91Ó°ÊÓ

Given system can be written in the following form:

$\begin{array}{rcl}& & \left({\mathrm{D}}^2+5\right)\left[\mathrm{x}\right]-2\left[\mathrm{y}\right]=0\\ & & -2\left[\mathrm{x}\right]+\left({\mathrm{D}}^2+2\right)\left[\mathrm{y}\right]=3\sin 2\mathrm{t}\end{array}$

By${\mathrm{L}}_1={\mathrm{D}}^2+5,{\mathrm{L}}_2=-2,{\mathrm{L}}_3=-2,{\mathrm{L}}_4={\mathrm{D}}^2+2,{\mathrm{f}}_1=0,{\mathrm{f}}_2=\sin 2\mathrm{t}$, using the elimination procedure,

One obtains$\left({\mathrm{L}}_1{\mathrm{L}}_4-{\mathrm{L}}_2{\mathrm{L}}_3\right)\left[\mathrm{x}\right]={\mathrm{L}}_4\left[{\mathrm{f}}_1\right]-{\mathrm{L}}_2\left[{\mathrm{f}}_2\right]$

i.e.

$$\begin{gathered} \left(\left({\mathrm{D}}^2+5\right)\left({\mathrm{D}}^2+2\right)-4\right)\left[\mathrm{x}\right]=\left({\mathrm{D}}^2+2\right)\left[0\right]-(-2)\left[3\sin 2\mathrm{t}\right] \\ \left({\mathrm{D}}^4+7{\mathrm{D}}^2+6\right)\left[\mathrm{x}\right]=0+6\sin 2\mathrm{t} \\ \left(\left({\mathrm{D}}^2+1\right)\left({\mathrm{D}}^2+6\right)\right)\left[\mathrm{x}\right]=6\sin 2\mathrm{t}⋯\left(1\right) \end{gathered}$$

The auxiliary equation is $\left({\mathrm{r}}^2+1\right)\left({\mathrm{r}}^2+6\right)=0$and its roots ${\mathrm{r}}_{1,2}=Â\pm \mathrm{i},{\mathrm{r}}_{3,4}=Â\pm \mathrm{i}\sqrt{6}$.

Therefore, the solution to the corresponding homogeneous equation is

${\mathrm{x}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}+{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}$.

Let's check if $\mathrm{x}\left(\mathrm{t}\right)=-\sin 2\mathrm{t}$ is a particular solution for (1).

First, one will find derivatives

$\begin{array}{rcl}& & \mathrm{x}\text{'}\left(\mathrm{t}\right)=-2\cos 2\mathrm{t}\\ & & \mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)=4\sin 2\mathrm{t}\\ & & \mathrm{x}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=8\cos 2\mathrm{t}\\ & & {\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)=-16\sin 2\mathrm{t}\end{array}$

Since

$$\begin{gathered} {\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)+7\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+6\mathrm{x}\left(\mathrm{t}\right)=-16\sin 2\mathrm{t}+7×4\sin 2\mathrm{t}+6(-\sin 2\mathrm{t}) \\ =6\sin 2\mathrm{t} \end{gathered}$$

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\sin 2\mathrm{t}$ is a particular solution to 1 and the general solution is

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ ={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}+{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}-\sin 2\mathrm{t} \end{gathered}$$.

From the first equation of the system,$\mathrm{x}\text{'}\text{'}+5\mathrm{x}-2\mathrm{y}=0$ one can find a general solution for $\mathrm{y}\left(\mathrm{t}\right)$ but the first one needs to find $\mathrm{x}\text{'}\text{'}$

$$\begin{gathered} \mathrm{x}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1\mathrm{sint}+{\mathrm{c}}_2\mathrm{cost}-\sqrt{6}{\mathrm{c}}_3\sin \sqrt{6}\mathrm{t}+\sqrt{6}{\mathrm{c}}_4\cos \sqrt{6}\mathrm{t}-2\cos 2\mathrm{t} \\ \mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1\mathrm{cost}-{\mathrm{c}}_2\mathrm{sint}-6{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}-6{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}+4\sin 2\mathrm{t} \end{gathered}$$

Then simplify,

$$\begin{gathered} 2\mathrm{y}=\mathrm{x}\text{'}\text{'}+5\mathrm{x} \\ =-{\mathrm{c}}_1\mathrm{cost}-{\mathrm{c}}_2\mathrm{sint}-6{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}-6{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}+4\sin 2\mathrm{t}+5{\mathrm{c}}_1\mathrm{cost}+5{\mathrm{c}}_2\mathrm{sint} \\ +5{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}+5{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}-5\sin 2\mathrm{t} \\ =4{\mathrm{c}}_1\mathrm{cost}+4{\mathrm{c}}_2\mathrm{sint}-{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}-{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}-\sin 2\mathrm{t} \end{gathered}$$

Therefore,$\mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{c}}_1\mathrm{cost}+2{\mathrm{c}}_2\mathrm{sint}-\frac{{\mathrm{c}}_3}{2}\cos \sqrt{6}\mathrm{t}-\frac{{\mathrm{c}}_4}{2}\sin \sqrt{6}\mathrm{t}-\frac{1}{2}\sin 2\mathrm{t}$

It remains to find constants from initial conditions.

Let's compute $\mathrm{y}\text{'}$

$\mathrm{y}\left(\mathrm{t}\right)=-2{\mathrm{c}}_1\mathrm{sint}+2{\mathrm{c}}_2\mathrm{cost}+\frac{{\mathrm{c}}_3\sqrt{6}}{2}\sin \sqrt{6}\mathrm{t}-\frac{{\mathrm{c}}_4\sqrt{6}}{2}\cos \sqrt{6}\mathrm{t}-\cos 2\mathrm{t}$

$$\begin{gathered} 0=\mathrm{x}\left(0\right)={\mathrm{c}}_1+{\mathrm{c}}_3 \\ 0=\mathrm{x}\text{'}\left(0\right)={\mathrm{c}}_2+\sqrt{6}{\mathrm{c}}_4-2 \\ 1=\mathrm{y}\left(0\right)=2{\mathrm{c}}_1-\frac{{\mathrm{c}}_3}{2} \\ 0=\mathrm{y}\text{'}\left(0\right)=2{\mathrm{c}}_2-\frac{\sqrt{6}}{2}{\mathrm{c}}_4-1 \end{gathered}$$

By solving this system, one obtains ${\mathrm{c}}_1=\frac{2}{5},{\mathrm{c}}_2=\frac{4}{5},{\mathrm{c}}_3=-\frac{2}{5},{\mathrm{c}}_4=\frac{\sqrt{6}}{5}$

Finally, $\mathrm{x}\left(\mathrm{t}\right)=\frac{2}{5}\mathrm{cost}+\frac{4}{5}\mathrm{sint}-\frac{2}{5}\cos \sqrt{6}\mathrm{t}+\frac{\sqrt{6}}{5}\sin \sqrt{6}\mathrm{t}-\sin 2\mathrm{t}$

And $\mathrm{y}\left(\mathrm{t}\right)=\frac{4}{5}\mathrm{cost}+\frac{8}{5}\mathrm{sint}+\frac{1}{5}\cos \sqrt{6}\mathrm{t}-\frac{\sqrt{6}}{10}\sin \sqrt{6}\mathrm{t}-\frac{1}{2}\sin 2\mathrm{t}$