質問<1175>
「「微分」」
日付 2003/4/19
質問者


(1) a(xdy + 2ydx)=xydx
(2) (xの2乗*y)y'+ x*yの2乗 - y=0  {put xy=u}

すいませんが、至急、解き方を教えてください。

お便り
日付 2003/4/20
回答者 phaos


(1) 所謂変数分離型
ax dy = (xy - 2y)dx
ax dy = y(x - 2) dx
a∫d\(\frac{y}{y}\) = ∫(x - 2)d\(\frac{x}{x}\)
a log y = ∫(1 - \(\frac{2}{x}\)) dx = x - 2 log x + log C
a log y = log \(e^{x}\) - log \(x^{2}\) + log C = log(C\(x^{2}\) \(e^{x}\))
log y = log(C\(x^{2}\) \(e^{x}\))^(\(\frac{1}{a}\))
y = (C\(x^{2}\) \(e^{x}\))^(\(\frac{1}{a}\)), C は任意の定数。

(2) u = xy と置くと d\(\frac{u}{d}\)x = y + xd\(\frac{y}{d}\)x
x・d\(\frac{y}{d}\)x = (d\(\frac{u}{d}\)x -y)
lhs = (xy)xy' + (xy)y - y
= u(d\(\frac{u}{d}\)x - y) + uy - y
= u d\(\frac{u}{d}\)x - y = 0
y = \(\frac{u}{x}\) だから
u d\(\frac{u}{d}\)x - \(\frac{u}{x}\) = 0
∫du = ∫d\(\frac{x}{x}\)
u = log(Cx), C は任意の定数。