I am trying to tutor some students in differential equations and I am a little rusty. They presented me with this problem a TA did and I'm trying to help them understand it. It says "A population of bacteria grows at a rate proportional to its size. Write and solve a differential equation that expresses this. If there are 1000 bacteria after one hour and 2000 bacteria after two hours, how many bacteria are there after 3 hours?" Any help with this I'm really lost on this one.
[Math] How to set up this differential equation?
calculusordinary differential equations
Related Solutions
The equation $$p'(t)=f(t)\ p(t)$$ says: The rate $p'$ by which the population $P$ grows is proportional to the current size $p$ of $P$; but the proportionality factor $f$ valid at time $t$ depends on time. If, e.g., the growth rate depends on the seasons, the function $t\mapsto f(t)$ would be a certain periodic function with period 365 days. At any rate, the size of $P$ at some later time $t$ will be proportional to the initial size $p(0)$.
The equation $$p'(t)=\alpha p(t)+h(t)$$ with constant $\alpha\in{\mathbb R}$ says: The population $P$ would rise (or decline) at the constant rate $\alpha$, and therefore would be given by $p(t)=p(0)\ e^{\alpha t}$, if it were not for an additional extraneous influx (or decrease) $h(t)$ dependent only on time $t$, but not on the current size $p(t)$ of the population. In this case it may very well be that for large $t>0$ the initial value $p(0)$ has almost no effect on the actual value $p(t)$. This is the case, e.g., if $\alpha<0$, and $h$ is some periodic function with period $T$. Then for large $t$ there will be a certain "stable" periodic behavior. This "limiting" behavior depends only on $h$, but not on the initial value $p(0)$.
Concerning your last question: When you write the solution in the "more correct" form $$p(t)=\int_0^t e^{\alpha(t-\tau)}\ h(\tau)\ d\tau + p(0)e^{\alpha t}$$ then you can verify by inspection that $p(0)$ plays no rôle for large $t$ when $\alpha<0$.
$\frac{dP}{dt}$ is the “instantaneous” rate of change of the population. Formally, it’s the derivative of the population size function with respect to time. So, the equation $\frac{dP}{dt}=kP$ just says that the growth rate at any instant is proportional to the population at that time, which you already knew.
Now that you’ve computed $k$, you just plug it into the growth rate formula above, setting $P$ to the population size at $3$ hours, i.e., $\frac{dP}{dt}(3 hrs) = kP(3 hrs) = \ln 4.2 \cdot 100e^{3 \ln{4.2}} = {4.2}^3\cdot100\cdot\ln{4.2}$.
Best Answer
Let $P$ be the population size. This grows with time and so can be thought of as a function of time. If the growth of the population is proportional to the size of the population then $$ \frac{\operatorname{d}\!P}{\operatorname{d}\!t} \propto P$$ What does this mean? Well, it means that there exists some fixed number, say $k$ for which $$\frac{\operatorname{d}\!P}{\operatorname{d}\!t} = kP$$ We can solve this differential equation by "seperation of variables". Dividing by $P$ and then multiplying throught by $\operatorname{d}\!t$ gives $$\frac{1}{P}\operatorname{d}\!P = k\operatorname{d}\!t$$ Does this remind you of anything? It reminds me of an integral: $$\int \frac{1}{P}\operatorname{d}\!P = \int k\operatorname{d}\!t $$ We can perform the integrations on both sides to give $$\ln \left|P\right| = kt+C$$ It follows that $\left|P\right| = \operatorname{e}^{kt+C} \equiv \operatorname{e}^{kt}\times \operatorname{e}^C$. Dropping the modulus gives $P = \pm\operatorname{e}^{kt}\times \operatorname{e}^C$. Since $C$ was a random constant, positive or negative, we can relabel $\pm\operatorname{e}^C$ as a random constant, say $\rho$. Hence $$P = \rho\operatorname{e}^{kt}$$ Use your initial conditions to set up two equation to solve for $\rho$ and $k$.