Question

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}+y\text{'}\text{'}-5y\text{'}+3y=e^{-x}+sinx$

Short answer

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

Step-by-step solution

Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

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

The solutions of the auxiliary equation are

$r=1,r=1andr=3$

Find particular solution

Let the particular solution be

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

Then

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

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

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

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

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