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

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

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y^'=(2-y)(1-y)

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d x}{d t} = 1 - \frac{x}{40}

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (5)

y^{^{'}} = e^{2 y} sin (x), y (0) = 0

\frac{d y}{y} = - \frac{2 x}{x ^{2} - 1} d x

2 x d x + sin (y) d y = 0

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