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

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

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

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

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

Solution steps

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

(ln (e))^{^{'}}

\frac{d}{d x} (e^{- (x - 1)^{2}})

\frac{d y}{d x} = (4 cos^{2} (y) tan (x))

(\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 7 y^{3})

\int - 8 sin (x) d x

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