Solve $y''-3y'+2y=x^2$ Solve $$y''-3y'+2y=x^2$$ My approach: Homogen solution: $$y = Ae^x +Be^{2x}$$ Particular solution: $$ y_p = x(Ax^2+Bx+C) = Ax^3+Bx^2+Cx $$ $$ y_p' = 3Ax^2 + 2Bx + C$$ $$y_p'' = 6Ax + 2B$$ Put his into the initial equartion to get A, B and C gives me: $A=0, B=1/2, C=3/2$ This leads me to the answer: $$y = Ae^x +Be^{2x} + x^2/2 + 3x/2$$ However the correct answer is $$y = Ae^x +Be^{2x} + x^2/2 + 3x/2+ 7/4$$ Where's my miss? Where comes the last term from?

It looks like your method for finding the particular solution is wrong.

You want $y_p = ax^2 + bx + c$.

I'm not sure why you introduce an extra factor of $x$.

Here's a list of trial functions for your particular solution.