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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{tan}\mathbf{2}\mathbf{t}$

The homogenous equation is ${\mathrm{r}}^2+4=0$.

Two independent solutions are $\mathrm{r}=\pm 2\mathrm{i}$.

Then${\mathrm{y}}_1=\cos 2\mathrm{t},{\mathrm{y}}_2=\sin 2\mathrm{t}$

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos 2\mathrm{t}+{\mathrm{c}}_2\sin 2\mathrm{t}$

The particular solution is${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right)\cos 2\mathrm{t}+{\mathrm{v}}_2\left(\mathrm{t}\right)\sin 2\mathrm{t}$

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right)\cos 2\mathrm{t}+{\mathrm{v}}_2\left(\mathrm{t}\right)\sin 2\mathrm{t}$

And referring to (9) and solve the system then

$\begin{array}{c}{\mathrm{v}}_1\text{'}=\frac{-\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ =\frac{-\tan 2\mathrm{t}.\sin 2\mathrm{t}}{1[2{\cos }^{2}2\mathrm{t}+2{\sin }^{2}2\mathrm{t}]}\\ =\frac{-\tan 2\mathrm{t}.\sin 2\mathrm{t}}{2[{\cos }^{2}2\mathrm{t}+{\sin }^{2}2\mathrm{t}]}\\ =-\frac{{\sin }^{2}2\mathrm{t}}{2\cos 2\mathrm{t}}\end{array}$

Now integrating this.

$\begin{array}{c}{\mathrm{v}}_1\left(\mathrm{t}\right)=\int -\frac{{\sin }^{2}2\mathrm{t}}{2\cos 2\mathrm{t}}\mathrm{dt}\\ =-\frac{1}{4}\int \frac{{\sin }^2\mathrm{x}}{\cos x}\mathrm{dx}\\ =\frac{1}{4}\int \frac{{\cos }^2\mathrm{x}-1}{\cos x}\mathrm{dx}\\ =\frac{1}{4}[\int \mathrm{cosxdx}-\int \mathrm{secx}\mathrm{dx}]\\ =\frac{1}{4}\left[\mathrm{sinx}-\ln (\mathrm{tanx}+\mathrm{secx})\right]+\mathrm{C}\\ =\frac{1}{4}\left[\sin 2\mathrm{t}-\ln (\tan 2\mathrm{t}+\sec 2\mathrm{t})\right]+\mathrm{C}\end{array}$

$\begin{array}{c}{\mathrm{v}}_2\text{'}=\frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ =\frac{\tan 2\mathrm{t}.\cos 2\mathrm{t}}{1[2{\cos }^{2}2\mathrm{t}+2{\sin }^{2}2\mathrm{t}]}\\ =\frac{-\tan 2\mathrm{t}.\sin 2\mathrm{t}}{2[{\cos }^{2}2\mathrm{t}+{\sin }^{2}2\mathrm{t}]}\\ =\frac{\sin 2\mathrm{t}}{2}\end{array}$

Now integrate this.

$\begin{array}{c}{\mathrm{v}}_2\left(\mathrm{t}\right)=\int \frac{\sin 2\mathrm{t}}{2}\mathrm{dt}\\ =-\frac{1}{4}\cos 2\mathrm{t}+\mathrm{C}\end{array}$

Thus, a particular solution is:

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\left(\frac{1}{4}\left[\sin 2\mathrm{t}-\ln (\tan 2\mathrm{t}+\sec 2\mathrm{t})\right]+\mathrm{C}\right)\cos 2\mathrm{t}+\left(-\frac{1}{4}\cos 2\mathrm{t}+\mathrm{C}\right)\sin 2\mathrm{t}\\ =-\frac{\cos 2\mathrm{t}}{4}\left(\left[\ln (\tan 2\mathrm{t}+\sec 2\mathrm{t})\right]\right)\end{array}$

Thus, the general solution is:

$\begin{array}{l}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos 2\mathrm{t}+{\mathrm{c}}_2\sin 2\mathrm{t}-\frac{\cos 2\mathrm{t}}{4}\left(\left[\ln (\tan 2\mathrm{t}+\sec 2\mathrm{t})\right]\right)\end{array}$