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

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

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

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

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

Solution steps

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

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}}}

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