What is the solution for y^'=-1/x y^2-1/x y ?

The solution for y^'=-1/x y^2-1/x y is y=(e^{c_{1}})/(x-e^{c_{1)}}

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y^'=-1/x y^2-1/x y

y^{^{'}} = - \frac{1}{x} y^{2} - \frac{1}{x} y

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

Solution steps

y^{'} (x) = - \frac{1}{x} y^{2} - \frac{1}{x} y

Solve separable ODE:

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

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

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 5, (- 8, - 1)

\int_{1}^{3} x^{4} ln (x) d x

\frac{d y}{d t} = 5 y + y^{4}

\int \frac{x}{100} d x

\frac{\partial}{\partial x} (y - 4 x^{2} y^{- 2})

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