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{t}{t ^{4} + 25} d t

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

\int \frac{ln ( x ^{5} )}{x} d x

\frac{2}{3} t^{2} + 5 e^{- 4 t}

x \to \infty lim (ln (3 x))

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))}}