What is the solution for y^'= 1/2 (y^2)+1 ?

The solution for y^'= 1/2 (y^2)+1 is y=tan((sqrt(2)(t+c_{1}))/2)sqrt(2)

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y^'= 1/2 (y^2)+1

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

y = tan (\frac{2 ( t + c _{1} )}{2}) 2

Solution steps

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

Solve separable ODE:

y = tan (\frac{2 ( t + c _{1} )}{2}) 2

y = tan (\frac{2 ( t + c _{1} )}{2}) 2

2 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = - 1

\frac{d x}{d t} = - x^{2} + 4 x

\frac{d x}{d t} + cos (t) x - cos (t) = 0

y^{^{'}} = y (1 - \frac{y}{2000})

\frac{d x}{d t} = 12 - \frac{36 x}{100}

What is the solution for y^'= 1/2 (y^2)+1 ?
The solution for y^'= 1/2 (y^2)+1 is y=tan((sqrt(2)(t+c_{1}))/2)sqrt(2)