91Ó°ÊÓ

Theauxiliary equationof corresponding homogeneous equation

$r^3+r^2-2=(r-1)\left(r^2+2r+2\right)=0$

The solutions of the auxiliary equation are

$r=-1+i,r=-1-i,r=1$

Therefore a general solution to the homogeneous equation is

$y_h\left(x\right)=c_1{e}^x+c_2{e}^{-x}\cos x+c_3{e}^{-x}\sin x$

Let the particular solution be

$y_p\left(x\right)=ax{e}^x+b{x}^2{e}^x+c$

Then

$$\begin{gathered} y_p^{\text{'}}\left(x\right)=a{e}^x+(a+2b)x{e}^x+b{x}^2{e}^x \\ {y}_p^{\text{'}\text{'}}\left(x\right)=(2a+2b)e^x+(a+4b)x{e}^x+b{x}^2{e}^x \\ {y}_p^{\text{'}\text{'}\text{'}}\left(x\right)=(3a+6b)e^x+(a+6b)x{e}^x+b{x}^2{e}^x \end{gathered}$$

Then

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}\left(x\right)+y_p^{\text{'}\text{'}}\left(x\right)-2y_p\left(x\right) \\ =(3a+6b)e^x+(a+6b)x{e}^x+b{x}^2{e}^x+(2a+2b)e^x+(a+4b)x{e}^x+b{x}^2{e}^x-2ax{e}^x-2b{x}^2{e}^x-2c \\ =(5a+8b)e^x+10bx{e}^x-2c \end{gathered}$$

If$(5a+8b)e^x+10bx{e}^x-2c=x{e}^x+1$

Then$5a+8b=0,10b=1and-2c=1$

Thenrole="math" localid="1663941222424" $a=-\frac{4}{25},b=\frac{1}{10}andc=-\frac{1}{2}$

Hence$y_p\left(x\right)=-\frac{4}{25}x{e}^x+\frac{1}{10}{x}^2{e}^x-\frac{1}{2}$

Then$y\left(x\right)=-\frac{4}{25}x{e}^x+\frac{1}{10}{x}^2{e}^x-\frac{1}{2}+c_1{e}^x+c_2{e}^{-x}\cos x+c_3{e}^{-x}\sin x$

Is the general solution of$y\text{'}\text{'}\text{'}+y\text{'}\text{'}-2y=x{e}^x+1$