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I am trying to understand how to find the solution for these two autonomous differential equations:

First Autonomous Differential Equation

$\frac{dX}{dt}=aX(1-X)$ which solution should be:

\begin{align}X_{t}=\frac{X_{0}}{X_{0} + (1 - X_{0})e^{-at}}&\end{align} under the assumption that $a$ is very small.

Second Autonomous Differential Equation

$\frac{dX}{dt}=aX(X-\frac{1}{2})$ which solution should be: \begin{align} X_{t}=\frac{1/2}{1 - (e^{at/2} \cdot (1 - \frac{1}{2X_{0}}))}&\end{align} under the assumption that $a$ is very small.

Could someone explain what are the steps required for obtaining the solutions I mentioned?

CafféSospeso
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1 Answers1

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As was mentioned in the comments, the way to solve these equations is by the method of separation of variables. A differential equation is `separable' if everything involving one of the variables can be moved to (say) the left-hand side of the equation, and everything involving the other variable can be moved to the right-hand side. Take your first equation, for example. Dividing both sides by $X(1-X)$, and multiplying both sides by $dt$, produces:

$$ \frac{dX}{X(1-X)}=a\,dt $$

We have succeeded in `separating' the equation - everything in $X$ is on the left-hand side, and everything in $t$ is on the right-hand side. If we now integrate both sides:

$$ \int \frac{dX}{X(1-X)}=a\int dt $$

we find (the integral on the left can be evaluated by the method of partial fractions):

$$ \ln\Bigg|\frac{X}{1-X}\Bigg| = at + C $$

where $C$ is a constant of integration. Exponentiating both sides:

$$ \frac{X}{1-X}=C\,e^{at} $$

(we have removed the absolute value bars by allowing $C$ to take on negative values). Solving for $X$:

$$ X(t)=\frac{C\,e^{at}}{C\,e^{at}+1} $$

Your second equation can be solved by following a similar procedure.

CW279
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