Okay, the "non-homogeneous"part is \(\displaystyle (x+ 1)e^x\). That is precisely the kind of function we would expect as a solution to a homogeneous differential equation with 1 as a double root to the characteristic equation. So we can expect to get a solution to the entire equation by "undetermined coefficients", using a function of the form \(\displaystyle y= (Ax+ B)e^x\).
So what are y' and y'' for that y? Put those into the differential equation. You will have an equation of the form (Px+ Q)e^x on the right side (where P and Q are functions of A and B) and (x+ 1)e^x
Set P= 1 and Q= 1 and solve those two equations for A and B.