Allright, I understand that we need two solutions to be able to apply the method like [tex]y_{1}[/tex] and [tex]y_{2}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Problem gives 1 of them or lets you find only that 1 solution. But I can't apply the method since I don't have the other solution. The method I know is:

[tex]u_{1}'(x)y_{1}(x)+u_{2}'(x)y_{2}=0[/tex]

[tex]u_{1}'(x)y_{1}'(x)+u_{2}'(x)y_{2}'=g(x)[/tex]

solve for the [tex]u_{1}'(x)[/tex] and [tex]u_{2}'(x)[/tex] and do the integrals, solve the problem.

This is the problem i'm tackling with:

Find a value of [tex]p[/tex] such that [tex]e^{px}[/tex] is a solution of

[tex]xy''+(x-1)y'-y=2x^{2}e^{-x}[/tex]

well i can find out that p=-1 and its correct i'm pretty sure. How can I handle the rest of it as i mentioned above?

Thanks.

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# Method of Variation of Parameters

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