Question

x'y+x=-yx^2 . . . . . ​

Answer 1

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