y=C1e^x+C2xe^x
y'=C1e^x+C2xe^x+C2e^x
y"=C1e^x+C2xe^x+2C2e^x
Require equation y"+ay'+by=0; C1e^x+C2xe^x+2C2e^x+aC1e^x+aC2xe^x+aC2e^x+bC1e^x+bC2xe^x=0.
Therefore, since e^x cannot be zero, we can divide through by e^x:
C1+C2x+2C2+aC1+aC2x+aC2+bC1+bC2x=0=C2(1+a+b)x+C1(1+a+b)+C2(2+a).
We need to make the coefficients of C1 and C2 zero by choosing suitable a and b.
From this 2+a=0, so a=-2, 1+a+b=0=-1+b, b=1.
y"-2y'+y=0 is a second order DE that doesn't contain C1 or C2 but satisfies y.