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

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

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y^'=((y-y^2))/x

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{1 + x}{1 - x} - 1

f (x) = sec (x^{2})

\frac{d}{d x} (9 + 3 x + sin (3 x))

f (x) = 2 x^{- 1}, at x = 5

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

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