# 1st Order Differential Equations

• Mar 14th 2006, 04:05 AM
guess
1st Order Differential Equations
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
• Mar 14th 2006, 05:04 AM
ThePerfectHacker
Quote:

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 $\displaystyle Q(t)$ be a function which show the population at any given time. By the rule of growth:
$\displaystyle \frac{dQ(t)}{dt}=kQ(t)$ where $\displaystyle k$ is some constant. This is a differencial equation with seperable terms, thus,
$\displaystyle \frac{dQ(t)}{Q(t)}=kdt$
Integrating,
$\displaystyle \ln Q(t)+C_1=kt+C_2$
Thus,
$\displaystyle \ln Q(t)=kt+C_3$ where $\displaystyle C_3=C_2-C_1$
$\displaystyle Q(t)=e^{kt+C_3}=Ce^{kt}$ where $\displaystyle C=e^{C_3}$
Notice when $\displaystyle t=0$ then $\displaystyle Q(t)$ is the initial amount (the starting populations) thus,
$\displaystyle A=Q(0)=Ce^{k\cdot 0}=Ce^0=C$
Thus, $\displaystyle A=C$, thus,
$\displaystyle Q(t)=Ae^{kt}$ for some $\displaystyle k$.
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Solve the problem, 4th day was 350 and 5th was 500.
Substitute that infromation into the exponential equation,
$\displaystyle \left\{\begin{array}{cc}350=&Ae^{4k}\\500=&Ae^{5k} \end{array}\right$
Divide the second equation by the first to get,
$\displaystyle \frac{Ae^{5k}}{Ae^{4k}}=\frac{500}{350}$
Thus,
$\displaystyle e^k\approx 1.43$ solve for $\displaystyle k$,
$\displaystyle k=\ln (1.43)\approx .357$
Thus, $\displaystyle A^{e^.357\dot 4}=350$
Thus, $\displaystyle A\approx 4.17$
Thus, the formula is, $\displaystyle 4.17e^{.357t}$
• Mar 14th 2006, 05:21 AM
guess
can you further solve the question because i still don't really understand
• Mar 14th 2006, 05:22 AM
guess
So sorry, i never see the last part.. thanks a lot
• Mar 14th 2006, 05:29 AM
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
• Mar 14th 2006, 06:08 AM
CaptainBlack
Quote:

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

Hacker, you managed to make that hard work :D

Observe:

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

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

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

and so the general solution is:

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

for some constant $\displaystyle A$.

RonL
• Mar 14th 2006, 07:57 AM
CaptainBlack
Quote:

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

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

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

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

RonL
• Mar 14th 2006, 09:17 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
No the increase on the fourth and fith days were 350 and 500 respectivly,
so:

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

These equations expand to:

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

Dividing the secon of these by the first gives:

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

so:

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

Now:

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

So:

$\displaystyle Q(t)=280.1165 e^{0.356675t}$
• Mar 14th 2006, 09:22 AM
guess
Thanks Bro.. you got it!!
• Mar 14th 2006, 09:22 AM
CaptainBlack
Quote:

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:

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

At the beginning of the fourth day the population is:

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

At the beginning of the first day the population is:

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

RonL
• Mar 17th 2006, 10:49 AM
ThePerfectHacker
Sorry bout that, the phrase "... an increase of..." got me.