Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz3 => Topic started by: Jialin Shang on October 11, 2019, 01:24:36 PM

Q: Find the solution of the given initial value problem.
y'' + y'  2y = 0, y(0) = 1, y'(0) = 1
Answer:
r^2+r2 = 0
(r1)(r2) = 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, C12C2 = 1
C1 = 1, C2 = 0
Thus, y = e^t