Solve logistic differential equation

Question

Task: $\frac{du}{dt} = u(1-\frac uA)$ with $u(0) = u_{0}, 0 < u_{0} < A$

I know that this is a variable separation type, however, I have no idea to solve it. what should be the next step?

Answer 1

Hint: we have $$-\frac{u'(t)}{\frac{u(t)(-A+u(t))}{A}}=1$$

Answer 2

You could also solve this logistic equation as a Bernoulli equation setting $v=A/u$, $$ \dot v=1-v. $$

Answer 3

We have $\frac{du}{dt} = u(1-\frac uA)$

$\frac{du}{dt} = u(\frac{A-u}{A})$

$\frac{1}{u(A-u)}\,du = \frac{1}{A}\,dt$

$\int\frac{1}{u(A-u)}\,du=\int\frac1A\,dt$

$\int\frac{1}{Au}+\frac{1}{A(A-u)}\,du=\int\frac 1A\,dt $

$\frac 1A\big[\ln(u)-\ln(A-u)\big]=\frac 1At+c $

$\ln(\frac{u}{A-u}) = t+C$

Answer 4

Hint $$\frac{du}{dt} = u(1-\frac uA)$$ $$\frac{du}{dt} = \frac 1 {A}u(A-u)$$ $$\int \frac{du}{u(A-u)} = \int \frac {dt} {A}$$ $$\int \frac{du}{u(A-u)} = \frac t {A}+K$$ Use decomposition of fraction and integrate $$\int \frac{du}{u(A-u)} = \frac t {A}+K$$ $$\frac 1 A(\int \frac{du}{u}-\int \frac{du}{(u-A)}) = \frac t {A}+K$$ $$\int \frac{du}{u}-\int \frac{du}{(u-A)} = t +K$$ $$......$$