1st Order Differential Equations

**guess** · March 14th 2006, 04:05 AM

A population of insects is allowed to grow in a experimental environment. The rate of increase of the population is proportional to the number, n, of insects, at any time t days after the start of the experiment. Regarding n and t as continuous variables, form a differential equation relating n and t, and solve it to show that n=Ae^kt, where A and K are constants. The net increases during the fourth and fifth days are 350 and 500 insects respectively. Determine the population at the beginning of the fourth say. Hence, or otherwise, determine the population at the beginning of the first day

**ThePerfectHacker** · March 14th 2006, 05:04 AM

Originally Posted by guess

A population of insects is allowed to grow in a experimental environment. The rate of increase of the population is proportional to the number, n, of insects, at any time t days after the start of the experiment. Regarding n and t as continuous variables, form a differential equation relating n and t, and solve it to show that n=Ae^kt, where A and K are constants. The net increases during the fourth and fifth days are 350 and 500 insects respectively. Determine the population at the beginning of the fourth say. Hence, or otherwise, determine the population at the beginning of the first day

This is a classical differencial equation. It is a mathematical model based on growth-which is exponential. Let $Q(t)$ be a function which show the population at any given time. By the rule of growth:
$\frac{dQ(t)}{dt}=kQ(t)$ where $k$ is some constant. This is a differencial equation with seperable terms, thus,
$\frac{dQ(t)}{Q(t)}=kdt$
Integrating,
$\ln Q(t)+C_1=kt+C_2$
Thus,
$\ln Q(t)=kt+C_3$ where $C_3=C_2-C_1$
$Q(t)=e^{kt+C_3}=Ce^{kt}$ where $C=e^{C_3}$
Notice when $t=0$ then $Q(t)$ is the initial amount (the starting populations) thus,
$A=Q(0)=Ce^{k\cdot 0}=Ce^0=C$
Thus, $A=C$ , thus,
$Q(t)=Ae^{kt}$ for some $k$ .
------------------
Solve the problem, 4th day was 350 and 5th was 500.
Substitute that infromation into the exponential equation,
$\left\{\begin{array}{cc}350=&Ae^{4k}\\500=&Ae^{5k} \end{array}\right$
Divide the second equation by the first to get,
$\frac{Ae^{5k}}{Ae^{4k}}=\frac{500}{350}$
Thus,
$e^k\approx 1.43$ solve for $k$ ,
$k=\ln (1.43)\approx .357$
Thus, $A^{e^.357\dot 4}=350$
Thus, $A\approx 4.17$
Thus, the formula is, $4.17e^{.357t}$

**guess** · March 14th 2006, 05:21 AM

can you further solve the question because i still don't really understand

**guess** · March 14th 2006, 05:22 AM

So sorry, i never see the last part.. thanks a lot

**guess** · March 14th 2006, 05:29 AM

However i still can't get the answer. The answer is: 817 for the population at the beginning of the fourth day and 280 for the population at the beginning of the first day

**CaptainBlack** · March 14th 2006, 06:08 AM

Originally Posted by ThePerfectHacker

This is a classical differencial equation. It is a mathematical model based on growth-which is exponential. Let $Q(t)$ be a function which show the population at any given time. By the rule of growth:
$\frac{dQ(t)}{dt}=kQ(t)$ where $k$ is some constant. This is a differencial equation with seperable terms, thus,
$\frac{dQ(t)}{Q(t)}=kdt$
Integrating,
$\ln Q(t)+C_1=kt+C_2$
Thus,
$\ln Q(t)=kt+C_3$ where $C_3=C_2-C_1$
$Q(t)=e^{kt+C_3}=Ce^{kt}$ where $C=e^{C_3}$
Notice when $t=0$ then $Q(t)$ is the initial amount (the starting populations) thus,
$A=Q(0)=Ce^{k\cdot 0}=Ce^0=C$
Thus, $A=C$ , thus,
$Q(t)=Ae^{kt}$ for some $k$ .
------------------

Hacker, you managed to make that hard work

Observe:

$\frac{dQ(t)}{dt}=kQ(t)$

this is virtualy the defining relation for the exponential function, so
obviously this has a particular integral:

$ \mathcal{Q}(t)=e^{kt} $

and so the general solution is:

$ Q(t)=Ae^{kt} $

for some constant $A$ .

RonL

**CaptainBlack** · March 14th 2006, 07:57 AM

Originally Posted by ThePerfectHacker

Solve the problem, 4th day was 350 and 5th was 500.
Substitute that infromation into the exponential equation,
$\left\{\begin{array}{cc}350=&Ae^{4k}\\500=&Ae^{5k} \end{array}\right$

No the increase on the fourth and fith days were 350 and 500 respectivly,
so:

$ 350=Q(4)-Q(3)$
$ 500=Q(5)-Q(4) $ ,

which should allow us to find $A$ and $k$

RonL

**CaptainBlack** · March 14th 2006, 09:17 AM

Originally Posted by CaptainBlack

No the increase on the fourth and fith days were 350 and 500 respectivly,
so:

$ 350=Q(4)-Q(3)$
$ 500=Q(5)-Q(4) $ ,

These equations expand to:

$ 350=A(e^{4k}-e^{3k})$
$ 500=A(e^{5k}-e^{4k}) $ ,

Dividing the secon of these by the first gives:

$ \frac{500}{350}={e^{4k}(e^k-1) \over e^{3k}(e^k-1) }=e^k $

so:

$ k=\ln\left(\frac{500}{350}\right)\approx 0.356675 $

Now:

$ A=\frac{350}{e^{4k}-e^{3k}} \approx 280.1165 $

So:

$ Q(t)=280.1165 e^{0.356675t} $

**guess** · March 14th 2006, 09:22 AM

Thanks Bro.. you got it!!

**CaptainBlack** · March 14th 2006, 09:22 AM

Originally Posted by guess

However i still can't get the answer. The answer is: 817 for the population at the beginning of the fourth day and 280 for the population at the beginning of the first day

We have:

$ Q(t)=280.1165 e^{0.356675t} $

At the beginning of the fourth day the population is:

$ Q(3)=816.666 \approx 817 $ .

At the beginning of the first day the population is:

$ Q(0)=280.1165 \approx 280 $ .

RonL

**ThePerfectHacker** · March 17th 2006, 10:49 AM

Sorry bout that, the phrase "... an increase of..." got me.

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