[[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 $y=ux,\quad y'=u'x+u$
 $u'x+u=\frac{u}{2}-\frac1{2u}$
 $u'x=-\frac{u}2-\frac1{2u}=\frac{-u^2-1}{2u}$