Question

(y^2-2*x*y)*d*x+x^2*d*y=0 Найти общее решение диф уравнения.
ответ y/(x-y)=Cx
Нужно решение

Answer 1

$(y^2-2xy)\, dx+x^2\, dy=0\\\\\\(y^2-2xy)+x^2\cdot \dfrac{dy}{dx}=0\ \ ,\ \ \ \dfrac{dy}{dx}=-\dfrac{y^2-2xy}{x^2}\ \ ,\ \ y'=\dfrac{2xy-y^2}{x^2}\ \ ,\\\\\\y'=\dfrac{2y}{x}-\dfrac{y^2}{x^2}\ \ ,\ \ \ \ u=\dfrac{y}{x}\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\\\u'x+u=2u-u^2\ \ ,\ \ u'x=u-u^2\ \ ,\ \ \dfrac{du}{dx}=\dfrac{u-u^2}{x}\\\\\\\int \dfrac{du}{u-u^2}=\int \dfrac{dx}{x}\ \ ,\ \ \ -\int \dfrac{du}{u(u-1)}=\int \dfrac{dx}{x}\ ,$

$\star \int \dfrac{du}{u(u-1)}=\int \Big(\dfrac{-1}{u}+\dfrac{1}{u-1}\Big)\, dx=-ln|u|+ln|u-1|+C_1\ \ \star \\\\\\ln|u|-ln|u-1|=ln|x|+lnC\\\\\\ln\dfrac{u}{u-1}=ln(Cx)\ \ ,\ \ \dfrac{u}{u-1}=Cx\ \ ,\ \ \ \dfrac{\dfrac{y}{x}}{\dfrac{y}{x}-1}=Cx\ \ ,\dfrac{y}{x\cdot \dfrac{y-x}{x}}=Cx\\\\\\\boxed{\ \dfrac{y}{y-x}=Cx\ }\\\\\\\\P.S.\\\\\dfrac{y}{y-x}=Cx\ \ \to \ \ \dfrac{y}{-(x-y)}=Cx\ \ \to \ \ \dfrac{y}{x-y}=-Cx\ \ \to \ \ \boxed{\ \dfrac{y}{x-y}=C_1\, x\ }\ ,\\\\C_1=-C$