a separable differential equation

Question

given this

$$\frac{d}{dx} = x(1-x)$$

where

$$x(0) = 0.1$$

is this correct:?

$$\frac{dx}{dt} = x(1-x)$$

$$t-t_0 = \int\frac{dx}{x(1-x)}$$

Where did the $t$ and $t_0$ come from?

Answer 1

From $\dfrac{dx}{dt} = x(1-x)$ you get $\dfrac{dx}{x(1-x)} = dt$ and then $\displaystyle\int \frac{dx}{x(1-x)} = \int dt = t-t_0.$

\begin{align} & \underbrace{\int \frac{dx}{x(1-x)} = \int \left( \frac 1 x + \frac 1 {1-x} \right) \, dx}_\text{partial fractions} = \log x - \log(1-x) + \text{constant} \\[10pt] = {} & \log \frac x {1-x} + \text{constant}, \text{ so } \\[10pt] & \log \frac x {1-x} = t - t_0 \text{ where } t_0 \text{ is the “constant''.} \\[10pt] & \frac x {1-x} = e^{t-t_0}. \text{ When $t=0$ then $x=0.1,$ so you have} \\[10pt] & \frac{0.1}{1-0.1} = e^{0-t_0}. \qquad \frac 1 9 = e^{-t_0}. \qquad e^{t-t_0} = \frac{e^t} 9. \qquad \frac x {1-x} = \frac{e^t} 9. \\[10pt] & \frac{9x}{1-x} = e^t. \\[10pt] & 9x = e^t - xe^t. \\[10pt] & 9x - xe^t = e^t. \\[10pt] & x(9-e^t) = e^t. \\[10pt] & x = \frac{e^t}{ 9 - e^t} = \frac 1 {9e^{-t} - 1}. \end{align} That last form makes it easy to find $t$ as a function of $x$ if you wanted that. Either of the last two forms is an explicit closed-form for $x$ as a function of $t,$ and shows $x\to-1$ as $t\to+\infty$ and $x\to 0 \text{ as } t\to-\infty.$

Normally I would leave most of this as an exercise, especially since this goes beyond what was asked, but the question seemed like something you wouldn't ask if all this weren't useful.

Answer 2

$$\frac{dx}{dt}=x(1-x) \implies \frac{dx}{x(1-x)}=dt $$

$$ \implies \int \frac{dx}{x(1-x)}= \int dt $$

$$\implies \int \frac{dx}{x(1-x)}=\int dt=t+C =t-t_0$$

(Here $t_0=-C$)

Answer 3

Using separation of variables, you get $$\int dt=\int \frac{dx}{x(1-x)}$$ So evaluating the lefthand side, you are left with $$t-t_{0}=\int\frac{dx}{x(1-x)}$$ Where $t_{0}$ is the constant of integration. I believe that you are solving a differential equation peratining to motion, so in this case $t_{0}$ would be the initial time or starting time.

Answer 4

Since the other users have answered to question in totality, I would just like to clear up some notational things for those who may have this same question in the future. I do a bit of differential geometry and so I am assuming that $\frac{d}{dx} = x(1-x)$ is to be thought of as a vector in $\mathbb{R}$. To make the notation a bit simpler, let $\textbf{v} = \frac{d}{dx}$.

Hence, we are trying to get a solution curve in $\mathbb{R}$ whose timed-derivative is $v(t)$ for each $t \in I$. Here $I$ is just an interval about $0$ in which the solution lives. If $\phi$ is our solution then mathematically this amounts to saying,

$$\phi' = \frac{d \phi}{dt} = \textbf{v}(\phi) = \phi(1-\phi)$$

i.e we have,

$$ \frac{d \phi}{dt} = \phi(1-\phi), t \in I$$

Now this clears up the notation and what was given in the problem. Let me know if there is anything else I can add to help.