Lecture Notes in Differential Equations - Bruce E. Shapiro

164 LESSON 19. METHOD OF UNDETERMINED COEFFICIENTS
Since the characteristic equation is r 2 − 1 = 0, with roots r = ±1, we know
that the homogeneous solution is
As a guess to the particular solution we try
Differentiating,
Substituting into (19.1),
y H = C 1 e t + C 2 e −t (19.4)
y P = At 2 + Bt + C (19.5)
y P ′ = 2At + B (19.6)
y P ′′ = 2A (19.7)
2A − At 2 − Bt − C = t 2 (19.8)
Since each side of the equation contains polynomials, we can equate the
coefficients of the power on each side of the equation. Thus
Coefficients of t 2 : − A = 1 (19.9)
Coefficients of t : B = 0 (19.10)
Coefficients of t 0 : 2A − C = 0 (19.11)
The first of these gives A = −1, the third C = 2A = −2. Hence
y P = −t 2 − 2 (19.12)
Substitution into (19.1) verifies that this is, in fact, a particular solution.
The complete solution is then
y = y H + y P = C 1 e r + C 2 e −t − t 2 − 2 (19.13)
Use of this method depends on being able to come up with a good guess.
Fortunately, in the case where f is a combination of polynomials, exponentials,
sines and cosines, a good guess is given by the following heuristic.
1. If f(t) is a polynomial of degree n, use
y P = a n t n + a n−1 t n−1 + · · · a 2 t 2 + a 1 t + a 0 (19.14)
2. If f(t) = e rt and r is not a root of the characteristic equation, try
y P = Ae rt (19.15)