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} (\frac{e ^{x}}{7 x ^{2} + 5})

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2}

x y^{^{''}} + 3 y^{^{'}} = 1984 (x^{19})^{82}

\int \frac{10800 x}{9 x ^{2} + 675} d x

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

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)