Method of Variation of Parameters

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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]

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.

g_edgar

So first you need to do something else. Perhaps your textbook has this also: in a second-order homogeneous linear DE, given one solution to find the other one. As you note, you need to have both solutions of the homogeneous DE before you can apply the method of variation of parameters to get a solution of the inhomogeneous DE.

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Should it be writing down characteristic equation ?? I see no way out with the [tex]x[/tex] and [tex]x-1[/tex] that are coefficients to [tex]y''[/tex] and [tex]y'[/tex]. As I write it like this:

[tex]xr^{2}+(x-1)r-1=0[/tex]
umm, 2 unknowns, 1 equation ? Maybe with the known solution helps here? How?

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Allright, I think i found a clue: Reduction of Order method will let us find the other solution ? Please someone confirm it whether wrong or right. Thanks

g_edgar

right