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

The solution for yy^'=-2x(1+y^2) is y=sqrt(e^{-2x^2+c_{1)}-1},y=-sqrt(e^{-2x^2+c_{1)}-1}

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

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

y = e^{- 2 x^{2} + c_{1}} - 1, y = - e^{- 2 x^{2} + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{- 2 x^{2} + c_{1}} - 1, y = - e^{- 2 x^{2} + c_{1}} - 1

y = e^{- 2 x^{2} + c_{1}} - 1, y = - e^{- 2 x^{2} + c_{1}} - 1

x \to 3 lim (2 f (x) - 3 g (x))

\int \frac{x}{x ^{2} - 5} d x

f (x) = x^{3}, a

\int (13 x + 11) d x

\int - 3 x \cdot e^{- 3 x} d x

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