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

The solution for y^'=ky(1-y) is y=-(e^{kt+c_{1}})/(1-e^{kt+c_{1)}}

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y^'=ky(1-y)

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (\frac{1}{x} + ln (x))

5 x^{2} - x^{3}

f (x) = \frac{1}{x ^{2} + 4}

f (x) = \frac{5 x}{x ^{2} - 4}, at x = 3

x \to 6 lim (x - 6)

What is the solution for y^'=ky(1-y) ?
The solution for y^'=ky(1-y) is y=-(e^{kt+c_{1}})/(1-e^{kt+c_{1)}}