Lecture Notes in Differential Equations - Bruce E. Shapiro

166 LESSON 19. METHOD OF UNDETERMINED COEFFICIENTS
Equating like coefficients,
−6A = 6 =⇒ A = −1 (19.28)
A + B = 0 =⇒ B = −A = 1 (19.29)
y p = −t + 1 (19.30)
hence the general solution is
y = C 1 e 2t + C 2 e −3t − t + 1 (19.31)
Example 19.3. Find the general solution to
y ′′ − 3y ′ − 4y = e −t (19.32)
The characteristic equation is
r 2 − 3r − 4 = (r − 4)(r + 1) = 0 (19.33)
so that
y H = C 1 e 4t + C 2 e −t (19.34)
From the form of the forcing function we are tempted to try
y P = Ae −t (19.35)
but that is already part of the homogeneous solution, so instead we try
y P = Ate −t (19.36)
Differentiating,
y ′ = Ae −t − Ate −t (19.37)
y ′′ = −2Ae −t + Ate −t (19.38)
Substituting into the differential equation,
−2Ae −t + Ate −t − 3(Ae −t − Ate −t ) − 4(Ate −t ) = e −t (19.39)
−2A + At − 3(A − At) − 4(At) = 1 (19.40)
−2A + At − 3A + 3At − 4At = 1 (19.41)
−5A = 1 (19.42)
A = − 1 5
(19.43)
hence
y P = − 1 5 te−t (19.44)
and the general solution of the differential equation is
y = C 1 e 4t + C 2 e −t − 1 5 te−t (19.45)