상미분방정식,ordinary_differential

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

$\displaystyle u'x+u=f(u)$ or
$\displaystyle u'x=y-u=f(u)-u$

만약 $\displaystyle f(u)-u\ne0$ 이면, 분리하여

$\displaystyle \frac{du}{f(u)-u}=\frac{dx}{x}$

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 $\displaystyle y=ux,\quad y'=u'x+u$

$\displaystyle u'x+u=\frac{u}{2}-\frac1{2u}$
$\displaystyle u'x=-\frac{u}2-\frac1{2u}=\frac{-u^2-1}{2u}$