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I have this second order ODE:

$$(x^2-1).y''-2xy'+2y=(x^2-1)^2$$

To solve it, I'm thinking of method of variation of parameters.

But, I'm not able to transform it into an equation with constant co-efficients. I'm stuck now.

Help is requested.

Thanks in advance.

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  • $\begingroup$ Solve the homogeneous equation first. $\endgroup$ – science Mar 16 '15 at 16:28
  • $\begingroup$ @science, can you help with that? I'm completely clueless. $\endgroup$ – dajoker Mar 16 '15 at 16:30
  • $\begingroup$ What techniques have you been taught to solve a second order homogeneous differential equation with non constant coefficients? $\endgroup$ – science Mar 16 '15 at 16:37
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you can see that $y = x$ is a solution to the homogeneous equation $$(x^2 -1)y'' -2xy' + 2y = 0.$$ so we will make a change of variable $$y = xu, y' = xu' + u, y'' = xu'' + 2u' $$ and $u$ satisfies $$(x^2 - 1)(xu'' +2u')-2x(xu' + u)+ 2xu=(x^2 -1)^2 \to x(x^2 - 1)u''+2u'(x^2 - 1-x^2)=(x^2 - 1)^2$$ therefore we have $$x(x^2 - 1)v'-2v=(x^2 -1)^2, v = u'.$$ can you take it from here.

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  • $\begingroup$ Got it. Thanks! $\endgroup$ – dajoker Mar 16 '15 at 17:25
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Just assume the solution to have the form $y=x^{\alpha}$ and substitute back in the homogeneous ode to get $y_1=x$ and then try to find $y_2$.

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  • $\begingroup$ Thanks. WIll try that. A doubt : What is the motivation behind choosing $y=x^a$ $\endgroup$ – dajoker Mar 16 '15 at 16:43
  • $\begingroup$ are you sure about your hint? this is not an euler equation. $\endgroup$ – abel Mar 16 '15 at 16:44
  • $\begingroup$ Think he is looking at solving $(x^2-1).y''-2xy'+2y=0$ first . $\endgroup$ – randomgirl Mar 16 '15 at 16:50

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