What is the solution for y^'=1-y^2 ?

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

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y^'=1-y^2

y^{^{'}} = 1 - y^{2}

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

Solution steps

y^{'} (t) = 1 - y^{2}

Solve separable ODE:

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

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

\int x + x^{2} d x

\int 5 x^{3} d x

\frac{d}{d x} ((1 + sin (x)) cos (2 x))

n = 0 \sum \infty \frac{5 n}{2 n - 1}

y^{^{'}} + 3 y = e^{- 3 x}

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