differential equation

**skystar** · June 3rd 2008, 11:48 AM

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.

$N(t)=N(0)e^{kt}$ .. 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.

**Sean12345** · June 3rd 2008, 12:00 PM

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,

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

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

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

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

**TheEmptySet** · June 3rd 2008, 12:03 PM

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.

$N(t)=N(0)e^{kt}$ .. 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.

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

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

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

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

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

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

Now we can use the inital condition to find A

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

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

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

Now we end up with

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

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

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

So the population would be 8 million.

**skystar** · June 3rd 2008, 12:11 PM

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

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

**TheEmptySet** · June 3rd 2008, 12:17 PM

Originally Posted by skystar

my lecture has the answer down however as 4.5 million?

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

here is my error I used N(2)=4

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

It should be

$<br /> N(2)=3=2e^{k(2)} \iff 3=2e^{2k} \iff \frac{3}{2}=e^{2k}$
$\ln\left( \frac{3}{2} \right)=2k \iff k=\frac{1}{2}\ln\left( \frac{3}{2}\right)<br />$

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

Sorry I wrote the other condition down wrong

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