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I have this differential equation

$$x(1-x^2)y''-(1-x^2)^2y'+5x^3y=0,$$

for $-1<x<1$. The hint claims that we should use the change of coordinates $t=-\frac{1}{2}\ln(1-x^2)$, and this will transform the ODE into a constant coefficient one. However, after the sustitution:

$$ \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{x}{1-x^2}\frac{dy}{dt} $$

and

$$ \frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left(\frac{dt}{dx}\right)^2+\frac{dy}{dt}\frac{d^2t}{dx^2}=\frac{x^2}{(1-x^2)^2}\frac{d^2y}{dt^2}+\frac{1+x^2}{(1-x^2)^2}\frac{dy}{dt},$$

and replacing into the original ODE, I get NOT a constant coefficient equation as desired.

I don't know what detail I have missed or if I took a wrong approach. Any suggestions?

Thanks

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Either you miscopied the differential equation or the hint is wrong.

FWIW, Maple solves your differential equation in terms of Whittaker functions.

EDIT: Working backwards, from the constant-coefficient differential equation $a \dfrac{d^2 y}{dt^2} + b \dfrac{dy}{dt} + c y = 0$ the suggested change of variables gives you $$ a x (1 - x^2)^2 y'' - (1-x^2)(a + (a-b) x^2) y' + c x^3 y = 0$$ so perhaps the intended differential equation was $$ x (1-x^2)^2 y'' - (1-x^2)^2 y' + 5 x^2 y = 0 $$

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  • $\begingroup$ I read the problem from the website of a university. Probably the hint is wrong, cause i did not miscopied the problem. I started to get suspicious when Wolfram Mathetica gave an odd answer. Anyway, a further hint is welcome. $\endgroup$
    – user139177
    May 12, 2014 at 6:05

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