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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (sin (x^{2} + e^{x}))

\int_{0}^{π} 3 sin (x) d x

\int \frac{2 x ^{2} + x - 4}{x ^{3} - x ^{2} - 2 x} d x

\frac{\partial}{\partial x} (7 sin (x) sin (y))

\frac{d}{d x} (\frac{6}{5})

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