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How can I solve this linear ode: $$y''+\dfrac{4x}{x^2-1 }y'+\dfrac{x^2+1}{x^2-1}y=0 $$ I tried few variables changing but I did not get any result. thanks

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Note that \begin{align} &(x^2-1)y''+4x y'+(x^2+1)y=\\ =&(x^2-1)y''+2(x^2-1)'y'+(x^2-1)''y+(x^2-1)y=\\ =&[(x^2-1)y]''+(x^2-1)y. \end{align} Therefore if we write $\displaystyle y(x)=\frac{f(x)}{x^2-1}$, the equation becomes $$f''+f=0.$$

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(+1) good observation. – Mhenni Benghorbal May 4 '14 at 23:34
That's very nice! – Paul Safier May 4 '14 at 23:56

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