In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathit{l}\mathit{n}\mathit{t}$

The homogenous equation is $r^2+4r+4=0$.

Two independent solutions are $r=-2,-2$.

Then$y_1=e^{-2t},y_2=t{e}^{-2t}$

$y_h\left(t\right)=c_1{e}^{-2t}+c_{2}t{e}^{-2t}$

The particular solution is$y_p=v_1\left(t\right)e^{-2t}+v_2\left(t\right)t{e}^{-2t}$

Here$y_p=v_1\left(t\right)e^{-2t}+v_2\left(t\right)t{e}^{-2t}$

And referring to (9) $y\left(t\right)=c_1{e}^{\alpha t}\cos \beta t{c}_2{e}^{\alpha t}\sin \beta t$and solve the system then put the value of$y_p=0$

$$\begin{gathered} v_1\text{'}{e}^{-2t}+v_2\text{'}t{e}^{-2t}=0 \\ -2v_1\text{'}{e}^{-2t}+2v_2\text{'}\left[e^{-2t}-2t{e}^{-2t}\right]=\frac{f}{a} \\ -2v_1\text{'}{e}^{-2t}+2v_2\text{'}\left[e^{-2t}-2t{e}^{-2t}\right]=\ln t{e}^{-2t} \\ -2v_1\text{'}+2v_2\text{'}\left[1-2t\right]=\ln t \end{gathered}$$

$$\begin{gathered} v_1\text{'}=\frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{-\ln t{e}^{-2t}.t{e}^{-2t}}{\left[e^{-2t}.\left[1-2t\right]e^{-2t}+2e^{-2t}t{e}^{-2t}\right]} \\ =-t\ln t \end{gathered}$$

Now integrating this.

$$\begin{gathered} v_1\left(t\right)=\int -t\ln tdt \\ =\frac{t^2}{2}\left(\frac{1}{2}-\ln t\right)+C \end{gathered}$$

$$\begin{gathered} v_2\text{'}=\frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{\ln t{e}^{-2t}.e^{-2t}}{\left[e^{-2t}.\left[1-2t\right]e^{-2t}+2e^{-2t}t{e}^{-2t}\right]} \\ =\ln t \end{gathered}$$

Integrate this.

$$\begin{gathered} v_2\left(t\right)=\int \ln tdt \\ =t\left(\ln t-1\right)+C \end{gathered}$$

Thus, a particular solution is

$$\begin{gathered} y_p=\left(\frac{t^2}{2}\left(\frac{1}{2}-\ln t\right)+C\right)e^{-2t}+\left(t(\ln t-1)+C\right)t{e}^{-2t} \\ {y}_p=\frac{t^2}{2}{e}^{-2t}\left(\ln t-\frac{3}{2}\right) \end{gathered}$$

Therefore, the general solution is

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^{-2t}+c_{2}t{e}^{-2t}+\frac{t^2}{2}{e}^{-2t}\left(lnt-\frac{3}{2}\right) \end{gathered}$$$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^{-2t}+c_{2}t{e}^{-2t}+\frac{t^2}{2}{e}^{-2t}\left(lnt-\frac{3}{2}\right) \end{gathered}$$