jisu61hbke
2022-03-23
Answered

Solute the equation:

$(1-{x}^{2})y{}^{\u2033}=x{y}^{\prime}$

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Avery Maxwell

Answered 2022-03-24
Author has **13** answers

The given equation is second order linear ordinary differential equation. This solution for the interval, |x|<1, there is another solution for |x|>1.

$(1-{x}^{2})y{}^{\u2033}=x{y}^{\prime}$

$\Rightarrow (1-{x}^{2})\frac{d}{dx}\left({y}^{\prime}\right)=x{y}^{\prime}$

$\Rightarrow \frac{d\left({y}^{\prime}\right)}{dn}=\frac{x{y}^{\prime}}{1-{x}^{2}}$

$\Rightarrow \frac{d{\left(y\right)}^{\prime}}{{y}^{\prime}}=\frac{x}{1-{x}^{2}}dx$

Integrating$\int \frac{{dy}^{\prime}}{{y}^{\prime}}=\int \frac{x}{1-{x}^{2}}dx$

${\mathrm{ln}y}^{\prime}=-\frac{1}{2}\mathrm{ln}(1-{x}^{2})+\mathrm{ln}C$ [$\mathrm{ln}c$ constant]

$\Rightarrow {y}^{\prime}=\frac{c\cdot 1}{\sqrt{1-{x}^{2}}}$

$\Rightarrow dy=\frac{c\left(dx\right)}{\sqrt{1-{x}^{2}}}$

$\Rightarrow \frac{1}{c}\int dy=\int \frac{dx}{\sqrt{1-{x}^{2}}}$

$\Rightarrow \frac{1}{c\cdot y}=\mathrm{sin}x+d$ [d is constant]

$\mathrm{sin}x=\frac{y}{c-d}$

$\Rightarrow x=\mathrm{sin}\left(\frac{y}{c-d}\right)$

Solution:$x=\mathrm{sin}(Ay+B),\text{}A=\frac{1}{c},\text{}B=-d$

$\left|n\right|>1$

$y}^{\prime}=\frac{c}{\sqrt{{n}^{2}-1}$

$dy=\frac{c\left(dx\right)}{\sqrt{{n}^{2}-1}}$

$\int dy=c\int \frac{dx}{\sqrt{{n}^{2}-1}}$

$y=c\cdot \mathrm{ln}|n+\sqrt{{n}^{2}-1}|+d$ c,d constant

Integrating

Solution:

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