91Ó°ÊÓ

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

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

Then$y_1=\cos 2t,y_2=\sin 2t$

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

The particular solution is$y_p=v_1\left(t\right)\cos 2t+v_2\left(t\right)\sin 2t$

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

And referring to (9)$y\left(t\right)=c_1{e}^{\mathrm{Î\pm }t}\cos \mathrm{β}t{c}_2{e}^{\mathrm{Î\pm }t}\sin \mathrm{β}t$ and solve the system then

Put the value of$y_p=0$

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

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

Now integrating this.

$$\begin{gathered} v_1\left(t\right)=∫-\frac{1}{2}\csc 2 tdt \\ =-\frac{1}{4}\ln \left[\csc 2t+\cot 2t\right]+C \end{gathered}$$

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

Integrate this.

$$\begin{gathered} v_2\left(t\right)=∫\frac{1}{2}{\csc }^{2}2t.\cos 2 tdt \\ =-\frac{1}{4}\csc 2t+C \end{gathered}$$

Thus, a particular solution is:

$$\begin{gathered} y_p=\left(-\frac{1}{4}\ln \left[\csc 2t+\cot 2t\right]+C\right)\cos 2t+\left(-\frac{1}{4}\csc 2t+C\right)\sin 2t \\ {y}_p=\frac{1}{4}\left(\cos 2t\ln \left[\csc 2t+\cot 2t\right]-1\right) \end{gathered}$$

Therefore, the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cos2t+c_{2}sin2t+\frac{1}{4}\left(cos2tln\left[csc2t+cot2t\right]-1\right) \end{gathered}$$