In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker?

$\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathit{e}}^{\mathbf{2}\mathbf{t}}$

The differential equation is$2x\text{'}\text{'}\left(t\right)-2x\text{'}\left(t\right)-4x\left(t\right)=2e^{2t}$

This can be written as$x\text{'}\text{'}\left(t\right)-x\text{'}\left(t\right)-2x\left(t\right)=e^{2t}$

The homogenous equation is $r^2-r-2=0$.

Two independent solutions are $r=2,-1$.

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

$y_h\left(t\right)=c_1{e}^{2t}+c_2{e}^{-t}$

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

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

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 by derivative then,

$$\begin{gathered} v_1\text{'}{e}^{2t}+v_2\text{'}{e}^{-t}=0 \\ 2v_1\text{'}{e}^{2t}-v_2\text{'}{e}^{-t}=\frac{f}{a} \\ 2v_1\text{'}{e}^{2t}-v_2\text{'}{e}^{-t}=e^{2t} \end{gathered}$$

Now for finding the values.

$$\begin{gathered} v_1\text{'}=\frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{-e^{2t}.e^{-t}}{\left[-e^{2t}.e^{-t}-2e^{-t}.e^{-t}\right]} \\ =\frac{1}{3} \end{gathered}$$

Now integrating this;

$$\begin{gathered} v_1\left(t\right)=\int \frac{1}{3} dt \\ =\frac{t}{3}+C \\ {v}_2\text{'}=\frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{e^{2t}.e^t}{\left[-e^{2t}.e^{-t}-2e^{-t}.e^{-t}\right]} \\ =-\frac{1}{3}{e}^{3t} \end{gathered}$$

Integrate this.

$$\begin{gathered} v_2\left(t\right)=\int -\frac{1}{3}{e}^{3t}dt \\ =-\frac{1}{9}{e}^{3t}+C \end{gathered}$$

Thus, the particular solution is when$C=0$

$$\begin{gathered} y_p=\frac{t}{3}{e}^{2t}+C-(\frac{1}{9}{e}^{3t}+C)e^{-t} \\ {y}_p=\frac{t}{3}{e}^{2t}-\frac{1}{9}{e}^{2t} \end{gathered}$$

Therefore, the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^{2t}+c_{2}2e^{-t}+\frac{t}{3}{e}^{2t}-\frac{1}{9}{e}^{2t} \end{gathered}$$