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

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

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

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

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}}

(x - 8)^{2} \frac{d y}{d x} = (7 - y)^{2}

y^{^{'}} = b - b \cdot y - a \cdot y

\frac{d x}{d y} = \frac{e ^{- 2 y}}{x ^{2} + 4}

\frac{d y}{d x} = \frac{x y}{ln ( y )}, y > 1

(2 y^{2} + 3 y) d x + 2 x y d y = 0

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