Solve the IVP d2ydt2 4 dydt 4y 3e2t y0 1 y0 0SolutionSi

Solve the IVP d^2y/dt^2 + 4 dy/dt + 4y = 3e^-2t, y(0) = 1, y\'(0) = 0

Solution

Since the homogeneous equation has solution y = (C1 x + C2) * e^(-2t),
we assume that the particular solution is y_p = A3x^2 e^(-2t).

Then, y\'\' + 4y\' + 4y = 3e^(-2t)
==> [(2A + 8At + 4At^2) e^(-2t)] + 4[(2At + 2At^2) e^(-2t)] + 4[At^2 e^(-2t)] = e^(-2t)

On solving we will get
==> 2A = 1
==> A = 1/2.

Hence, y_p = (3/2)t^2 e^(-2t).

y = (c1t + c2 ) e^(-2t)

So, using initial values,

At x=0, y=1

1 = c2

Differentiating y,

y\' = c1e^(-2t) - 2 c1te^(-2t) - 2c2e^(-2t)

At t = 0 , y=0

0 = c1 - 0 - 2c2

c1 = 2

So,

y = (t + 2) e^(-2t) + 3/2 t^2 e^(-2t)