find a general solution to the given equation. $y\text{'}\text{'}\text{'}+y\text{'}\text{'}-5y\text{'}+3y=e^{-x}+sinx.$

Theauxiliary equationof corresponding homogeneous equation

$r^3+r^2-5r+3=(r-1{)}^2(r+3)=0$

The solutions of the auxiliary equation are

$r=1,r=1,r=-3$

Therefore a general solution to the homogeneous equation is

$y_h\left(x\right)=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-3x}$

Let the particular solution be

$y_p\left(x\right)=a{e}^{-x}+b\cos x+c\sin x$

Then

role="math" localid="1663939601921" $$\begin{gathered} y_p^{\text{'}}\left(x\right)=-a{e}^{-x}-b\sin x+c\cos x \\ {y}_p^{\text{'}}\left(x\right)=a{e}^{-x}-b\cos x-c\sin x \\ {y}_p^{\text{'}}\left(x\right)=-a{e}^{-x}+b\sin x-c\cos x \end{gathered}$$

Then

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}\left(x\right)+y_p^{\text{'}\text{'}}\left(x\right)-5y_p^{\text{'}}\left(x\right)+3y_p\left(x\right) \\ =-a{e}^{-x}+b\sin x-c\cos x+a{e}^{-x}-b\cos x-c\sin x+5a{e}^{-x}+5b\sin x-5c\cos x+3a{e}^{-x}+3b\cos x+3c\sin x \\ =8a{e}^{-x}+(2b-6c)\cos x+(6b+2c)\sin x \end{gathered}$$

If$8a{e}^{-x}+(2b-6c)\cos x+(6b+2c)\sin x=e^{-x}+\sin x$

Then$8a=1,2b-6c=0and6b+2c=1$

Then

$a=-\frac{1}{8},b=\frac{{\displaystyle 3}}{{\displaystyle 20}}andc=\frac{{\displaystyle 1}}{{\displaystyle 20}}$

Hence

$y_p\left(x\right)=\frac{1}{8}{e}^{-x}+\frac{3}{20}\cos x+\frac{1}{20}\sin x$

Then$y\left(x\right)=\frac{1}{8}{e}^{-x}+\frac{3}{20}\cos x+\frac{1}{20}\sin x+c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-3x}$

Is the general solution of$y\text{'}\text{'}\text{'}+y\text{'}\text{'}-5y\text{'}+3y=e^{-x}+\sin x$