Page 79 - Advanced engineering mathematics

2.3 The Nonhomogeneous Equation 59

EXAMPLE 2.15
x
x
Revisit Example 2.14. Since f (x) = 8e , our first impulse was to try Y p (x) = Ae . But this is a
x
solution of the associated homogeneous equation, so multiply by x and try Y p (x) = Axe .Now
x
x
x
x

Y = Ae + Axe and Y = 2Ae + Axe .

p p
Substitute these into the differential equation to get
x
x
x
x
x
x
2Ae + Axe + 2(Ae + Axe ) − 3Axe = 8e .
x
x
x
This reduces to 4Ae = 8e ,so A = 2, yielding the particular solution Y p (x) = 2xe . The general
solution is
x
x
y(x) = c 1 e −3x + c 2 e + 2xe .
EXAMPLE 2.16
3x

Solve y − 6y + 9y = 5e .
2
The associated homogeneous equation has the characteristic equation (λ − 3) = 0 with
repeated roots λ = 3. The general solution of this associated homogeneous equation is
3x
3x
c 1 e + c 2 xe .
3x
For a particular solution, we might first try Y p (x)= Ae , but this is a solution of the homoge-
3x
neous equation. Multiply by x and try Y p (x)= Axe . This is also a solution of the homogeneous
2 3x
equation, so multiply by x again and try Y p (x)= Ax e . If this is substituted into the differential
2 3x
equation, we obtain A =5/2, so a particular solution is Y p (x)=5x e /2. The general solution is
5
y = c 1 e + c 2 xe + x e .
2 3x
3x
3x
2
The method of undetermined coefficients is limited by our ability to “guess” a particular
solution from the form of f (x), and unlike variation of parameters, requires that the coefficients
of y and y be constant.

Here is a summary of the method. Suppose we want to find the general solution of

y + ay + by = f (x).
Step 1. Write the general solution
y h (x) = c 1 y 1 (x) + c 2 y 2 (x)
of the associated homogeneous equation

y + ay + by = 0

with y 1 and y 2 linearly independent. We can always do this in the constant coefficient
case.
Step 2. We need a particular solution Y p of the nonhomogeneous equation. This may require
several steps. Make an initial attempt of a general form of a particular solution using
f (x) and perhaps Table 2.1 as a guide. If this is not possible, this method cannot be
used. If we can solve for the constants so that this first guess works, then we have Y p .
Step 3. If any term of the first attempt is a solution of the associated homogeneous equation,
multiply by x. If any term of this revised attempt is a solution of the homogeneous
equation, multiply by x again. Substitute this final general form of a particular solution
into the differential equation and solve for the constants to obtain Y p .
Step 4. The general solution is
y = y 1 + y 2 + Y p .
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October 14, 2010 14:12 THM/NEIL Page-59 27410_02_ch02_p43-76