Answer
$y=y_c+y_p=e^{x}(c_{1}\cos x+c_{2} \sin x)+2+x+e^x$
or,
$y=y_c+y_p=e^{x}(1+c_{1}\cos x+c_{2} \sin x)+2+x$
Work Step by Step
The auxiliary equation.
$r^{2}-2r+2=0 \implies r=1\pm i$
Now, $y_c=e^{x}(c_{1}\cos x+c_{2} \sin x)$
$y_{p}=A+Bx+Ce^x$
$y_{p}'=B+Ce^x$ and $y_{p}''=Ce^x$
$(A-2B)+Bx+Ce^x=x+e^x$
$C=1$
Compare the coefficients of $e^x$ and $B=1$
Here, we have $A-2B=0 \implies A=2$
$y=y_c+y_p=e^{x}(c_{1}\cos x+c_{2} \sin x)+2+x+e^x$
or,
$y=y_c+y_p=e^{x}(1+c_{1}\cos x+c_{2} \sin x)+2+x$
