Question: Find the solution of the initial value problem.
y' - 2y = e^{2t}, y(0) = 2
Solution:
p(t) = -2, g(t) = e^{2t}, then μ(t) = e^{∫-2 dt} = e^{-2t}
Multiply both sides by μ(t) and we get:
e^{-2t}y' - 2e^{-2t}y = e^{-2t}e^{2t} = e^{0} = 1
Integral both sides:
e^{-2t}y = ∫1dt
e^{-2t}y = t + c, where c is constant
y = e^{2t}t + e^{2t}c
Substitue y(0) = 2 into the equation above:
2 = 0 + c
c = 2
Then, we obtain the solution y = e^{2t}(t+2)