What is the solution for (dx)/(dt)=kx(k-x) ?

The solution for (dx)/(dt)=kx(k-x) is x= k/(e^{-k^2t+c_{1)}+1}

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(dx)/(dt)=kx(k-x)

\frac{d x}{d t} = k x (k - x)

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

Solution steps

\frac{d x}{d t} = k x (t) (k - x)

Solve Bernoulli ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (- 1) = 0, y^{^{'}} (- 1) = 1

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

\int e^{3 x} sin (5 x) d x

t^{3} e^{t} t

n \to \infty lim (c^{n})

What is the solution for (dx)/(dt)=kx(k-x) ?
The solution for (dx)/(dt)=kx(k-x) is x= k/(e^{-k^2t+c_{1)}+1}