Find the general solution of the given equation y'' + 4y' + 4y = t^{-2}e-^{2t}, t>0

First, solve y'' + 4y' + 4y = 0.

So r^{2} + 4r + 4 = 0, r = -2 and r = -2

We have Y_{c}(t) = ce-^{2t} +c_{2}te^{-2t}

y_{1} = e-^{2t}

y_{2} = te^{-2t}

Determine the Wronskian as follows:

W(y_{1}, y_{2})(t) = e^{-4t}

Since Y(t) = u_{1}(t)y_{1}(t) + u_{2}(t)y_{2}(t)

u_{1} = -ln* t*

u_{2} = -t^{-1}

Hence Y(t) = u_{1}(t)y_{1}(t) + u_{2}(t)y_{2}(t)

= -ln* t*e-^{2t} + -t^{-1}te^{-2t}

= -e^{-2t}ln*t*-e^{-2t}

So the general solution is y = y_{c} + Y(t) = c_{1}e^{-2t} + c_{2}te^{-2t}-e^{-2t}ln*t*

with c-1 = c_{1}