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

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

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

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

y = \frac{1}{x ( 1 + x ln ( x ) + c _{1} x )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{x ( 1 + x ln ( x ) + c _{1} x )}

y = \frac{1}{x ( 1 + x ln ( x ) + c _{1} x )}

\frac{\partial}{\partial x} (arctan (x y^{2}))

\int \frac{1 - cos ( 2 x )}{2} d x

\int + x^{2} sec^{2} (x^{3} + 1) d x

\frac{\partial}{\partial x} (x^{3} (1 + y) + z^{2})

\frac{d}{d x} (\frac{x ^{2}}{( 5 + x )})

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