Method of Undetermined Coefficient or Guessing Method

This method is based on a guessing technique. That is, we will guess the form of y_p and then plug it in the equation to find it. However, it works only under the following two conditions:
Condition 1: the associated homogeneous equations has constant coefficients;
Condition 2: the nonhomogeneous term g(x) is a special form

where P(x) and L(x) are polynomial functions.
Note that we may assume that g(x) is a sum of such functions (see the remark below for more on this).

Assume that the two conditions are satisfied. Consider the equation

where a, b and c are constants and

where P_n(x) is a polynomial function with degree n. Then a particular solution y_p is given by

where

where the constants A_k and B_k have to be determined. The power s is equal to 0 if α + iβ is not a root of the characteristic equation. If α + iβ is a simple root, then s = 1 and s = 2 if it is a double root.

Example

Find the general solution of the differential equation y'' + y' - 2y = e^-t sin t

First find the solution to the homogeneous differential equation

y'' + y' - 2y = 0

We have

r² + r - 2 = (r - 1)(r + 2) = 0
r = -2 or r = 1

Thus

y_h = c₁e^-2t + c₂e^t

Next notice that e^-t sin t and all of its derivatives are of the form

A e^-t sin t + B e^-t cos t

We set y_p = A e^-t sin t + B e^-t cos t

and find

y_p' = A (-e^-t sin t + e^-t cos t) + B (-e^-t cos t - e^-t sin t) = -(A + B)e^-t sin t + (A - B)e^-t cos t

and

y_p'' = -(A + B)(-e^-t sin t + e^-t cos t) + (A - B)(-e^-t cos t - e^-t sin t)
= [(A + B) - (A - B)]e^-t sin t + [-(A + B) - (A - B)]e^-t cos t
= 2B e^-t sin t - 2A e^-t cos t

Now put these into the original differential equation to get

2B e^-t sin t - 2A e^-t cos t + -(A + B)e^-t sin t + (A - B)e^-t cos t - 2(A e^-t sin t + B e^-t cos t) = e^-t sin t

Combine like terms to get

(2B - A - B - 2A) e^-t sin t + (-2A + A - B - 2B) e^-t cos t = e^-t sin t

or

(-3A + B) e^-t sin t + (-A - 3B) e^-t cos t = e^-t sin t

Equating coefficients, we get

-3A + B = 1
-A - 3B = 0

This system has solution

A = -3/10, B = 1/10

The particular solution is

y_p = -3/10 e^-t sin t + 1/10 e^-t cos t

Adding the particular solution to the homogeneous solution gives

y = y_h + y_p = c₁e^-2t + c₂e^t + -3/10 e^-t sin t + 1/10 e^-t cos t