Find the general solution of the given differential equation:
$y"+4y'+4y=t^{-2}e^{-2t},t>0.$
Let $y"+4y'+4y=0.$
$r^{2}+4r+4=0,r_{1}=-2=r_{2}.$
Thus,
$y_{c}(t)=C_{1}e^{-2t}+C_{2}te^{-2t}.$
$W=\begin{vmatrix}e^{-2t} & te^{-2t}\\
-2e^{-2t} & -2te^{-2t}+e^{-2t}
\end{vmatrix}=-2te^{4t}+e^{4t}+2te^{4t}=e^{4t}.$
$W_{1}=\begin{vmatrix}0 & te^{-2t}\\
1 & -2te^{-2t}+e^{-2t}
\end{vmatrix}=-te^{-2t}.$
$W_{2}=\begin{vmatrix}e^{-2t} & 0\\
-2e^{-2t} & 1
\end{vmatrix}=e^{-2t}.$
Let $\mu_{1}=\int\frac{W_{1}(t)g(t)}{W(t)}dt=\int\frac{-te^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-ln(t). $
$\mu_{2}=\int\frac{W_{2}(t)g(t)}{W(t)}dt=\int\frac{e^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-\frac{1}{t}.$
Therefore,
$y(t)=y_{1}(t)\mu_{1}+y_{2}(t)\mu_{2}+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$
