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$(1-2x^2)y''+2y'+4y=0$

I have tried using $z(x) = y'/y$ which leads to $z'+z^2 = y''/y$

That led me to:

$z'(1-2x^2) + z^2(1-2x^2) + 2z + 4$

and I have no idea what to from this point, I don't even know whether I should have used the substitution $z(x)=y'/y$.

Can anybody help me out here please?

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    $\begingroup$ A solution is $y(x) = b\left(x-x^2\right)$ where b is a constant. Also, this is certainly not true for all cases, but when i approach equations like these i tend to start with a "trail" polynomial and depart from there. $\endgroup$ – Chinny84 May 29 '14 at 11:06
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    $\begingroup$ It was a good idea to guess a quadratic, now I believe you should be able to find another solution via variation of parameters. $\endgroup$ – Bennett Gardiner May 29 '14 at 11:10
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Hint

Using Frobenius method, I found that $$y(x)=a_0+a_1 x-(2 a_0+a_1) x^2+a_0 \Big(\frac{4}{3} x^3-\frac{2}{3} x^4+\frac{4}{5} x^5-\frac{32}{45} x^6+\frac{8}{9} x^7+...\Big)$$ If you impose $a_0=0$, you find the result given by Chinny84.

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