Question

Найти общее решение дифференциального уравнения
( y^2-2xy)dx-x^2dy=0

Answer 1

$y^2-2xy-x^2y'=0\\ \dfrac{1}{x^4}-\dfrac{2}{x^3y}-\dfrac{y'}{x^2y^2}=0\\ -\dfrac{2}{x^3y}-\dfrac{y'}{x^2y^2}=- \dfrac{1}{x^4}\\ (*)\;\;\left(\dfrac{1}{y}\right)'=-\dfrac{1}{y^2}y'\\ \left(\dfrac{1}{x^2}\right)'=-\dfrac{2}{x^3}\;\;(*)\\ \left(\dfrac{1}{x^2}\right)'\dfrac{1}{y}+\dfrac{1}{x^2}\left(\dfrac{1}{y}\right)'=- \dfrac{1}{x^4}\\\left(\dfrac{1}{yx^2}\right)'=- \dfrac{1}{x^4}\\ \dfrac{1}{yx^2}=-\int\dfrac{1}{x^4}dx\\ \dfrac{1}{yx^2}=\dfrac{1}{3x^3}+C_1=>\dfrac{1}{y}=\dfrac{1}{3x}+C_1x^2$

$y=\dfrac{1}{\dfrac{1}{3x}+C_1x^2}=\dfrac{3x}{1+Cx^3}$