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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

y = x^{3} - x

\frac{d}{d x} (\frac{2 ( x + 8 ) ( x - 6 )}{x ^{2} - 7 ^{2}})

\frac{d}{d x} ((\frac{2 x + 4}{3 x - 1})^{4})

\int \frac{5 sin ( x )}{x} d x

y^{^{'}} - \frac{2}{t} y = t^{3}

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