Answer
The general solution of $y{''}+4y'+4y = t^{-2}e^{-2t}$ is:
$$y(t) = c_0e^{-2t}+c_2te^{-2t} -e^{-2t}\ln(t)$$
Work Step by Step
$$y{''}+4y'+4y = t^{-2}e^{-2t}$$
$$g(t) = t^{-2}e^{-2t}$$
To find the complementary function $y_c(t)$, consider the homogeneous part
$$y{''}+4y'+4y=0$$
Corresponding characteristic equation is:
$$m^2+4m+4=0$$
$$ m = -2, -2 $$
Hence, Complementary function is:
$$y_c(t) = c_1e^{-2t}+c_2te^{-2t}$$
Hence Independent solutions $y_1(t)$ and $y_2(t)$ of homogeneous equation will be
$$y_1(t) = e^{-2t} \text{ and } y_2(t) = te^{-2t}$$
Particular Solution $y_p(t)$ is given by:
$$y_p(t) = u_1(t)y_1(t)+u_2(t)y_2(t)$$
where $$u_1(t) = -\int \frac{y_2(t)g(t)}{W(y_1,y_2)(t)}dt + c_3$$
$$u_2(t) = \int \frac{y_1(t)g(t)}{W(y_1,y_2)(t)}dt+c_4$$
Since, $y_p(t)$ is a particular solution so we can choose $c_3=c_4=0$ such that both the constant term vanishes.
$$W(y_1, y_2)(t) =
\begin{vmatrix}
e^{-2t} & te^{-2t} \\
-2e^{-2t} & -2te^{-2t}+e^{-2t}
\end{vmatrix}=e^{-4t}$$
$$u_1(t) = -\int \frac{te^{-2t} \times t^{-2}e^{-2t}}{e^{-4t}}dt = -\int \frac{1}{t}dt$$
$$u_1(t) = -\ln(t)$$
$$u_2(t) = \int \frac{e^{-2t}\times t^{-2}e^{-2t}}{e^{-4t}}dt = \int \frac{1}{t^2}dt$$
$$u_2(t) = -\frac{1}{t}$$
$$\therefore y_p(t) = -\ln(t)e^{-2t}-\frac{1}{t}te^{-2t}$$
$$\therefore y_p(t) = -\ln(t)e^{-2t}-e^{-2t}$$
Therefore, the general solution is:
$$y(t) = y_c(t)+y_p(t)$$
$$y(t) = c_1e^{-2t}+c_2te^{-2t} -\ln(t)e^{-2t}-e^{-2t}$$
$$y(t) = (c_1-1)e^{-2t}+c_2te^{-2t} -e^{-2t}\ln(t)$$
Say $c_0 = c_1-1$
$$y(t) = c_0e^{-2t}+c_2te^{-2t} -e^{-2t}\ln(t)$$
