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

The solution for y^'=2xy^2,y(0)=-1/(x^2+c) is y=-1/(x^2+c_{1)}

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d V}{d t} = 90 - \frac{5 V}{100}

(y^{2} + 1) d x = (x^{2} + 1) d y

\frac{d y}{d x} = 15 x sec (y)

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

\frac{d x}{d t} = \frac{1}{x e ^{t + 11 x}}

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