[[VG:상미분방정식,ordinary_differential_equation,ODE]]

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From Kreyszig 10e p18

 $y'=f\left(\frac{y}{x}\right)$ 
꼴은,
 $y=ux$ 그리고 그 곱의 미분인
 $y'=u'x+u$
로 치환하면
 $u'x+u=f(u)$ or
 $u'x=y-u=f(u)-u$
만약 $f(u)-u\ne0$ 이면, 분리하여
 $\frac{du}{f(u)-u}=\frac{dx}{x}$

== ex 1 ==
Q: Solve
 $2xyy'=y^2-x^2$
Sol.
 $y'=\frac{y^2-x^2}{2xy}=\frac{y}{2x}-\frac{x}{2y}$
치환 $y=ux,\quad y'=u'x+u$
 $u'x+u=\frac{u}{2}-\frac1{2u}$
 $u'x=-\frac{u}2-\frac1{2u}=\frac{-u^2-1}{2u}$
 $\frac{xdu}{dx}=\frac{-u^2-1}{2u}$
 $\frac{2udu}{1+u^2}=-\frac{dx}{x}$
적분하면
 $\ln(1+u^2)=-\ln|x|+c_1=\ln\left|\frac1{x}\right|+c_1$
take exponents:
 $1+u^2=\frac{c}{x}$
 $1+\left(\frac{y}{x}\right)^2=\frac{c}{x}$
$x^2$ 을 곱하면
 $x^2+y^2=cx$
 $\left(x-\frac{c}2\right)^2+y^2=\frac{c^2}{4}$

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Up: [[미분방정식,differential_equation]]

https://everything2.com/title/Ordinary+differential+equation
https://freshrimpsushi.github.io/categories/상미분방정식/