The ODE to solve via variation of parameters is ##(1-x)y''+xy'-y=(1-x)^2##.(adsbygoogle = window.adsbygoogle || []).push({});

Knowing that ##e^x## and ##x## are solutions to the homogeneous ODE.

Now if I call ##y_1=x## and ##y_2=e^x##, the Wronskian is ##W(y_1,y_2)=e^{x}(x-1)##.

According to http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx, the particular solution of the non homogeneous ODE should be of the form ##-y_1 \int \frac{y_2 (1-x)^2}{W(y_1,y_2)} dx+y_2 \int \frac{y_1(1-x)^2}{W(y_1,y_2)}##. This gave me ##y_p(x)=-\frac{x^3}{2}-x-1## I've even checked out with wolfram alpha the evaluation of the integrals, that is the result. However the answer is ##y_p=x^2+1##.

I'm clueless on what's going on.

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# Variation of parameters applied to an ODE

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