What is the solution for x^'=ax(1-x/k) ?

The solution for x^'=ax(1-x/k) is x=-(ke^{at+c_{1}})/(1-e^{at+c_{1)}}

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x^'=ax(1-x/k)

x^{^{'}} = a x (1 - \frac{x}{k})

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

Solution steps

x^{'} (t) = a x (t) (1 - \frac{x}{k})

Solve separable ODE:

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

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

y^{^{'}} sin^{2} (x) = cos (x) (1 - 3 y)

y^{^{'}} = \frac{( cos ^{2} ( x ) )}{y}

y^{^{'}} = \frac{y + 5}{7 t + 50}, y (2) = 0

4 x y^{^{'}} - 3 = y^{2} + 2

(1 + x) \frac{d y}{d x} = e^{- y} - 1

What is the solution for x^'=ax(1-x/k) ?
The solution for x^'=ax(1-x/k) is x=-(ke^{at+c_{1}})/(1-e^{at+c_{1)}}