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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d P}{d t} = Pt, P (1) = 5

\frac{d y}{d t} = 10 y - 2 y^{2}

y^{- 1} d y + y \cdot e^{c o s (x)} \cdot sin (x) d x = 0

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

\frac{d y}{d x} = x y^{2}, y (1) = 1

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