Question

x'y+x=-yx^2 . . . . . ​

Answer 1

$x'y+x=-yx^2\ -dfrac{x'}{yx^2}-dfrac{1}{y^2x}=dfrac{1}{y}\ left[(dfrac{1}{x})'=-dfrac{x'}{x^2}right]\ -dfrac{x'}{x^2}cdot dfrac{1}{y}+dfrac{1}{x}cdot(-dfrac{1}{y^2})=dfrac{1}{y}\ (dfrac{1}{xy})'=dfrac{1}{y}\ dfrac{1}{xy}=lny+C\ x=dfrac{1}{ycdot (lny+C)}$