TUT0602 Quiz1

[Vickyyy]

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^-2ty' - 2e^-2ty = e^-2te^2t = e⁰ = 1

Integral both sides:

e^-2ty = ∫1dt

e^-2ty = t + c, where c is constant

y = e^2tt + e^2tc

Substitue y(0) = 2 into the equation above:

2 = 0 + c

c = 2

Then, we obtain the solution y = e^2t(t+2)