What is the solution for (y^2+y)dx-(x^2+x)dy=0 ?

The solution for (y^2+y)dx-(x^2+x)dy=0 is y=-x/(x-e^{c_{1)}x-e^{c_{1}}}

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

(y^{2} + y) d x - (x^{2} + x) d y = 0

y = - \frac{x}{x - e ^{c_{1}} x - e ^{c_{1}}}

Solution steps

(y^{2} + y) d x - (x^{2} + x) d y = 0

Solve separable ODE:

y = - \frac{x}{x - e ^{c_{1}} x - e ^{c_{1}}}

y = - \frac{x}{x - e ^{c_{1}} x - e ^{c_{1}}}

\frac{d y}{d x} + \frac{y}{4} = 0

\frac{d x}{d y} = x^{2}

\frac{d}{d y} (x (y) (arctan (y))) = 0

y^{^{'}} = \frac{t y}{2}

\frac{d}{d x} ((f (x)^{- 1}) (11)) = 3 x^{2} - 6 x + 2

What is the solution for (y^2+y)dx-(x^2+x)dy=0 ?
The solution for (y^2+y)dx-(x^2+x)dy=0 is y=-x/(x-e^{c_{1)}x-e^{c_{1}}}