1. ## Growth/Decay problem

I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let $\displaystyle P(t)$ denote the proportion of the population that is infected. Then

$\displaystyle \frac{dP}{dt} = k(1-P)$

(i) Solve the differential equation to get an expression for P(t).
(ii) Find the value of the constant k.
(iii) How long will it take for 75% of the population to become infected?

My attempt:

(i)

$\displaystyle \integral \frac{dP}{dt} = \integral k(1-P)$

$\displaystyle dP=k(1-P)~dt$

$\displaystyle \int \frac{1}{1-P}~dP = \int k~dt$

$\displaystyle -ln|1-P| = kt~dt$

$\displaystyle -1+P = e^{kt}$

$\displaystyle P(t) = e^{kt} +1$

This seems wrong as at $\displaystyle P(0)$, no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

Also, why don't I have a coefficient in front of the $\displaystyle e^{kt}$ term? I have also never seen a constant next to the normal $\displaystyle P_0e^{kt}$ term; is that normal?

I will wait till we have figured out part (i) before I attempt the next two parts.

Thanks.

2. ## Re: Growth/Decay problem

Originally Posted by terrorsquid
I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let $\displaystyle P(t)$ denote the proportion of the population that is infected. Then

$\displaystyle \frac{dP}{dt} = k(1-P)$

(i) Solve the differential equation to get an expression for P(t).
(ii) Find the value of the constant k.
(iii) How long will it take for 75% of the population to become infected?

My attempt:

(i)

$\displaystyle \integral \frac{dP}{dt} = \integral k(1-P)$

$\displaystyle dP=k(1-P)~dt$

$\displaystyle \int \frac{1}{1-P}~dP = \int k~dt$

$\displaystyle -ln|1-P| = kt~dt$

$\displaystyle -1+P = e^{kt}$

$\displaystyle P(t) = e^{kt} +1$

This seems wrong as at $\displaystyle P(0)$, no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

Also, why don't I have a coefficient in front of the $\displaystyle e^{kt}$ term? I have also never seen a constant next to the normal $\displaystyle P_0e^{kt}$ term; is that normal?

I will wait till we have figured out part (i) before I attempt the next two parts.

Thanks.
When integrating you have to include the arbitrary constant of integration '+ C'. Get its value from the initial condition.

3. ## Re: Growth/Decay problem

Is this what you mean?

$\displaystyle -ln|1 -P| = kt + C$

$\displaystyle -1+P=e^{kt}e^{C}$

$\displaystyle let~e^C = P_0$

$\displaystyle P(t) = P_0e^{kt}+1$

$\displaystyle P(0) = P_0 + 1$

Not sure how to proceed... Initially 0 are infected, so that means $\displaystyle P_0 = -1 ?$

4. ## Re: Growth/Decay problem

$\displaystyle \int \frac{dP}{1-P} = \int k \, dt$

$\displaystyle \ln|1-P| = -kt + C$

$\displaystyle 1 - P = e^{-kt+C}$ ... let $\displaystyle A = e^C$

$\displaystyle 1 - P = Ae^{-kt}$

$\displaystyle P = 1 - Ae^{-kt}$

when $\displaystyle t = 0$ , $\displaystyle P = 0$ (which is very unrealistic for a problem like this)

$\displaystyle A = 1$

$\displaystyle P = 1 - e^{-kt}$

at $\displaystyle t = 10$, $\displaystyle P = 0.25$

$\displaystyle 0.25 = 1 - e^{-10k}$

solve for $\displaystyle k$ and finish it.

5. ## Re: Growth/Decay problem

Just out of curiosity, how did you spot to switch the sides of kt and ln|1-P|? After considering the situation presented by the question, it seems we are dealing with an exponential growth problem which would suggest I need to find a formula with a +k value. Typically I would try to get it into the form of +k for growth and -k for decay. I can see now with the final answer that it makes sense, but how would you spot it before getting to the end?

6. ## Re: Growth/Decay problem

Originally Posted by terrorsquid
Just out of curiosity, how did you spot to switch the sides of kt and ln|1-P|? After considering the situation presented by the question, it seems we are dealing with an exponential growth problem which would suggest I need to find a formula with a +k value. Typically I would try to get it into the form of +k for growth and -k for decay. I can see now with the final answer that it makes sense, but how would you spot it before getting to the end?
$\displaystyle \int \frac{1}{1-P} \, dP = \int k \, dt$

$\displaystyle \int \frac{-1}{1-P} \, dP = \int -k \, dt$

... since the derivative of (1-P) is -1

7. ## Re: Growth/Decay problem

So, whenever I have a problem where the term being multiplied by the constant k in the instantaneous growth/decay rate equation: $\displaystyle \frac{dP}{dt}=k(term)$
is negative, I look to find a P(t) term with a -k exponent and vice-versa, regardless of whether the question is about growth or decay?

Continuing:

(i)

$\displaystyle 0.25 = 1-e^{-10k}$

$\displaystyle -0.75 = -e^{-10k}$

$\displaystyle e^{-10k}=0.75$

$\displaystyle -10k = ln(0.75)$

$\displaystyle k = 0.02876820725$

(ii)

$\displaystyle 0.75 = 1-e^{-0.02876820725t}$

$\displaystyle t = \frac{ln(0.25)}{-0.0287682072}$

$\displaystyle =48.19~days$

8. ## Re: Growth/Decay problem

you're asking me "do I always?" ... understand that I cannot say yes. each problem is different. I just took advantage of being able to multiply both sides by -1 to make the antiderivative easier to find.