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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathbf{+}\mathbf{16}\mathit{y}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathbf{4}\mathit{\theta }$

The homogenous equation is $r^2+16=0$.

Two independent solutions are $r=\pm 4i$.

Then$y_1=cos4\theta ,y_2=sin4\theta$

${\text{y}}_h\left(t\right)=c_{1}cos4\theta +c_{2}sin4\theta$

The particular solution is$y_p=v_1\left(t\right)cos4\theta +v_2\left(t\right)sin4\theta$

Here$y_p=v_1\left(t\right)\cos 4\theta +v_2\left(t\right)\sin 4\theta$

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 $y_p=0$.

$$\begin{gathered} v_1\text{'}\cos 4\theta +v_2\text{'}\sin 4\theta =0 \\ -4v_1\text{'}\sin 4\theta +4v_2\text{'}\cos 4\theta =\frac{f}{a} \\ -4v_1\text{'}\sin 4\theta +4v_2\text{'}\cos 4\theta =\sec 4\theta \end{gathered}$$

_{$$\begin{gathered} v_1\text{'}=\frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{-\sec 4\theta .\sin 4\theta }{4\left[{\cos }^{2}4\theta .+{\sin }^{2}4\theta \right]} \\ =\frac{-1}{\tan 4\theta } \end{gathered}$$}

Now integrating this.

$$\begin{gathered} v_1\left(t\right)=\int \frac{-1}{\tan 4\theta } d\theta \\ =\frac{1}{16}\ln \left|\cos 4\theta \right|+C \end{gathered}$$

$$\begin{gathered} v_2\text{'}=\frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{\sec 4\theta .\cos 4\theta }{4\left[{\cos }^{2}4\theta .+{\sin }^{2}4\theta \right]} \\ =\frac{1}{4} \end{gathered}$$

Integrate this.

$$\begin{gathered} v_2\left(t\right)=\int \frac{1}{4}d\theta \\ =\frac{1}{4}\theta +C \end{gathered}$$

Thus, a particular solution is:

$$\begin{gathered} y_p=\left(\frac{1}{16}\ln \left|\cos 4\theta \right|+C\right)\cos 4\theta +\left(\frac{1}{4}\theta +C\right)\sin 4\theta \\ {y}_p=\left(\frac{1}{16}\ln \left|\cos 4\theta \right|\right)\cos 4\theta +\left(\frac{1}{4}\theta \right)\sin 4\theta \end{gathered}$$

Therefore, the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cos4\theta +c_{2}sin4\theta +(\frac{1}{16}ln\left|cos4\theta \right|)cos4\theta +(\frac{1}{4}\theta )sin4\theta \end{gathered}$$