What is the solution for xy^'=y^2-1 ?

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

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xy^'=y^2-1

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{x ^{3} + 1}{x ^{2}} \frac{d y}{d x} = \frac{1}{y}

y y^{^{'}} = - 5 x

\frac{d y}{d x} - \frac{2 x}{x ^{2} + 1} y = - \frac{2 x}{x ^{2} + 1}

(- s^{2} - s) y^{^{'}} (s) + (- 3 s - 1) y (s) = 0

x d y + y d x = 2 x^{2} 2 y d x

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