Page 77 - Advanced engineering mathematics

2.3 The Nonhomogeneous Equation 57

2.3.2 Undetermined Coefficients
We will discuss a second method for finding a particular solution of the nonhomogeneous

equation, which, however, applies only to the constant coefficient case y + ay + by = f (x).

The idea behind the method of undetermined coefficients is that sometimes we can guess a
general form for Y p (x) from the appearance of f (x).

EXAMPLE 2.11
We will find the general solution of y − 4y = 8x − 2x.
2

2x
It is routine to find the general solution c 1 e + c 2 e −2x of the associated homogeneous
equation. We need a particular solution Y p (x) of the nonhomogeneous equation.
2
Because f (x) = 8x − 2x is a polynomial and derivatives of polynomials are polynomials,
it is reasonable to think that there might be a polynomial solution. Furthermore, no such solution
3
can include a power of x higher than 2. If Y p (x) had an x term, this term would be retained by
2
the −4y term of y − 4y, and 8x − 2x has no such term.

2
This reasoning suggests that we try a particular solution Y p (x) = Ax + Bx + C. Compute

y (x) = 2Ax + B and y (x) = 2A. Substitute these into the differential equation to get

2
2
2A − 4(Ax + Bx + C) = 8x − 2x.
Write this as
2
(−4A − 8)x + (−4B + 2)x + (2A − 4C) = 0.
This second-degree polynomial must be zero for all x if Y p is to be a solution. But a second-
degree polynomial has only two roots, unless all of its coefficients are zero. Therefore,
−4A − 8 = 0,
−4B + 2 = 0,
and
2A − 4C = 0.
Solve these to get A =−2, B = 1/2, and C =−1. This gives us the particular solution

1
2
Y p (x) =−2x + x − 1.
2
The general solution is
1
2
2x
y(x) = c 1 e + c 2 e −2x − 2x + x − 1.
2
EXAMPLE 2.12
2x

Find the general solution of y + 2y − 3y = 4e .

x
The general solution of y + 2y − 3y = 0is c 1 e −3x + c 2 e .

2x
2x
Now look for a particular solution. Because derivatives of e are constant multiples of e ,
2x
2x
we suspect that a constant multiple of e might serve. Try Y p (x) = Ae . Substitute this into the
differential equation to get
2x
2x
2x
2x
2x
4Ae + 4Ae − 3Ae = 5Ae = 4e .
2x
This works if 5A = 4, so A = 4/5. A particular solution is Y p (x) = 4e /5.
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October 14, 2010 14:12 THM/NEIL Page-57 27410_02_ch02_p43-76