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

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

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

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

y = \frac{1}{t + c _{1}} + \frac{t}{t + c _{1}} + \frac{c _{1}}{t + c _{1}}

Solution steps

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

Solve separable ODE:

y = \frac{1}{t + c _{1}} + \frac{t}{t + c _{1}} + \frac{c _{1}}{t + c _{1}}

y = \frac{1}{t + c _{1}} + \frac{t}{t + c _{1}} + \frac{c _{1}}{t + c _{1}}

y^{^{'}} = 0.1 \cdot y \cdot (1 - \frac{y}{80})

5 \frac{d v}{d t} = - 2 v^{2}

y^{^{'}} = 3 x 2 y

\frac{d}{d x} (f (x)^{- 1}) f (x) = \frac{1}{5} x + 5, x = 6

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

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