Question

решить дифференциальное уравнение!

Answer 1

$(y^2-2xy)\, dx-x^2\, dy=0\\\\\displaystyle \frac{dy}{dx}=\frac{y^2-2xy}{x^2}\ \ ,\\\\\frac{y}{x}=u\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\u'x+u=u^2-2u}\ \ ,\ \ u'x=u^2-3u\ \ ,\ \ \frac{du}{dx}=\frac{u^2-3u}{x}\ \ ,\\\\\int \frac{du}{u(u-3)}=\int \frac{dx}{x}\ \ ,\ \ \int \Big(-\frac{1}{3u}+\frac{1}{3(u-3)}\Big)\, dx=\int \frac{dx}{x}\ \ ,\\\\\\-\frac{1}{3}\, ln|u|+\frac{1}{3}\, ln|u-3|=ln|x|+lnC_1$

$\displaystyle \frac{1}{3}\cdot ln\frac{u-3}{u}=ln(C_1x)\ \ ,\ \ \ \sqrt[3]{\frac{\frac{y}{x}-3}{\frac{y}{x}}}=C_1x\ \ ,\ \ \sqrt[3]{\frac{y-3x}{y}}=C_1x\ \ ,\\\\\\\frac{y-3x}{y}=Cx^3\ \ ,\ \ (\ C=C_1^3\ )\\\\\\y-3x=Cx^3y\ \ ,\ \ y(1-Cx^3)=3x\ \ ,\\\\\\\boxed{\ y=\frac{3x}{1-Cx^3}\ }$