What is the solution for y^'= 1/2 (y^2-1),y(0)=6 ?

The solution for y^'= 1/2 (y^2-1),y(0)=6 is y=(5e^t)/(7-5e^t)+7/(7-5e^t)

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y^'= 1/2 (y^2-1),y(0)=6

y^{^{'}} = \frac{1}{2} (y^{2} - 1), y (0) = 6

y = \frac{5 e ^{t}}{7 - 5 e ^{t}} + \frac{7}{7 - 5 e ^{t}}

Solution steps

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

Solve separable ODE:

y = \frac{5 e ^{t}}{7 - 5 e ^{t}} + \frac{7}{7 - 5 e ^{t}}

y = \frac{5 e ^{t}}{7 - 5 e ^{t}} + \frac{7}{7 - 5 e ^{t}}

y^{^{'}} = \frac{1}{2} (y^{2} - 1), y (0) = 4

y^{^{'}} = \frac{1}{2} (y^{2} - 1), y (0) = 2

\frac{d V}{d t} = 50 - \frac{5}{100} V

e^{y} \cdot y^{^{'}} = t + t^{3}

\frac{( x ^{2} - 9 ) d y}{d x} + x y = 0

What is the solution for y^'= 1/2 (y^2-1),y(0)=6 ?
The solution for y^'= 1/2 (y^2-1),y(0)=6 is y=(5e^t)/(7-5e^t)+7/(7-5e^t)