Author Topic: TUT0602 Quiz1  (Read 480 times)

Vickyyy

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TUT0602 Quiz1
« on: September 28, 2019, 10:50:17 PM »
Question: Find the solution of the initial value problem.
y' - 2y = e2t, y(0) = 2
Solution:
p(t) = -2, g(t) = e2t, then μ(t) = e∫-2 dt = e-2t
Multiply both sides by μ(t) and we get:
e-2ty' - 2e-2ty = e-2te2t = e0 = 1
Integral both sides:
e-2ty = ∫1dt
e-2ty = t + c, where c is constant
y = e2tt + e2tc
Substitue y(0) = 2 into the equation above:
2 = 0 + c
c = 2
Then, we obtain the solution y = e2t(t+2)