In Problems 11–18, find a general solution to the differential equation.

$\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{=}{\mathbf{sec}}^{\mathbf{4}}\mathbf{\left(}\mathbf{2}\mathbf{t}\mathbf{\right)}$

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

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

Then$v_1=\cos 2t,v_2=\sin 2t$

$v_h\left(t\right)=c_1\cos 2t+c_2\sin 2t$

The particular solution is$v_p=u_1\left(t\right)cos2t+u_2\left(t\right)sin2t$.

Here$y_p=v_1\left(t\right)\cos 2t+v_2\left(t\right)\sin 2t$

Here$f={\sec }^4\left(2t\right)$ androle="math" localid="1664109539986" $a=1$

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:

$\begin{array}{rcl}{u}_1\text{'}\cos 2t+u_2\text{'}\sin 2t& =& 0\\ -2u_1\text{'}\sin 2t+2u_2\cos 2t& =& \frac{f}{a}\\ -2u_1\text{'}\sin 2t+2u_2\cos 2t& =& {\sec }^{4}2t\end{array}$

$\begin{array}{rcl}{u}_1\text{'}& =& \frac{-f\left(t\right)v_2\left(t\right)}{a\left[v_1\left(t\right)v{\text{'}}_2\left(t\right)-v{\text{'}}_1\left(t\right)v_2\left(t\right)\right]}\\ & =& \frac{-\left({\sec }^{4}2t\right).\sin 2t}{2\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& -\frac{1}{2}\left({\sec }^{4}t\right).\sin 2t\end{array}$

Now integrating this.

$\begin{array}{rcl}{u}_1\left(t\right)& =& \int -\frac{1}{2}\left({\sec }^{4}t\right).\sin 2 tdt\\ & =& -\frac{1}{12}{\sec }^{3}2t+C\end{array}$

$\begin{array}{rcl}{u}_2\text{'}& =& \frac{f\left(t\right)y_1\left(t\right)}{a\left[v_1\left(t\right)v{\text{'}}_2\left(t\right)-v{\text{'}}_1\left(t\right)v_2\left(t\right)\right]}\\ & =& \frac{\left({\sec }^{4}2t\right).\cos 2t}{2\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& \frac{1}{2}{\sec }^{3}2t\end{array}$

Integrate this.

$\begin{array}{rcl}{u}_2\left(t\right)& =& \int \frac{1}{2}{\sec }^{3}2tdt\\ & =& \frac{1}{8}\left(\sec 2t\tan 2t+\ln \left|\sec 2t+\tan 2t\right|+C\right)\end{array}$

Thus, the particular solution is:

$$\begin{gathered} v_p=\left(-\frac{1}{12}{\sec }^{3}2t+C\right)\cos 2t+\frac{1}{8}\left(\sec 2t\tan 2t+\ln \left|\sec 2t+\tan 2t\right|+C\right)\sin 2t \\ {v}_p=\frac{1}{24}{\sec }^{3}2t-\frac{1}{8}+\frac{1}{8}\sin 2t\ln \left|\sec 2t+\tan 2t\right| \end{gathered}$$

Therefore, the general solution is:

$$\begin{gathered} v\left(t\right)=v_h\left(t\right)+v_p\left(t\right) \\ v\left(t\right)=c_{1}cos2t+c_{2}sin2t+\frac{1}{24}se{c}^{3}2t-\frac{1}{8}+\frac{1}{8}sin2tln\left|sec2t+tan2t\right| \end{gathered}$$