Resolver la ecuación $xy^2+3y^2-x^2\frac{dy}{dx}=0$
$$xy^2+3y^2=\frac{dy}{dx}x^2$$ $$dx(xy^2+3y^2)-x^2 dy=0$$ $$dx\frac{x+3}{x^2}-\frac{1}{y^2}dy=0$$ $$\int \frac{x+3}{x^2}dx=\frac{1}{y^2}dy$$ $$\int \frac{x+3}{x^2}dx= (-x^{-1})(x+3)-\int\frac{1}{x^2}dx=-\frac{1}{x}(x+3)+\frac{1}{x}$$ $$=-1-\frac{3}{x}+\frac{1}{x}$$ $$=-1-\frac{2}{x}$$ $$\int \frac{1}{y^2}dy=-y^{-1}$$
Entonces, desde $$\int \frac{x+3}{x^2}dx=\frac{1}{y^2}dy$$ $$-1-\frac{2}{x}=-y^{-1}$$ $$1+\frac{2}{x}=y^{-1}$$ $$\frac{1}{1+2/x}=y$$