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y''-4y'+y=x^2-2x+3
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This is fairly longish. First you have to solve this: y''-4y'+y=0 Do you know how to do that?
the auxillary function for the above given equation is D^2-4D+1 =0 solving we have for D we have D =( 4+-sqrt(16-4))/2 = 2+- sqrt(3) hence the CF of the equation is given by y = e^2x (Ae^(sqrt(3)x) +Be^(-sqrt(3)x) )...
now we need to find the PI for x^2-2x+3 which is given by 1/(1+ D^2-4D ) * ( x^2-2x+3) = (1+4D-D^2 +16D^2) *( x^2-2x+3) =(1+4D+15D^2)*( x^2-2x+3) =(( x^2-2x+3) +4(2x-2) +15*2)
hence complete solution is y=CF +PI
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