From Kreyszig 10e p18
$\displaystyle y'=f\left(\frac{y}{x}\right)$
꼴은,$\displaystyle y=ux$ 그리고 그 곱의 미분인
$\displaystyle y'=u'x+u$
로 치환하면$\displaystyle y'=u'x+u$
$\displaystyle u'x+u=f(u)$ or
$\displaystyle u'x=y-u=f(u)-u$
만약 $\displaystyle f(u)-u\ne0$ 이면, 분리하여$\displaystyle u'x=y-u=f(u)-u$
$\displaystyle \frac{du}{f(u)-u}=\frac{dx}{x}$
1. ex 1 ¶
Q: Solve
$\displaystyle 2xyy'=y^2-x^2$
Sol.$\displaystyle y'=\frac{y^2-x^2}{2xy}=\frac{y}{2x}-\frac{x}{2y}$
치환 $\displaystyle y=ux,\quad y'=u'x+u$$\displaystyle u'x+u=\frac{u}{2}-\frac1{2u}$
$\displaystyle u'x=-\frac{u}2-\frac1{2u}=\frac{-u^2-1}{2u}$
$\displaystyle \frac{xdu}{dx}=\frac{-u^2-1}{2u}$
$\displaystyle \frac{2udu}{1+u^2}=-\frac{dx}{x}$
적분하면$\displaystyle u'x=-\frac{u}2-\frac1{2u}=\frac{-u^2-1}{2u}$
$\displaystyle \frac{xdu}{dx}=\frac{-u^2-1}{2u}$
$\displaystyle \frac{2udu}{1+u^2}=-\frac{dx}{x}$
$\displaystyle \ln(1+u^2)=-\ln|x|+c_1=\ln\left|\frac1{x}\right|+c_1$
take exponents:$\displaystyle 1+u^2=\frac{c}{x}$
$\displaystyle 1+\left(\frac{y}{x}\right)^2=\frac{c}{x}$
$\displaystyle x^2$ 을 곱하면$\displaystyle 1+\left(\frac{y}{x}\right)^2=\frac{c}{x}$
$\displaystyle x^2+y^2=cx$
$\displaystyle \left(x-\frac{c}2\right)^2+y^2=\frac{c^2}{4}$
$\displaystyle \left(x-\frac{c}2\right)^2+y^2=\frac{c^2}{4}$