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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d θ} (6 cos (θ) cos (θ))

n = 1 \sum \infty \frac{4}{n ( n + 1 )}

\frac{d}{d x} (ln (x) + \frac{1}{ln ( x )})

x \to 0 lim (1 + 2 x)

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