19.6. Finding a Particular Integral. Introduction. Prerequisites. Learning Outcomes. Learning Style

Finding a Particular Integral



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19.6 ✑

Introduction

We stated in Block 19.5 that the general solution of an inhomogeneous equation is the sum of the complementary function and a particular integral. We have seen how to find the comple- mentary function in the case of a constant coefficient equation. We shall now deal with the problem of finding a particular integral. Recall that the particular integral is any solution of the inhomogeneous equation. There are a number of advanced techniques available for finding such solutions but we shall adopt a simpler strategy. Since any solution will do we shall try to find such a solution by a combination of educated guesswork and trial and error.

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Prerequisites

Before starting this Block you should . . .

① understand what is meant by a differential equation; (Block 19.1)

② be familiar with the terminology associated with differential equations: order,

dependent variable and independent variable; (Block 19.1)

③ be able to integrate; (Block 14)

④ have completed Block 19.5 on Constant Coefficient Equations

Learning Outcomes

After completing this Block you should be able to . . .

✓ understand what is meant by a particular integral

✓ find particular integrals by trial solution

✓ find general solutions of inhomogeneous equations by adding the complementary function to the particular integral

Learning Style

To achieve what is expected of you . . .

☞ allocate sufficient study time

☞ briefly revise the prerequisite material

☞ attempt every guided exercise and most

of the other exercises

1. What is meant by a particular integral?

Given a second order o.d.e.

a d ² y

dx ² + b dy

dx + c y = f (x)

a particular integral is any function, y p (x), which satisfies the equation. That is, any function which when substituted into the left hand side and simplified, results in the function on the right.

We denote a particular integral by y p (x).

Try each part of this exercise Show that

y = − ¹ ₄ e ^2x is a particular integral of d ² y dx ² − dy

dx − 6y = e ^2x (1) Part (a) Starting with y = − ¹ ₄ e ^2x , find ^dy _dx and ^d _dx

^y

:

Answer Part (b) Now substitute these into (1):

Answer

2. Finding a particular integral

In the previous section we explained what is meant by a particular integral. Now we look at how one is actually found. In fact our method is rather crude. It involves trial and error and educated guesswork. We try solutions which are of the same general form as the f (x) on the right hand side. As a guide, use Table 1.

Table 1. Trial solutions to find the particular integral

f (x) Trial solution

constant term c constant term γ polynomial in x polynomial in x of degree r: of degree r:

ax ^r + · · · + bx + c αx ^r + · · · + βx + γ a cos kx α cos kx + β sin kx a sin kx α cos kx + β sin kx

ae ^kx αe ^kx

Example Find a particular integral of the equation d ² y

dx ² − dy

dx − 6y = e ^2x (2)

Solution

We shall attempt to find a solution of the inhomogeneous problem by trying a function of the same form as that on the right-hand side. In particular, let us try y(x) = αe ^2x , where α is a constant that we shall now determine. If y(x) = αe ^2x then

dy

dx = 2αe ^2x and d ² y

dx ² = 4αe ^2x . Substitution in (2) gives:

4αe ^2x − 2αe ^2x − 6αe ^2x = e ^2x that is,

−4αe ^2x = e ^2x

so that y will be a solution if α is chosen so that −4α = 1, that is, α = − ¹ ₄ . Therefore the particular integral is y p (x) = − ¹ ₄ e ^2x .

Now do this exercise

By trying a solution of the form y = αe ^−x find a particular integral of the equation d ² y

dx ² + dy

dx − 2y = 3e ^−x

Substitute y = αe ^−x into the given equation to find α, and hence the particular integral. Answer

Example Obtain a particular integral of the equation:

d ² y

dx ² − 6 dy

dx + 8y = x

Solution

In the last example, we found that a fruitful approach was to assume a solution in the same form as that on the right-hand side. Suppose we assume a solution y(x) = αx and proceed to determine α. This approach will actually fail, but let us see why. If y(x) = αx then ^dy _dx = α and ^d _dx

^y

= 0. Substitution into the differential equation yields 0 − 6α + 8αx = x and α ought now to be chosen so that this expression is true for all x. If we equate the coefficients of x we find 8α = 1 so that α = ¹ ₈ , but with this value of α the constant terms are inconsistent (that is − ⁶ ₈ on the left, but zero on the right). Clearly a particular integral of the form αx is not possible. The problem arises because differentiation of the term αx produces constant terms which are unbalanced on the right-hand side. So, we try a solution of the form y(x) = αx + β with α, β constants. This is consistent with the recommendation in Table 1. Proceeding as

dy d

y

Solution

Substitution in the differential equation now gives:

0 − 6α + 8(αx + β) = x

Equating coefficients of x and then equating constant terms we find:

8α = 1 ( ∗) − 6α + 8β = 0 ( ∗∗) From ( ∗), α = ¹ ₈ and then from ( ∗∗)

−6  ₁

8

 + 8β = 0

so that, 8β = ³ ₄ that is, β = ₃₂ ³ . The required particular integral is y p (x) = ¹ ₈ x + ₃₂ ³ .

Try each part of this exercise

Find a particular integral for the equation:

d ² y

dx ² − 6 dy

dx + 8y = 3 cos x

Part (a) First try to decide on an appropriate form for the trial solution. Refer to Table 1 if necessary

Answer Part (b) Equate coefficients of cos x in your previous answer:

Answer Part (c) Also, equate coefficients of sin x in your previous answer:

Answer Part (d) Solve these simultaneously to find α and β, and hence the particular integral:

Answer

3. Finding the general solution of a second-order inhomogeneous equation

The general solution of a second-order linear inhomogeneous equation is the sum of its particular

integral and the complementary function. In Block 19.5 you learned how to find a complementary

function, and in the previous section you learnt how to find a particular integral. We now put

these together to find the general solution.

Now do this exercise

Find the general solution of

d ² y

dx ² + 3 dy

dx − 10y = 3x ²

The complementary function was found in Block 19.5 page 6 to be y cf = Ae ^2x + Be ^−5x . The particular integral is found by trying a solution of the form y = ax ² + bx + c. Substitute into the homogeneous equation to find a, b and c, and hence y p (x). Answer

Key Point

The general solution of a constant coefficient ordinary differential equation a d ² y

dx ² + b dy

dx + cy = f (x) is y = y p + y cf

being the sum of the particular integral and the complementary function. y p contains no arbi- trary constants; y cf contains two arbitrary constants.

Example An LC circuit with sinusoidal input. The differential equation governing the flow of current in a series LC circuit when subject to an applied voltage v(t) = V 0 sin ωt is

L d ² i dt ² + 1

C i = ωV 0 cos ωt L C

i

v

Obtain its general solution.

Solution

The homogeneous equation is

L d ² i cf

dt ² + i cf

C = 0.

Letting i cf = e ^kt we find the auxiliary equation is Lk ² + _C ¹ = 0 so that k = ±i/ √

LC. Therefore, the complementary function is:

i cf = A cos t

√ LC + B sin t

√ LC where A and B arbitrary constants

Solution

To find a particular integral try i p = E cos ωt + F sin ωt, where E, F are constants. We find:

di p

dt = −ωE sin ωt + ωF cos ωt d ² i p

dt ² = −ω ² E cos ωt − ω ² F sin ωt Substitution into the inhomogeneous equation yields:

L( −ω ² E cos ωt − ω ² F sin ωt) + 1

C (E cos ωt + F sin ωt) = ωV 0 cos ωt Equating coefficients of sin ωt gives: −ω ² LF + (F/C) = 0.

Equating coefficients of cos ωt gives: −ω ² LE + (E/C) = ωV 0 .

Therefore F = 0 and E = CV 0 ω/(1 − ω ² LC). Hence the particular integral is i p = CV 0 ω

1 − ω ² LC cos ωt.

Finally, the general solution is:

i = i cf + i p = A cos t

√ LC + B sin t

√ LC + CV 0 ω

1 − ω ² LC cos ωt

4. Inhomogeneous term appearing in the complementary function

Occasionally you will come across a differential equation a ^d _dx

^y

+ b dy

dx + cy = f (x) for which the inhomogeneous term, f (x), forms part of the complementary function. One such example is the equation

d ² y dx ² − dy

dx − 6y = e ^3x

It is straightforward to check that the complementary function is y cf = Ae ^3x + Be ^−2x . Note that the first of these terms has the same form as the inhomogeneous term, e ^3x , on the right-hand side of the differential equation.

You should verify for yourself that trying a particular integral of the form y p (x) = αe ^3x will not work in a case like this. Can you see why?

Instead, try a particular integral of the form y p (x) = αxe ^3x . Verify that dy p

dx = αe ^3x (3x + 1) and d ² y p

dx ² = αe ^3x (9x + 6).

Substitute these expressions into the differential equation to find α = ¹ ₅ . Finally, the particular integral is y p (x) = ¹ ₅ xe ^3x and so the general solution to the differential equation is:

y = Ae ^3x + Be ^−2x + ¹ ₅ xe ^3x

More exercises for you to try

1. Find the general solution of the following equations:

(a) d ² x

dt ² − 2 dx

dt − 3x = 6. (b) d ² y

dx ² + 5 dy

dx + 4y = 8 (c) d ² y

dt ² + 5 dy

dt + 6y = 2t (d) d ² x

dt ² +11 dx

dt +30x = 8t (e) d ² y dx ² +2 dy

dx +3y = 2 sin 2x (f) d ² y dt ² + dy

dt +y = 4 cos 3t (g) d ² y

dx ² + 9y = 4e ^8x (h) d ² x

dt ² − 16x = 9e ^6t 2. Find a particular integral for the equation d ² x

dt ² − 3 dx

dt + 2x = 5e ^3t 3. Find a particular integral for the equation d ² x

dt ² − x = 4e ^−2t 4. Obtain the general solution of y ^ − y ^ − 2y = 6.

5. Obtain the general solution of the equation d ² y

dx ² + 3 dy

dx + 2y = 10 cos 2x.

Find the particular solution satisfying y(0) = 1, dy

dx (0) = 0.

6. Find a particular integral for the equation d ² y dx ² + dy

dx + y = 1 + x 7. Find the general solution of

(a) d ² x

dt ² − 6 dx

dt + 5x = 3 (b) d ² x

dt ² − 2 dx

dt + x = e ^t

Answer

5. Computer Exercise or Activity

For this exercise it will be necessary for you to access the com- puter package DERIVE. To solve a second-order differential equation using DERIVE it is necessary to load what is called a Utility File named ode2. To do this is simple. Proceed as follows: In DERIVE, choose File:Load:Math and select the file (double click) on the ode2 icon. This will load a number of commands which enable you to solve second-order differential equations. You can use the Help facility to learn more about these if you wish.

Of particular relevance here is the command

Dsolve2(p, q, r, x, c1, c2)

which finds the general solution (containing two arbitrary constants c1, c2) to the second order differential equation

d ² y

dx ² + p(x) dy

dx + q(x)y = r(x)

For the examples in this Block both p(x) and q(x) are given constants.

The general solution is the sum of the complementary function (the part containing the arbitrary constants) and the particular integral. Hence by inspecting the solution given by DERIVE the particular integral can be obtained. For example the general solution to

d ² y dx ² − dy

dx − 6y = e ^2x can be obtained by keying Author:Expression

Dsolve2( −1, −6, exp(2x), c1, c2) followed by Simplify and DERIVE responds with

c1 · ˆe ^{3 ·x} − ˆe ^{2 ·x}

4 + c2 · ˆe ^−2·x from which we deduce the particular integral:

− ˆe ^{2 ·x} 4

As an exercise use DERIVE to check the correctness of the particular integrals requested in the

examples and guided exercises of this Block.

MAPLE will solve a wide range of ordinary differential equations including systems of differential equations using the command

dsolve(deqns,vars,eqns) where:

deqns − ordinary differential equation in vars, or set of equations and/or initial conditions.

vars − variable or set of variables to be solved for eqns − optional equation of the form keyword=value For example to solve

d ² y

dt ² + 2 dy

dt + 2y = e ^−t y(0) = 0, y ^ (0) = 0 we would key in

> dsolve( {diff(y(t),t$2)+2*diff(y(t),t)+2*y(t)=exp(-t),y(0)=0, D(y)(0)=0},y(t),type=exact);

MAPLE responds with

1 − cos(t) e ^t

If the initial conditions are omitted MAPLE will present the solution with the correct number of arbitrary constants denoted by C1, C2 . . . . Thus the general solution of

d ² y

dt ² + 2 dy

dt + 2y = e ^−t is obtained by keying in

> dsolve( {diff(y(t),t$2)+2*diff(y(t),t)+2*y(t)=exp(-t)},y(t),type=exact);

and MAPLE responds with

y(t) = exp( −t) + C1 ∗ exp(−t) ∗ cos(t) + C2 ∗ exp(−t) ∗ sin(t)

As with the DERIVE response the particular integral can be deduced from this general solution

as being that part not multiplied by an arbitrary constant.

End of Block 19.6

dy

dx = − ¹ ₂ e ^2x , ^d _dx

^y

= −e ^2x

Back to the theory

Substitution into (1) yields −e ^2x − 

− ¹ ₂ e ^2x 

− 6 

− ¹ ₄ e ^2x 

which simplifies to e ^2x , the same as the right hand side. Therefore y = − ¹ ₄ e ^2x is a particular integral and we write (attaching a subscript p)

y p (x) = − ¹ ₄ e ^2x

Back to the theory

α = − ³ ₂ ; y p (x) = − ³ ₂ e ^−x

Back to the theory

y = α cos x + β sin x in which α, β are constants to be found. We shall try a solution of the form y(x) = α cos x + β sin x. Differentiating, we find:

dy

dx = −α sin x + β cos x d ² y

dx ² = −α cos x − β sin x Substitution into the differential equation gives:

(−α cos x − β sin x) − 6(−α sin x + β cos x) + 8(α cos x + β sin x) = 3 cos x

Back to the theory

7α − 6β = 3

Back to the theory

7β + 6α = 0

Back to the theory

α = ²¹ ₈₅ , β = − ¹⁸ ₈₅ , y p (x) = ²¹ ₈₅ cos x − ¹⁸ ₈₅ sin x

Back to the theory

a = − ₁₀ ³ , b = − ₅₀ ⁹ , c = − ₅₀₀ ⁵⁷ , y p (x) = − ₁₀ ³ x ² − ₅₀ ⁹ x − ₅₀₀ ⁵⁷ . Thus the general solution is y = y p (x) + y cf (x) = − ₁₀ ³ x ² − ₅₀ ⁹ x − ₅₀₀ ⁵⁷ + Ae ^2x + Be ^−5x

Back to the theory

1. (a) x = Ae ^−t + Be ^3t − 2 (b) y = Ae ^−x + Be ^−4x + 2 (c) y = Ae ^−2t + Be ^−3t + ¹ ₃ t − ₁₈ ⁵ (d) x = Ae ^−6t + Be ^−5t + 0.267t − 0.0978

(e) y = e ^−x [A sin √

2x + B cos √

2x] − ₁₇ ⁸ cos 2x − ₁₇ ² sin 2x

(f) y = e ^−0.5t (A cos 0.866t + B sin 0.866t) − 0.438 cos 3t + 0.164 sin 3t (g) y = A cos 3x + B sin 3x + 0.0548e ^8x (h) x = Ae ^4t + Be ^−4t + ₂₀ ⁹ e ^6t 2. x p = 2.5e ^3t

3. x p = ⁴ ₃ e ^−2t

4. y = Ae ^2x + Be ^−x − 3

5. y = Ae ^−2x + Be ^−x + ³ ₂ sin 2x − ¹ ₂ cos 2x, ³ ₂ e ^−2x + ³ ₂ sin 2x − ¹ ₂ cos 2x 6. y p = x

7. (a) x = Ae ^t + Be ^5t + ³ ₅ (b) x = Ae ^t + Bte ^t + ¹ ₂ t ² e ^t

Back to the theory