0
$\begingroup$

So I have this differential equation which is a little bit more complicated than what I'm used to and I'm just struggling with solving it.

$$\frac{dx}{dt} = ax(400-x)$$

From what I would usually do, I would bring the $ax(200-x)$ over to the left-hand side and the dt on the right-hand side, however, when I get to that stage I'm unsure of how I take the integral or whether that's actually correct. Just as a side note, a is a constant so I believe that can be taken out the front when doing an intergral but again I'm unsure. Any help on the topic would be greatly appreciated. Thank you.

$\endgroup$

3 Answers 3

2
$\begingroup$

This equations is seperable, however it may look more difficult than what you're used to. You can still "bring over" the $ax(400-x)$ to the other side. So we have $$\frac{dx}{ax(400-x)}=dt\implies \int\frac{1}{ax(400-x)}dx=\int dt+C$$ Now to solve this integral, we can use the method of partial fraction decomposition. The integrand on the left side has a partial fraction decomposition as follows: $$\frac{1}{ax(400-x)}=\frac{A}{ax}+\frac{B}{400-x}\implies 1=A(400-x)+B(ax)= x(aB-A)+400A$$ from this we obtain $$\begin{cases}aB-A=0\\400A=1\end{cases}$$ Hence, $$\frac{1}{ax(400-x)}=\frac{1}{400ax}+\frac{1}{400a(400-x)}$$ I'll let you continue, while remembering that $$\int\frac{1}{x}dx=\ln|x|+C$$

$\endgroup$
2
  • $\begingroup$ Great answer thanks or your help! $\endgroup$
    – User0015
    Commented Sep 7, 2017 at 21:34
  • $\begingroup$ My pleasure. Any further questions about the answer just let me know. $\endgroup$
    – Dave
    Commented Sep 7, 2017 at 22:01
1
$\begingroup$

hint

$$\frac {400}{x (400-x)}=\frac {1}{x}+\frac {1}{400-x}$$

$\endgroup$
1
  • $\begingroup$ Hi thanks for the reply, I'm just still a little unsure as to how you got the stage of 400/x(x-400) and how exactly that would actually help me? Again sorry it's just I'm relatively new to solving differnetial equations. $\endgroup$
    – User0015
    Commented Sep 7, 2017 at 21:16
0
$\begingroup$

At first find stationary solution:

$$u(t)=0$$ and $$u(t)=400$$ are solutions of the differential equation.

Now separate variables and integrate to obtain general solutions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .