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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

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

\int sec^{5} (2 x) tan^{3} (2 x) d x

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 3

y^{^{''}} - y^{^{'}} - 2 y = e^{2 t}

5 x^{5}

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