln a (a > 0)

sin x cos x sinh x cosh xcos x sin x cosh x sinh xtan x ln |cos x | tanh x ln cosh xcosec x ln tan x2 cosech x ln tanh

x2

sec x ln

|sec x + tan x

|sech x 2tan 1 ex

sec2 x tan x sech2 x tanh xcot x ln |sin x | coth x ln |sinh x |sin2 x x2 sin 2 x4 sinh 2 x sinh 2 x4 x2cos2 x x2 +

sin 2 x4 cosh

2 x sinh 2 x4 +x2

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Section 5: Standard integrals 11

f (x ) f (x ) dx f (x ) f (x ) dx1a 2 + x 2

1a tan

1 xa

1a 2 x 2

12a ln

a + xa x (0 < |x | 0) 1x 2 a 21

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

1 a 2 x 2 sin

1 xa

1 a 2 + x 2 ln

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

(a < x < a ) 1 x 2 a 2 lnx + x 2 a 2a (x>a> 0)

a 2 x 2 a2

2 sin1 xa a 2 + x 2 a 22 sinh1 xa + x a 2 + x 2a 2+ x a 2 x 2a 2 x 2 a 2 a

2

2 cosh1 xa + x x 2a 2a 2

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Section 6: Tips on using solutions 12

6. Tips on using solutions

q When looking at the THEORY, ANSWERS, INTEGRALS orTIPS pages, use the Back button (at the bottom of the page) toreturn to the exercises.

q 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.

q Try to make less use of the full solutions as you work your way

through the Tutorial.

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Solutions to exercises 13

Full worked solutionsExercise 1.

We need: dydx = 2 2e2x = 4 e2xd 2 ydx 2 = 2 4e2x = 8 e2x .

d 2 ydx 2 dydx 2y = 8 e2x 4e2x 2 e2x

= (8 8)e2x= 0

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Solutions to exercises 14

Exercise 2.

To show that y = 7 cos 3 x

2sin2x is a particular solution of

d2

ydx 2 + 2 y = 49cos3x + 4 sin 2 x , work out the following:dydx

= 21sin3x 4cos2xd2y

dx2 =

63cos3x + 8 sin 2 x

d 2 ydx 2 + 2 y = 63cos3x + 8 sin 2 x + 2(7 cos 3 x 2sin2x )= ( 63 + 14) cos 3 x + (8 4)sin2x= 49cos3x + 4 sin 2 x

= RHSNotes The equation is second order, so the general solutionwould have two arbitrary (undetermined) constants.

Notice how similar the particular solution is to theRight-Hand-Side of the equation. It involves thesame functions but they have different coefficients i.e.

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Solutions to exercises 15

y is of the form

a cos3x + b sin2xa = 7

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Solutions to exercises 16

Exercise 3.

We need: dydx = A cos x + B (sin x )d 2 ydx 2 = A sin x B cos x

d 2 ydx 2 + y = ( A sin x B cos x ) + ( A sin x + B cos x )

= 0

= RHS

Since the differential equation is second order and the solution hastwo arbitrary constants, this solution is the general solution.

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Solutions to exercises 17

Exercise 4.

This is an equation of the formdy

dx= f (x ), and it can be solved by

direct integration.

Integrate both sides with respect to x :

dy

dxdx =

(2x + 3) dx

i.e. dy = (2x + 3) dxi.e. y = 2

12

x 2 + 3 x + C

i.e. y = x 2 + 3 x + C,where C is the (combined) arbitrary constant that results fromintegrating both sides of the equation. The general solution musthave one arbitrary constant since the differential equation is rst

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Solutions to exercises 18

Exercise 5.

This is of the form d2 y

dx 2 = f (x ) , so we can solve for y by directintegration.

Integrate both sides with respect to x :

dy

dx = sinxdx

= (cos x ) + AIntegrate again:

y = sin x + Ax + B

where A, B are the two arbitrary constants of the general solution(the equation is second order).

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Solutions to exercises 19

Exercise 6.

Integrate both sides with respect to t :

dydt

= a dti.e.

dydt

= at + C

Integrate again:

y = (at + C )dti.e. y =

1

2at 2 + Ct + D ,

where C and D are the two arbitrary constants required for thegeneral solution of the second order differential equation.

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Solutions to exercises 20

Exercise 7.

Integrate both sides with respect to x :

d2ydx 2

= 3x 2dxi.e.

d2y

dx2 = 3

1

3

x 3 + C

i.e.d2ydx 2

= x 3 + C

Integrate again:

dydx

= (x 3 + C )dxi.e.

dydx

=x 4

4+ Cx + D

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Solutions to exercises 21

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Solutions to exercises 21

Integrate again: y = x4

4+ Cx + D dx

i.e. y =14

15x 5 + C

12x 2 + Dx + E

i.e. y =120

x 5 + C x 2 + Dx + E

where C =C 2 , D and E are the required three arbitrary constantsof the general solution of the third order differential equation.

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Solutions to exercises 22

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Solutions to exercises 22

Exercise 8.

Multiplying both sides of the equation by ex gives:

ex exd2ydx 2

= ex 3

i.e.d2ydx 2

= 3 ex

This is now of the formd 2 ydx 2 = f (x ), where f (x ) = 3 ex , and the

solution y can be found by direct integration.

Integrating both sides with respect to x :

dydx = 3ex dxi.e.

dydx

= 3 ex + C .

Integrate again:y = (3ex + C )dx

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Solutions to exercises 23

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Solutions to exercises 23

i.e. y = 3 ex + Cx + D ,where C and D are the two arbitrary constants of the general