# Solving a differential equation about rates: $\frac{dx}{dt} = ax(400-x)$

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.

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 $$\int\frac{1}{x}dx=\ln|x|+C$$

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

hint

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

• 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. Commented Sep 7, 2017 at 21:16

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.