What is the solution for x^'= 1/2 (x^2-1) ?

The solution for x^'= 1/2 (x^2-1) is x=(e^{t+c_{1}})/(1-e^{t+c_{1)}}+1/(1-e^{t+c_{1)}}

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x^'= 1/2 (x^2-1)

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

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

Solution steps

x^{'} (t) = \frac{1}{2} (x^{2} - 1)

Solve separable ODE:

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

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

\frac{d t}{d x} = 3 x t^{2}

\frac{d x}{d t} = - 0.6364 x

y^{^{'}} tan (x) = y ln (y)

\frac{d y}{d t} = \frac{[ 3 t ^{2} + 4 t + 2 ]}{[ 2 y - 2 ]}, y (0) = - 1

y^{^{'}} = \frac{8 ( 1 + x )}{x y}

What is the solution for x^'= 1/2 (x^2-1) ?
The solution for x^'= 1/2 (x^2-1) is x=(e^{t+c_{1}})/(1-e^{t+c_{1)}}+1/(1-e^{t+c_{1)}}