1. ## differential equation

Population increased from 2mill in year 2000 to 3mill in 2002.Rate of increase of population directly proportional to number of people present,predict the population in 2004.

$\displaystyle N(t)=N(0)e^{kt}$ .. this is the general form i take it?

then N(2)=3 and N(0)=2

and now im lost as ever, i have the answer i just dont understand the method shown. thanks alot.

2. Yes that is the general form but it has to be shown from the basics of the question;

"Rate of increase of population directly proportional to number of people present"

Thus we can say,

$\displaystyle \frac{dN}{dt}~\propto~N$

$\displaystyle \implies~\frac{dN}{dt}~=~kN$

Now this is a straight forward "seperate the variables and integrate" differential equation which gets to that general form.

Then sub in the conditions $\displaystyle N(2)~=~3~~,~~N(0)~=~2$ to work out the unknown constants: $\displaystyle k$ and the constant of integration which rightly said is $\displaystyle N(0)~=~2$ and from that sub in $\displaystyle t~=~4$ to get the population in 2004.

3. Originally Posted by skystar
Population increased from 2mill in year 2000 to 3mill in 2002.Rate of increase of population directly proportional to number of people present,predict the population in 2004.

$\displaystyle N(t)=N(0)e^{kt}$ .. this is the general form i take it?

then N(2)=3 and N(0)=2

and now im lost as ever, i have the answer i just dont understand the method shown. thanks alot.
Hello skystar,

Since the change in population is directly proportional to the number of people we get the equation.

$\displaystyle \frac{dN}{dt}=kN$

Where N is the poplulation and k is the constant of proportionality.

To solve this ODE we seperate the variables and integrate...

$\displaystyle \frac{dN}{dt}=kN \iff \frac{1}{N}dN=kdt \iff \int \frac{1}{N}dN= \int kdt$

$\displaystyle \ln|N|=kt+c \iff N=e^{kt+c}=e^{kt}\cdot e^{c}=Ae^{kt}$

Now we have $\displaystyle N(t)=Ae^{kt}$

Now we can use the inital condition to find A

$\displaystyle N(0)=2=Ae^{k(0)} \iff A=2$

$\displaystyle N(t)=2e^{kt}$ Now we can use the other condition to solve for k.

$\displaystyle N(2)=4=2e^{k(2)} \iff 4=2e^{2k} \iff 2=e^{2k} \iff \ln(2)=2k \iff k=\frac{1}{2}\ln(2)$

Now we end up with

$\displaystyle N(t)=2e^{\frac{t}{2}\ln(2)}=2[e^{\ln(2)}]^{\frac{t}{2}}=2[2]^{\frac{t}{2}}=2^{\frac{t+2}{2}}$

$\displaystyle N(t)=2^{\frac{t+2}{2}}$

$\displaystyle N(4)=2^{\frac{4+2}{2}}=2^3=8$

So the population would be 8 million.

4. my lecture has the answer down however as 4.5 million? N(2)=3 not 4,nevertheless thanks a million methods what matters

$\displaystyle N(0)e^{4k}=2(e^{2k})^{2}=2 \cdot (\frac{3}{2})^{2} =2 \cdot\frac{9}{4}$

5. Originally Posted by skystar
my lecture has the answer down however as 4.5 million?

$\displaystyle N(0)e^{4k}=2(e^{2k})^{2}=2 \cdot \frac{{3}{2}}^{2} =2 x\cdot\frac{9}{4}$
here is my error I used N(2)=4

$\displaystyle N(2)=4=2e^{k(2)} \iff 4=2e^{2k} \iff 2=e^{2k} \iff \ln(2)=2k \iff k=\frac{1}{2}\ln(2)$

It should be

$\displaystyle N(2)=3=2e^{k(2)} \iff 3=2e^{2k} \iff \frac{3}{2}=e^{2k}$
$\displaystyle \ln\left( \frac{3}{2} \right)=2k \iff k=\frac{1}{2}\ln\left( \frac{3}{2}\right)$

$\displaystyle N(t)=2e^{\frac{t}{2}\ln\left( \frac{3}{2}\right)}=2\left( \frac{3}{2}\right)^{\frac{t}{2}}$

Sorry I wrote the other condition down wrong