# Solute the equation: $$\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}={x}{y}'$$

Solute the equation:
$\left(1-{x}^{2}\right)y{}^{″}=x{y}^{\prime }$
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Avery Maxwell
The given equation is second order linear ordinary differential equation. This solution for the interval, |x|<1, there is another solution for |x|>1.
$\left(1-{x}^{2}\right)y{}^{″}=x{y}^{\prime }$
$⇒\left(1-{x}^{2}\right)\frac{d}{dx}\left({y}^{\prime }\right)=x{y}^{\prime }$
$⇒\frac{d\left({y}^{\prime }\right)}{dn}=\frac{x{y}^{\prime }}{1-{x}^{2}}$
$⇒\frac{d{\left(y\right)}^{\prime }}{{y}^{\prime }}=\frac{x}{1-{x}^{2}}dx$
Integrating $\int \frac{{dy}^{\prime }}{{y}^{\prime }}=\int \frac{x}{1-{x}^{2}}dx$
${\mathrm{ln}y}^{\prime }=-\frac{1}{2}\mathrm{ln}\left(1-{x}^{2}\right)+\mathrm{ln}C$ [$\mathrm{ln}c$ constant]
$⇒{y}^{\prime }=\frac{c\cdot 1}{\sqrt{1-{x}^{2}}}$
$⇒dy=\frac{c\left(dx\right)}{\sqrt{1-{x}^{2}}}$
$⇒\frac{1}{c}\int dy=\int \frac{dx}{\sqrt{1-{x}^{2}}}$
$⇒\frac{1}{c\cdot y}=\mathrm{sin}x+d$ [d is constant]
$\mathrm{sin}x=\frac{y}{c-d}$
$⇒x=\mathrm{sin}\left(\frac{y}{c-d}\right)$
Solution:
$|n|>1$
${y}^{\prime }=\frac{c}{\sqrt{{n}^{2}-1}}$
$dy=\frac{c\left(dx\right)}{\sqrt{{n}^{2}-1}}$
$\int dy=c\int \frac{dx}{\sqrt{{n}^{2}-1}}$
$y=c\cdot \mathrm{ln}|n+\sqrt{{n}^{2}-1}|+d$ c,d constant