# RATE OF CHANGE differential equation

• Aug 7th 2011, 10:52 AM
NFS1
RATE OF CHANGE differential equation
Hi, need help answering a question in a past exam paper. Below is the question:

5 mice in a stable population of 500 are intentionally infected with a disease. The rate of change in the infected population is proportional to the product of the number of mice who have the disease with the number that are disease free. If N(t) denotes the number of mice with the disease at time, t, the theory predicts that:

dN/dt = kN(500 - N)

where k is a constant. Show that the time taken for half the population to contract the disease is

t= 0.00919/k

I have substitued N=5 to give dN/dt = k x 5 x (495) but i dont understand what to do next.
• Aug 7th 2011, 11:05 AM
e^(i*pi)
Re: RATE OF CHANGE differential equation
You are working with the rate of change (dN/dt) rather than the population itself (N). You need to integrate this equation - this one is separable so it's not too hard as far as DEs go.

$\displaystyle \int \dfrac{dN}{N(500-N)} = \int k dt$

I suggest using partial fractions to integrate the LHS
• Aug 7th 2011, 11:10 AM
MSUMathStdnt
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by NFS1
Hi, need help answering a question in a past exam paper. Below is the question:

5 mice in a stable population of 500 are intentionally infected with a disease. The rate of change in the infected population is proportional to the product of the number of mice who have the disease with the number that are disease free. If N(t) denotes the number of mice with the disease at time, t, the theory predicts that:

dN/dt = kN(500 - N)

where k is a constant. Show that the time taken for half the population to contract the disease is

t= 0.00919/k

I have substitued N=5 to give dN/dt = k x 5 x (495) but i dont understand what to do next.

Don't substitute yet. $\displaystyle N(t)=5$ only when $\displaystyle t=0$ (that is, $\displaystyle N(0)=5$).

Try to solve the original DE for $\displaystyle N(t)$ (which may not be so easy - I'm learning this stuff, too):

$\displaystyle \frac {dN}{dt} = k N (500 -N) = 500 k N - k N^2$.

Once you've solved that, you will have an equation for $\displaystyle N(t)$. Then, just plug $\displaystyle N(t)=250$ into the equation and solve for $\displaystyle t$.
• Aug 7th 2011, 11:48 AM
NFS1
Re: RATE OF CHANGE differential equation
Can someone solve this step by step so i can see how its done?

Using MSUMathStdnt method i got K=0.00202020...what do i do from here?

Sorry to sound like a novice but I have no examples to look at or too learn from which is why i am finding it difficult.
• Aug 7th 2011, 11:58 AM
e^(i*pi)
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by NFS1
Can someone solve this step by step so i can see how its done?

Using MSUMathStdnt method i got K=0.00202020...what do i do from here?

Sorry to sound like a novice but I have no examples to look at or too learn from which is why i am finding it difficult.

You're not asked to solve for k so why bother? What don't you understand about integrating by partial fractions?

$\displaystyle \dfrac{dN}{N(500-N)} = \dfrac{A}{N} + \dfrac{B}{500-N}$

Spoiler:
$\displaystyle 1 = A(500-N) + BN$

Let $\displaystyle N = 0$ - giving us: $\displaystyle 1 = 500A \Leftrightarrow A = \dfrac{1}{500}$

Let $\displaystyle N = 500$ giving: $\displaystyle 1 = 500B \Leftrightarrow B = \dfrac{1}{500}$

$\displaystyle \dfrac{dN}{N(500-N)} = \dfrac{1}{500N} + \dfrac{1}{500(500-N)}$

Using partial fractions we arrive at

$\displaystyle \int \dfrac{dN}{N(500-N)} = \dfrac{1}{500} \left(\int \dfrac{dN}{N} + \int \dfrac{dN}{500-N}\right)$

These are much simpler to integrate. It seems surprising that you're doing differential equations without covering integration by partial fractions
• Aug 7th 2011, 12:20 PM
NFS1
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by e^(i*pi)
You're not asked to solve for k so why bother? What don't you understand about integrating by partial fractions?

$\displaystyle \dfrac{dN}{N(500-N)} = \dfrac{A}{N} + \dfrac{B}{500-N}$

Spoiler:
$\displaystyle 1 = A(500-N) + BN$

Let $\displaystyle N = 0$ - giving us: $\displaystyle 1 = 500A \Leftrightarrow A = \dfrac{1}{500}$

Let $\displaystyle N = 500$ giving: $\displaystyle 1 = 500B \Leftrightarrow B = \dfrac{1}{500}$

$\displaystyle \dfrac{dN}{N(500-N)} = \dfrac{1}{500N} + \dfrac{1}{500(500-N)}$

Using partial fractions we arrive at

$\displaystyle \int \dfrac{dN}{N(500-N)} = \dfrac{1}{500} \left(\int \dfrac{dN}{N} + \int \dfrac{dN}{500-N}\right)$

These are much simpler to integrate. It seems surprising that you're doing differential equations without covering integration by partial fractions

ok so ive got...1/500 (ln(n) + ln(500-n))...now do i substitute in n=5??
• Aug 7th 2011, 01:41 PM
MSUMathStdnt
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by NFS1
ok so ive got...1/500 (ln(n) + ln(500-n))...now do i substitute in n=5??

You've only integrated the left hand side. Now, integrate on the right hand side. Remeber RHS will have a constant, C.

Use (t,N)=(0,5) in the integrated equation to determine the value of C. Then you can substitute N=250 to find t when half the population is infected.

If my math is correct t=11.02729384/k. Maybe e^{i*pi) (should we call you "One") can confirm (better yet, see if you can confirm).
• Aug 7th 2011, 03:44 PM
NFS1
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by MSUMathStdnt
You've only integrated the left hand side. Now, integrate on the right hand side. Remeber RHS will have a constant, C.

Use (t,N)=(0,5) in the integrated equation to determine the value of C. Then you can substitute N=250 to find t when half the population is infected.

If my math is correct t=11.02729384/k. Maybe e^{i*pi) (should we call you "One") can confirm (better yet, see if you can confirm).

ok so:

>> 1/500 (ln n - ln 500-n) = kt + C
>> 1/500 (ln 5 - ln 495) = kt + C
>> -0.00919 = kt + C
>> -0.00919/k = t ... but i have a -ve but in the question it says half the population with disease must equate to t=0.00919/k

so wat am i still doing wrong??
• Aug 7th 2011, 04:01 PM
MSUMathStdnt
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by NFS1
ok so:

>> 1/500 (ln n - ln 500-n) = kt + C
>> 1/500 (ln 5 - ln 495) = kt + C
>> -0.00919 = kt + C
>> -0.00919/k = t ... but i have a -ve but in the question it says half the population with disease must equate to t=0.00919/k

so wat am i still doing wrong??

First, you should be adding ln 5 to ln 495 (you're subtracting).
Second, you should be solving for C. Given that N=5 when t=0, you can solve for C.

Once you have C, only then do you substitute N=250 to find a value of t. You solve for t until you know C.
• Aug 8th 2011, 03:41 AM
NFS1
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by MSUMathStdnt
First, you should be adding ln 5 to ln 495 (you're subtracting).
Second, you should be solving for C. Given that N=5 when t=0, you can solve for C.

Once you have C, only then do you substitute N=250 to find a value of t. You solve for t until you know C.

again it does not work out...

>>1/500 (ln5 + ln495) = k(0) + C
>>0.01562799 = C
so now i sub in 250 this time
>>1/500 (ln250 + ln250) = kt + 0.01562799
>>0.02200858 = kt + 0.01562799
>>t = 0.00645785/K

this is not the answer i should be getting as in the question it says i should get t=0.00919/K
• Aug 8th 2011, 09:55 AM
Siron
Re: RATE OF CHANGE differential equation
You got:
$\displaystyle \frac{1}{500}\left(\int \frac{dN}{N}+\int \frac{dN}{500-N}\right)=\int k\cdot dt$
$\displaystyle \Leftrightarrow \frac{1}{500}\left(\ln|N|-\ln|500-N|\right)=k\cdot t$
$\displaystyle \Leftrightarrow \frac{1}{500}\left(\ln\left|\frac{N}{500-N}\right|\right)=k\cdot t$

Enter $\displaystyle N=5$ ...
• Aug 8th 2011, 01:09 PM
NFS1
Re: RATE OF CHANGE differential equation
Quote:

Originally Posted by Siron
You got:
$\displaystyle \frac{1}{500}\left(\int \frac{dN}{N}+\int \frac{dN}{500-N}\right)=\int k\cdot dt$
$\displaystyle \Leftrightarrow \frac{1}{500}\left(\ln|N|-\ln|500-N|\right)=k\cdot t$
$\displaystyle \Leftrightarrow \frac{1}{500}\left(\ln\left|\frac{N}{500-N}\right|\right)=k\cdot t$

Enter $\displaystyle N=5$ ...

yes ofcourseeeee....ln5-ln500-5 is ln5/ln500-5 and since its modulus there is no negatives!! How could i forget. Your a genius mate.