# 17.2

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## 17.2 Nonhomogeneous Linear

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### Nonhomogeneous Linear Equations

In this section we learn how to solve second-order

nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form

ay″ + by′ + cy = G(x)

where a, b, and c are constants and G is a continuous function. The related homogeneous equation

ay″ + by′ + cy = 0

is called the complementary equation and plays an important role in the solution of the original

nonhomogeneous equation (1).

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### Nonhomogeneous Linear Equations

There are two methods for finding a particular solution:

The method of undetermined coefficients is straightforward but works only for a restricted class of functions G.

The method of variation of parameters works for every function G but is usually more difficult to apply in practice.

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### The Method of Undetermined Coefficients

We first illustrate the method of undetermined coefficients for the equation

ay″ + by′ + cy = G(x) where G(x) is a polynomial.

It is reasonable to guess that there is a particular solution yp that is a polynomial of the same degree as G because if y is a polynomial, then ay″ + by′ + cy is also a polynomial.

We therefore substitute yp(x) = a polynomial (of the same degree as G) into the differential equation and determine the coefficients.

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### Example 1

Solve the equation y″ + y′ – 2y = x2. Solution:

The auxiliary equation of y″ + y′ – 2y = 0 is r2 + r – 2 = (r – 1)(r + 2) = 0 with roots r = 1, –2.

So the solution of the complementary equation is yc = c1ex + c2e –2x

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### Example 1 – Solution

Since G(x) = x2 is a polynomial of degree 2, we seek a particular solution of the form

yp(x) = Ax2 + Bx + C

Then = 2Ax + B and = 2A so, substituting into the given differential equation, we have

(2A) + (2Ax + B) – 2(Ax2 + Bx + C) = x2 or

–2Ax2 + (2A – 2B)x + (2A + B – 2C) = x2

Polynomials are equal when their coefficients are equal.

cont’d

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### Example 1 – Solution

Thus

–2A = 1 2A – 2B = 0 2A + B – 2C = 0

The solution of this system of equations is

A = B = C =

A particular solution is therefore yp(x) =

and, by Theorem 3, the general solution is y = yc + yp =

cont’d

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### Example 3

Solve y″ + y′ – 2y = sin x.

Solution:

We try a particular solution

yp(x) = A cos x + B sin x Then

= –A sin x + B cos x

= –A cos x – B sin x

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### Example 3 – Solution

So substitution in the differential equation gives

or

This is true if

–3A + B = 0 and –A – 3B = 1 The solution of this system is

A = B =

cont’d

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### Example 3 – Solution

So a particular solution is yp(x) =

In Example 1 we determined that the solution of the complementary equation is

yc = c1ex + c2e–2x.

Thus the general solution of the given equation is

y(x) = c1ex + c2e –2x(cos x + 3 sin x)

cont’d

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### The Method of Undetermined Coefficients

If G(x) is a product of functions of the preceding types, then we take the trial solution to be a product of functions of the same type.

For instance, in solving the differential equation y″ + 2y′ + 4y = x cos 3x

we would try

yp(x) = (Ax + B) cos 3x + (Cx + D) sin 3x

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### The Method of Undetermined Coefficients

If G(x) is a sum of functions of these types, we use the easily verified principle of superposition, which says that if

and are solutions of

ay″ + by′ + cy = G1(x) ay″ + by′ + cy = G2(x)

respectively, then is a solution of ay″ + by′ + cy = G1(x) + G2(x)

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### Example 6

Determine the form of the trial solution for the differential equation

y″ – 4y + 13y = e2x cos 3x.

Solution:

Here G(x) has the form of part 2 of the summary, where k = 2, m = 3, and P(x) = 1.

So, at first glance, the form of the trial solution would be yp(x) = e2x(A cos 3x + B sin 3x)

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### Example 6 – Solution

But the auxiliary equation is r2 – 4r + 13 = 0, with roots

r = 2 ± 3i, so the solution of the complementary equation is

yc(x) = e2x(c1cos 3x + c2sin 3x)

This means that we have to multiply the suggested trial solution by x.

yp(x) = xe2x(A cos 3x + B sin 3x)

cont’d

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### The Method of Variation of Parameters

Suppose we have already solved the homogeneous equation ay″ + by′ + cy = 0 and written the solution as

y(x) = c1y1(x) + c2y2(x)

where y1 and y2 are linearly independent solutions.

Let’s replace the constants (or parameters) c1 and c2 in Equation 4 by arbitrary functions and u1(x) and u2(x).

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### The Method of Variation of Parameters

We look for a particular solution of the nonhomogeneous equation ay″ + by′ + cy = G(x) of the form

yp(x) = u1(x) y1(x) + u2(x) y2(x)

(This method is called variation of parameters because we have varied the parameters c1 and c2 to make them functions.)

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### The Method of Variation of Parameters

Differentiating Equation 5, we get

Since u1 and u2 are arbitrary functions, we can impose two conditions on them.

One condition is that yp is a solution of the differential equation; we can choose the other condition so as to simplify our calculations.

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### The Method of Variation of Parameters

In view of the expression in Equation 6, let’s impose the condition that

Then

Substituting in the differential equation, we get

or

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### The Method of Variation of Parameters

But y1 and y2 are solutions of the complementary equation, so

and and Equation 8 simplifies to

Equations 7 and 9 form a system of two equations in the unknown functions and .

After solving this system we may be able to integrate to find u1 and u2 then the particular solution is given by Equation 5.

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### Example 7

Solve the equation

y″ + y = tan x, 0 < x < π/2.

Solution:

The auxiliary equation is r2 + 1 = 0 with roots ±i, so the solution of y″ + y = 0 is y(x) = c1 sin x + c2 cos x.

Using variation of parameters, we seek a solution of the form

yp(x) = u1(x) sin x + u2(x) cos x Then

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### Example 7 – Solution

Set

Then

For yp to be a solution we must have

Solving Equations 10 and 11, we get

(sin2x + cos2x) = cos x tan x

cont’d

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### Example 7 – Solution

= sinx u1(x) = –cos x

(We seek a particular solution, so we don’t need a constant of integration here.)

Then, from Equation 10, we obtain

cont’d

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### Example 7 – Solution

= cos x – sec x So

u2(x) = sin x – ln(sec x + tan x)

(Note that sec x + tan x > 0 for 0 < x < π/2.) Therefore

yp(x) = –cos x sin x + [sin x – ln(sec x + tan x)] cos x

= –cos x ln(sec x + tan x) and the general solution is

y(x) = c1sin x + c2cos x – cos x ln(sec x + tan x)

cont’d

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