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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

    q Table of contentsq Begin Tutorial

    c 2004 [email protected]

    http://www.cse.salford.ac.uk/mailto:[email protected]:[email protected]://www.cse.salford.ac.uk/
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    Table of contents1. Introduction

    2. Theory

    3. Exercises

    4. Answers

    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

    q Theory q Answers q Integrals q Tips

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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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    Section 4: Answers 9

    4. Answers

    1. HINT: Work out d2 y

    dx2 and dy

    dxand then substitute your results,

    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

    = RHSReturn to Exercise 1

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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

    b = 2Return to Exercise 2

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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.

    Return to Exercise 3

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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

    order. Return to Exercise 4Toc Back

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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).

    Return to Exercise 5

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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.

    Return to Exercise 6

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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

    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.

    Return to Exercise 7

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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

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

    solution of the original second order differential equation.Return to Exercise 8

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