In my differential equations class, this question came up:
$\frac{dx}{dt} = -ax\ln(\frac{x}{b})$ where $x=x(t)$ is the number of cells in the tumor, $t$ is time, and $a,b > 0$ are parameters. We study the dynamics of the model
and we were required to find $a)$ fixed point, $b)$ stability of each fixed points, $c)$ phase diagram, $d)$ explicit solution for $x(t)$ given an initial condition $x(0)$
what i've done so far is:
$\it a)$ To find the fixed points, set $$\frac{dx*}{dt} = -ax^*\ln(\frac{x^*}{b}) = 0$$ $$x^*\ln(\frac{x^*}{b}) = 0$$ $$x* = 0, b$$ $\it b)$ to find stability for these fixed points i set (stable if $f'(x^*)<0)$,$$f(x) = -ax\ln(\frac{x}{b})$$ $$f'(x) = -a\ln(\frac{x}{b})-a$$ then we have, $$f'(0) = -a\ln(\frac{0}{b})-a = -a < 0?$$ $$f'(b) = -a\ln(\frac{b}{b})-a < 0?$$ so both are stable fixed points? im not so sure. Also, I dont know how to do draw a phase diagram for this, can i get some help? and for $\it d)$ any ideas?
Since I haven't learned anything much about parts $a,b,c$, but have learned enough calculus to answer part $d$, here is my take:
$$\frac{dx}{dt}=-ax\ln\frac xb$$
$$\frac{d\ln x}{dt}=\frac{\frac{dx}{dt}}x=-a\ln\frac xb=a\ln b-a\ln x$$
We'll call $\ln x=y$
$$\frac{dy}{dt}=a\ln b-ay$$
$$\int\frac{dy}{a\ln b-ay}=\int1dt=t+c$$
$$\frac{-1}a\ln\left(a\ln b-ay\right)=t+c$$
$$\ln x=y=\ln b-\frac1a\exp\left(c-\frac ta\right)$$
$$x(t)=\exp\left[\ln b-\frac1a\exp\left(c-\frac ta\right)\right]$$