$y^{\text{'}\text{'}}+2y^{\text{'}}-3y=1+x{e}^x$

$y_p=Ax{e}^x+B{e}^x+C$

$y_p^{\text{'}}=A{e}^x+Ax{e}^x+B{e}^x+0$

$y_p^{\text{'}\text{'}}=A{e}^x+A{e}^x+Ax{e}^x+B{e}^x$

$A{e}^x+A{e}^x+Ax{e}^x+B{e}^x+2[A{e}^x+Ax{e}^x+B{e}^x]-3[Ax{e}^x+B{e}^x+C]=1+x{e}^x$

$2A{e}^x+Ax{e}^x+B{e}^x+2A{e}^x+2Ax{e}^x+2B{e}^x-3Ax{e}^x-3B{e}^x-3C=1+x{e}^x$

$4A{e}^x-3C=1+x{e}^x,C=\frac{-1}{3}$

$4A=x,-3C=1,A=\frac{x}{4}$

$y_p(x)=-\frac{1}{3}+\frac{1}{16}(2x^2-x)e^x$ Complete the solution steps At the end is the final answer