# 変数分離形微分方程式に関する問題

## ■問題

${y}^{\prime }=\frac{x{y}^{3}+x{y}^{2}}{{x}^{3}y-xy}$

## ■答

$\frac{{y}^{2}\left(x+1\right)}{{\left(y+1\right)}^{2}\left(x-1\right)}=A$

（ただし$A$は任意定数）

## ■解き方

${y}^{\prime }=\frac{x{y}^{3}+x{y}^{2}}{{x}^{3}y-xy}$

${y}^{\prime }=\frac{dy}{dx}$ より

$\frac{dy}{dx}$$=\frac{x{y}^{2}\left(y+1\right)}{xy\left({x}^{2}-1\right)}$

$=\frac{y\left(y+1\right)}{{x}^{2}-1}$

$=\frac{y\left(y+1\right)}{\left(x-1\right)\left(x+1\right)}$

$\because {x}^{2}-1=\left(x-1\right)\left(x+1\right)$

$y\ne 0$$y\ne -1$ の場合

$\frac{1}{y\left(y+1\right)}dy=\frac{1}{\left(x-1\right)\left(x+1\right)}dx$

⇒詳しくはこちら

$\left(\frac{1}{y}-\frac{1}{y+1}\right)dy$$=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx$

ここで両辺を積分すると

$\int \left(\frac{1}{y}-\frac{1}{y+1}\right)dy$$=\frac{1}{2}\int \left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx$$+C$

(ただし $C$は任意定数)

$\int \frac{1}{y}dy-\int \frac{1}{y+1}dy$$=\frac{1}{2}\int \frac{1}{x-1}dx$$-\frac{1}{2}\int \frac{1}{x+1}dx$$+C$

$\mathrm{log}|y|-\mathrm{log}|y+1|$$=\frac{1}{2}\mathrm{log}|x-1|$$-\frac{1}{2}\mathrm{log}|x+1|$$+C$

$2\mathrm{log}|y|-2\mathrm{log}|y+1|$$=\mathrm{log}|x-1|$$-\mathrm{log}|x+1|$$+2C$

$\mathrm{log}{y}^{2}-\mathrm{log}{\left(y+1\right)}^{2}$$=\mathrm{log}|x-1|$$-\mathrm{log}|x+1|$$+2C$

$\mathrm{log}\left\{\frac{{y}^{2}}{{\left(y+1\right)}^{2}}\right\}=\mathrm{log}\frac{|x-1|}{|x+1|}+2C$

この式を整理して

$\mathrm{log}\left\{\frac{{y}^{2}}{{\left(y+1\right)}^{2}}\right\}-\mathrm{log}\frac{|x-1|}{|x+1|}=2C$

$\mathrm{log}\left\{\frac{\frac{{y}^{2}}{{\left(y+1\right)}^{2}}}{\frac{|x-1|}{|x+1|}}\right\}=2C$

$\mathrm{log}\left\{\frac{{y}^{2}|x+1|}{{\left(y+1\right)}^{2}|x-1|}\right\}=2C$

この式の右辺は$\mathrm{log}{e}^{2C}$ と変形できるので

⇒詳しくはこちら

$\mathrm{log}\left\{\frac{{y}^{2}|x+1|}{{\left(y+1\right)}^{2}|x-1|}\right\}=\mathrm{log}{e}^{2C}$

したがって

$\frac{{y}^{2}|x+1|}{{\left(y+1\right)}^{2}|x-1|}={e}^{2C}$

$\frac{{y}^{2}\left(x+1\right)}{{\left(y+1\right)}^{2}\left(x-1\right)}=±{e}^{2C}$

$±{e}^{2C}=A$ ($A\ne 0$ )とおくと

$\frac{{y}^{2}\left(x+1\right)}{{\left(y+1\right)}^{2}\left(x-1\right)}=A$

$y=0$$y=-1$ の場合

$y=0$$y=-1$ は微分方程式を満たす．このとき，$A=0$ となる．

$\frac{{y}^{2}\left(x+1\right)}{{\left(y+1\right)}^{2}\left(x-1\right)}=A$

（ただし$A$は任意定数）

となる．

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