I am having some trouble with the following separable differential equation
$$\frac{dx}{dt} = x(x-1)(x-3)$$
with initial condition $x(0) = 2$. What is $\displaystyle\lim_{t \to \infty} x(t)$?
I am having some trouble with the logarithmic laws when solving for $x(t)$.
On
I agree with the answer from Mohammad Riazi-Kermani : $x(t\to\infty)\to 1$.
The solving of the ODE leads to the same result : $$t=\int\frac{dx}{x(x-1)(x-3)}=\int \left(\frac{1}{3x}-\frac{1}{2(x-1)}+\frac{1}{6(x-3)} \right)dx$$ $$6t=2\ln|x|-3\ln|x-1|+\ln|x-3|+\text{constant}$$
$$e^{-6t}=c\frac {(x-1)^3}{x^2(x-3)}$$ The condition $x(0)=2$ or $t(2)=0$ implies
$e^{0}=c \frac{2^2(2-3)}{(2-1)^3} \implies c=-4$ . $$e^{-6t}=-4\frac{(x-1)^3}{x^2(x-3)} $$ For $t\to\infty\quad:\quad e^{-6t}\to 0.\quad$ Thus $x\to 1.$
You do not have to solve the differential equation $$ \frac{dx}{dt} = x(x-1)(x-3)$$ to answer the question.
Note that you have three equilibrium points, namely $$ x=0,1,3 $$
Qualitative analysis of these equilibrium points show that $x=1$ is asymptotically stable.
Thus starting at $x(0)=2$ the solution will tend to $x=1.$