Lecture Notes in Differential Equations - Bruce E. Shapiro

168 LESSON 19. METHOD OF UNDETERMINED COEFFICIENTS
The second initial condition is y ′ (0) = −1 so that
− 1 = 2C 2 (19.58)
hence
y = − 1 2 sin 2t + 3 t sin 2t (19.59)
4
Example 19.5. Solve the initial value problem
⎫
y ′′ − y = t + e 2t ⎪⎬
y(0) = 0
⎪⎭
y ′ (0) = 1
(19.60)
The characteristic equation is r 2 − 1 = 0 so that
y H = C 1 e t + C 2 e −t (19.61)
For a particular solution we try a linear combination of particular solutions
for each of the two forcing functions,
Substituting into the differential equation,
y P = At + B + Ce 2t (19.62)
y ′ P = A + 2Ce 2t (19.63)
y ′′
P = 4Ce 2t (19.64)
4Ce 2t − At − B − Ce 2t = t + e 2t (19.65)
3Ce 2t − At − B = t + e 2t (19.66)
A = −1 (19.67)
B = 0 (19.68)
C = 1 3
(19.69)
Hence
y = C 1 e t + C 2 e −t − t + 1 3 e2t (19.70)
From the first initial condition, y(0) = 0,
0 = C 1 + C 2 + 1 3
(19.71)
From the second initial condition y ′ (0) = 1,
1 = C 1 − C 2 + 2 3
(19.72)