What is the solution for y^'=0.25-y^2 ?

The solution for y^'=0.25-y^2 is y=-(e^{t+c_{1}})/(2(1-e^{t+c_{1)})}-1/(2(1-e^{t+c_{1)})}

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y^'=0.25-y^2

y^{^{'}} = 0.25 - y^{2}

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

Solution steps

y^{'} (t) = 0.25 - y^{2}

Solve separable ODE:

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

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

2 x \frac{d x}{d t} - (x^{2} + 1) = 0

\frac{d y}{d x} = \frac{2 y + 3}{x ^{2}}

(x y - 2 x + 4 y - 8) d y - (x y + 3 x - y - 3) d x = 0

y y^{^{'}} + 6 x = 0, y^{2} (0) = - 6 x^{2} + c, y (0) = 4

y^{^{'}} = - \frac{3}{7} e^{3 y}

What is the solution for y^'=0.25-y^2 ?
The solution for y^'=0.25-y^2 is y=-(e^{t+c_{1}})/(2(1-e^{t+c_{1)})}-1/(2(1-e^{t+c_{1)})}