Question

$(x^2-y{x}^2)dy+(y^2+x{y}^2)dx=0$

Answer 1

$(x^2-y{x}^2)dy+(y^2+x{y}^2)dx=0$

$\frac{dy}{dx}+\frac{(y^2+x{y}^2)}{x^2-y{x}^2}=0$

$\frac{dy}{dx}+\frac{y^2}{x^2}\left(\frac{1+x}{1-y}\right)=0$

$\frac{1-y}{y^2}dy=\frac{(1+x^2)}{x^2}dx$

$\Rightarrow \int \frac{1-y}{y^2}dy=-\int \frac{1+x^2}{x^2}dx$

$\Rightarrow \int \frac{1}{y^2}dy-\Rightarrow \int \frac{1}{y^2}dy=-\Rightarrow \int \frac{1}{x^2}dx-\Rightarrow \int \frac{x^2}{x^2}dx$

$\Rightarrow \frac{-1}{y}-\log y=\frac{1}{x}-x+c$

$x=1y=1\Rightarrow -1-0=\perp -\perp +c$

$c=-1$

$-1/y-\log y=\frac{1}{x}-x-1$

$(x^2+x-1)y=(1+y\log y)x$