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

The solution for yy^'=-2x(1+y^2),y(0)=1 is y=sqrt(2e^{-2x^2)-1}

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yy^'=-2x(1+y^2),y(0)=1

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

y = 2 e^{- 2 x^{2}} - 1

Solution steps

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

Solve separable ODE:

y = 2 e^{- 2 x^{2}} - 1

y = 2 e^{- 2 x^{2}} - 1

\int 1. 8^{0.045 \cdot x} - 1 d x

\int \frac{1}{a - b x} d x

\int \frac{1}{x ^{3} x ^{2} - 36} d x

y^{^{'}} sin (5 π x) = π y cos (5 π x)

\frac{d}{d x} (2000)

What is the solution for yy^'=-2x(1+y^2),y(0)=1 ?
The solution for yy^'=-2x(1+y^2),y(0)=1 is y=sqrt(2e^{-2x^2)-1}