Question

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

Answer 1

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

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