What is the solution for y^'+2xy^2=0,y(1/2)=-4 ?

The solution for y^'+2xy^2=0,y(1/2)=-4 is y=-2/(-2x^2+1)

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y^'+2xy^2=0,y(1/2)=-4

y^{^{'}} + 2 x y^{2} = 0, y (\frac{1}{2}) = - 4

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

Solution steps

y^{'} (x) + 2 x y^{2} = 0

Solve separable ODE:

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

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

y^{^{'}} csc (t) = \frac{- y ^{15}}{14}

\frac{d x}{d t} = \frac{1}{x e ^{t + 7 x}}

(ln (y))^{6} d y = x^{6} y d x, y (1) = e^{2}

\frac{d y}{d x} = 10 e^{x} \cdot y^{2}

x^{2} d y = - (x^{2} - 1) y^{3} d x

What is the solution for y^'+2xy^2=0,y(1/2)=-4 ?
The solution for y^'+2xy^2=0,y(1/2)=-4 is y=-2/(-2x^2+1)