What is the solution for (dy}{dx}=\frac{y^2-1)/2 ?

The solution for (dy}{dx}=\frac{y^2-1)/2 is y=(e^{x+c_{1}})/(1-e^{x+c_{1)}}+1/(1-e^{x+c_{1)}}

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(dy}{dx}=\frac{y^2-1)/2

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

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

Solution steps

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

Solve separable ODE:

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

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

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

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

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

sin (x) \frac{d y}{d x} + y cos (x) = 0, y (\frac{π}{2}) = 1

\frac{d y}{d x} = x^{3} 6 y

What is the solution for (dy}{dx}=\frac{y^2-1)/2 ?
The solution for (dy}{dx}=\frac{y^2-1)/2 is y=(e^{x+c_{1}})/(1-e^{x+c_{1)}}+1/(1-e^{x+c_{1)}}