What is the solution for 2y^'-y^2+4=0 ?

The solution for 2y^'-y^2+4=0 is y=(2e^{2t+2c_{1}})/(1-e^{2(t+c_{1))}}+2/(1-e^{2(t+c_{1))}}

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2y^'-y^2+4=0

2 y^{^{'}} - y^{2} + 4 = 0

y = \frac{2 e ^{2 t + 2 c_{1}}}{1 - e ^{2 (t + c_{1})}} + \frac{2}{1 - e ^{2 (t + c_{1})}}

Solution steps

2 y^{'} (t) - y^{2} + 4 = 0

Solve separable ODE:

y = \frac{2 e ^{2 t + 2 c_{1}}}{1 - e ^{2 (t + c_{1})}} + \frac{2}{1 - e ^{2 (t + c_{1})}}

y = \frac{2 e ^{2 t + 2 c_{1}}}{1 - e ^{2 (t + c_{1})}} + \frac{2}{1 - e ^{2 (t + c_{1})}}

\int \frac{x ^{3}}{1 - 2 x ^{2}} d x

3 x e^{3 x}

\frac{d}{d y} (y e^{x y})

6 \frac{d y}{d t} - 2 y = t y^{4}, y (0) = - 2

\frac{d}{d t} (7 sin (t))

What is the solution for 2y^'-y^2+4=0 ?
The solution for 2y^'-y^2+4=0 is y=(2e^{2t+2c_{1}})/(1-e^{2(t+c_{1))}}+2/(1-e^{2(t+c_{1))}}