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

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

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y^'=4-y^2

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

y = - \frac{2 e ^{4 t + 4 c_{1}}}{1 - e ^{4 (t + c_{1})}} - \frac{2}{1 - e ^{4 (t + c_{1})}}

Solution steps

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

Solve separable ODE:

y = - \frac{2 e ^{4 t + 4 c_{1}}}{1 - e ^{4 (t + c_{1})}} - \frac{2}{1 - e ^{4 (t + c_{1})}}

y = - \frac{2 e ^{4 t + 4 c_{1}}}{1 - e ^{4 (t + c_{1})}} - \frac{2}{1 - e ^{4 (t + c_{1})}}

2 y^{2} y^{^{'}} = 3 e^{2 x}

\frac{d y}{d x} = 0.02 y + 40000, y (0) = 2000000

y (25 - x^{2})^{\frac{1}{2}} d y = (25 + y^{2})^{\frac{1}{2}} d x

d y = 2 x y e^{x^{2}} d x

\frac{d y}{d x} = y (x^{2} + 6)

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