solve the following initial value problem?

Question

y' + 2y = 4 , y(0) = 1

y(t) = ?

gp4rts, looks as if your answer is incorrect...

Answer 1

Using the formula y'(t) + p(t)y(t) = g(t) -> y(t) = {e^-int(p(t))][int(g(t) * e^int(p(t)))], we get
y(t) = [e^-2t][2e^2t + C] = 2 +Ce^-2t
y(0)= 1 = 2+ C gives C = -1 so,
y(t) = 2 - e^-2t

Answer 2

Solve the diff eq by integrating both sides w.r.t t and then substitute t for 0 and y(t) for 1 to solve for the constant of integration C.

Answer 3

dy/dt = 4 - 2y

dy/(4-2y) = dt

-1/2*ln(4-2y) = t + K

ln(4 - 2y) = -2t + K'

4 - 2y = K''e^-2t

y = 4 - K''e^-2t

at t=0 y=1, 1 = 4 - K''

K'' = 3

y = 4 - 3e^-2t