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The method of undetermined coefficients requires that the operator L have constant coefficients and that the function g be of a particular class. It works for ...
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Some Preliminary Examples
Example 1
Suppose we want to find a particular solution for y′^ + y = 8e^3 x. Since L simply adds the derivative and the function, it makes sense that the function we want should be a multiple of e^3 x. Since we don’t know which
one, we can use the generic form y = Ae^3 x. Then y′^ = 3Ae^3 x^ and Ly = 4Ae^3 x. We want Ly to be 8e^3 x,
which is accomplished by taking A = 2. The particular solution is yp = 2e^3 x^ and the general solution is y = 2e^3 x^ + c 1 e−x.
Example 2 Suppose we want to find a particular solution for y′^ + y = x. This time it makes sense that we should try a
first degree polynomial, but we don’t know which one. Using the generic form y = A + Bx, we get y′^ = B and Ly = A + B + Bx. We want Ly to be x, which requires A + B = 0 and B = 1. Thus, A = −1,
yp = −1 + x, and the general solution is y = −1 + x + c 1 e−x.
Example 3
To find a particular solution for y′^ + y = e−x, we can try y = Ae−x, as we did in Example 1. However, we then get Ly = 0 and we can’t choose A to get Ly = e−x. The correct trial solution turns out to be
Axe−x. Then y′^ = Ae−x^ − Axe−x, so Ly = Ae−x. This works, because A = 1 gives us the desired g. So the particular solution is yp = xe−x^ and the general solution is y = xe−x^ + c 1 e−x^ = (c 1 + x)e−x.
Example 4
To find a particular solution for y′^ + y = 2 cos x, our first thought is that we could use y = A cos x. But then y′^ = −A sin x and Ly = A cos x − A sin x. We can get the right cosine coefficient by choosing
A = 2, but this does not work because Ly = 2 cos x − 2 sin x and g = 2 cos x are not the same function. Instead, the correct trial solution turns out to be A cos x + B sin x. Then y′^ = −A sin x + B cos x and
Ly = (A + B) cos x + (B − A) sin x. We can get cos x from this family by choosing A + B = 2 and B − A = 0. The result is the particular solution yp = cos x + sin x and the general solution is y = cos x + sin x + c 1 e−x.
Example 6 The family of generalized exponential functions with characteristic value pair 1 ± 2 i and degree 1 can be written as y 1 ± 2 i, 1 = (Ax + B)ex^ cos 2x + (Cx + D)ex^ sin 2x,
where A and C are not both zero, in which case the degree would be 0 instead of 1.
Solving Ly = g
Example 7 Let Ly = y′^ + y. We now apply our method to Examples 1–4.
Example 8 To solve y′′^ − 4 y′^ + 13y = xe^2 x^ cos 3x + 3e^2 x^ sin x, we first identify g as a generalized exponential function
with c = 2 ± 3 i and d = 1. The family of such functions is yc,d = (A + Bx)e^2 x^ cos 3x + (C + Dx)e^2 x^ sin 3x.
The operator L has one complex pair of characteristic values, which happens to be the same as c. Thus, s = 1 and the trial solution is y = (Ax + Bx^2 )e^2 x^ cos 3x + (Cx + Dx^2 )e^2 x^ sin 3x.