Q: Find the solution of the given initial value problem.
y'' + y' - 2y = 0, y(0) = 1, y'(0) = 1
Answer:
r^2+r-2 = 0
(r-1)(r-2) = 0
r = 1 or r = -2
y = C1e^t + C2e^(-2t)
hence, y'(t) = C1e^t -2C2e^(-2t)
Because y(0)= 1, C1+ C2 = 1
y'(0) = 1, C1-2C2 = 1
C1 = 1, C2 = 0
Thus, y = e^t
