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**JML2618** For our practice final one of the questions is

Find the general solution to the following differential equation: y'' - 4y' + 4y =x^{2 }- 3x + 2

I understand that one would add the particular solution and general solution. For the particular solution I got 1/4x^{2}-1/4x+1/8 and for the general solution I got Ae^{2x}+Be^{2x}. However he gave us the solution and it was Ae^{-2x}+Bxe^{-2x}+1/4x^{2}-1/4x+1/8. Does anyone see where I went wrong? I can show more work if thats helpful.