find a general solution to the given equation.

$y\text{'}\text{'}\text{'}+4y\text{'}\text{'}+y\text{'}-26y=e^{-3x}sin2x+x$

The auxiliary equation ofthe corresponding homogeneous equation:

$r^2(r-2){\left(r^2+6r+13\right)}^2=0$

The solutions of the auxiliary equation are:

$r_1=r_2=0,r_3=2,r_4=r_5=-3+2i,r_6=r_7=-3-2i$

Therefore a general solution to the homogeneous equation is:

$y_h\left(x\right)=c_3{e}^{2x}{c}_4{e}^{-3x}\cos 2x+c_6{e}^{-3x}\sin 2x$

Let the particular solution be;

$y_p\left(x\right)=c_1+c_{2}x+c_{5}x{e}^{-3x}\cos 2x+c_{7}x{e}^{-3x}\sin 2x$

Then,

$y_p^{\text{'}}\left(x\right)=c_2+c_5{e}^{-3x}\cos 2x+c_7{e}^{-3x}\sin 2x+\left(-3c_5+2c_7\right)x{e}^{-3x}\cos 2x+\left(-2c_5-3c_7\right)x{e}^{-3x}\sin 2x$

$$\begin{gathered} y_p^{\text{'}\text{'}}\left(x\right)=\left(-6c_5+4c_7\right)e^{-3x}\cos 2x+\left(-4c_5-6c_7\right)e^{-3x}\sin 2x+\left(5c_5-12c_7\right)x{e}^{-3x}\cos 2x \\ +\left(12c_5+5c_7\right)x{e}^{-3x}\sin 2x \end{gathered}$$

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}\left(x\right)=\left(15c_5-36c_7\right)e^{-3x}\cos 2x+\left(36c_5+15c_7\right)e^{-3x}\sin 2x+\left(9c_5+46c_7\right)x{e}^{-3x}\cos 2x \\ +\left(-46c_5+9c_7\right)x{e}^{-3x}\sin 2x \end{gathered}$$

Then,

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}+4y_p^{\text{'}\text{'}}+y_p^{\text{'}}-36y_p=\left(-8c_5-20c_7\right)e^{-3x}\cos 2x+\left(20c_5-8c_7\right)e^{-3x}\sin 2x \\ +c_2-26c_1-26c_{2}x \\ =e^{-3x}\sin 2x+x \end{gathered}$$

Then we have:

$$\begin{gathered} -8c_5-20c_7=0c_2-26c_1=0 \\ 20c_5+8c_7=1-26c_2=1 \end{gathered}$$

Therefore,

$c_1=-\frac{1}{676},c_2=-\frac{1}{26},c_5=\frac{1}{29},c_7=-\frac{1}{58}$

Hence,

$y_p\left(x\right)=-\frac{1}{676}-\frac{1}{26}x+\frac{1}{29}x{e}^{-3x}\cos 2x-\frac{1}{58}x{e}^{-3x}\sin 2x.$

Thus, the required general solution is:

$y\left(x\right)=c_3{e}^{2x}+c_4{e}^{-3x}\cos 2x+c_6{e}^{-3x}\sin 2x-\frac{1}{676}-\frac{x}{26}+\frac{1}{29}x{e}^{-3x}\cos 2x-\frac{1}{58}x{e}^{-3x}\sin 2x$