# Thread: Models of separable equations (population growth, half life, interest)

1. ## Models of separable equations (population growth, half life, interest)

Hey all,

I'm working the odd problems in section 3.1 of Christian Constanda's "Differential Equations: A Primer for Scientists and Engineers" ISBN 978-1-4614-7296-4. It happens to present a number of problems that do not show up in Google, and I'm having trouble in it. There's also no math lab/free tutoring at my university. So I figured I would start putting together some stuff for Googlers to land on - with your help. All italic text is a direct quotation from page 44 of the aforementioned text.

I have two incorrect problems and one correct problem, in that order. So I do need help!

Here are the problems:

1. "The birth and death rates of two non-competing populations are Beta_1 = 1/(t+1), delta_1 = 1/10 and Beta_2 = 2/(t+2), delta_2 = 1/10 respectively. If their starting sizes are p_10 = 2 and p_20 = 1 units, find out how long it takes the second population to 1. become larger than the first one, and (ii) be twice as large as the first population."

The accepted answer is t=2(1+sqrt(2)) ~ 4.83 and t=2(3+2*sqrt(3)) ~ 12.93

3. "The half life of a radioactive isotope is 8. Compute the initial amount (number of atoms) of substence if the amount is 6000 at time t=4"

The accepted answer is N_0 = 6000*sqrt(2) ~ 8485

5. "A sum of money is invested at an annual interest rate of 3%. Two years later, an equal sum is invested at an annual rate of 5%. Determine after how many years the second investment starts being more profitable than the first one."

The accepted answer is t=3 years.

My work:

1.
P(t) = P_0 * e^((beta)-(delta)*t) (This equation is given in the book)
P_1 = 2*e^((1/(t+1) - 1/10)*t)
P_2 = 1*e^((2/(t+2) - 1/10)*t)

( http://www.wolframalpha.com/input/?i=2*e^%28%281%2F%28t%2B1%29+-+1%2F10%29*t%29+%3D+1*e^%28%282%2F%28t%2B2%29+-+1%2F10%29*t%29 )

P_1 = P_2 should be when one population surpasses another, since P_2 is growing faster than P_1 this should be the critical point.

However, I have set this up incorrectly.

3.
Let k = half life. Half life is decreasing so constant k is negative.
Let t = elapsed time.
Let N_0 = the number of atoms at t=0
Let N(t) = the number of atoms total.

Given equations:
1/2 * N_0 = N_0 * e^(-k*t)
N(t) = N_0 * e^(-k*t)

My work:
N(4) = 6000
6000 = N_0 * e^(-8*4)
N_0 = 6000/(e^(-8*4))

This is not equal to 6000 * sqrt(2). My issue appears to be in my setup.

5.
Let S_0 be the starting amount in the account.
Let k be the interest rate.
Let t be the time elapsed.
Let S(t) be the current amount in the account.

S(t) = s_0*e^(k*t) (given equation)

S(t)_1 = s_0*e^(.03*2 + .03*t) (We're starting with two years elapsed)
S(t)_2 = s_0*e^(.05*t)

Setting S(t)_1 = S(t)_2,

e^(.03*2 + .03*t) = e^(.05*t), t = 3.

2. ## Re: Models of separable equations (population growth, half life, interest)

Originally Posted by AnotherGeek
Hey all,

I'm working the odd problems in section 3.1 of Christian Constanda's "Differential Equations: A Primer for Scientists and Engineers" ISBN 978-1-4614-7296-4. It happens to present a number of problems that do not show up in Google, and I'm having trouble in it. There's also no math lab/free tutoring at my university. So I figured I would start putting together some stuff for Googlers to land on - with your help. All italic text is a direct quotation from page 44 of the aforementioned text.

I have two incorrect problems and one correct problem, in that order. So I do need help!

Here are the problems:

1. "The birth and death rates of two non-competing populations are Beta_1 = 1/(t+1), delta_1 = 1/10 and Beta_2 = 2/(t+2), delta_2 = 1/10 respectively. If their starting sizes are p_10 = 2 and p_20 = 1 units, find out how long it takes the second population to 1. become larger than the first one, and (ii) be twice as large as the first population."

The accepted answer is t=2(1+sqrt(2)) ~ 4.83 and t=2(3+2*sqrt(3)) ~ 12.93

3. "The half life of a radioactive isotope is 8. Compute the initial amount (number of atoms) of substence if the amount is 6000 at time t=4"

The accepted answer is N_0 = 6000*sqrt(2) ~ 8485

5. "A sum of money is invested at an annual interest rate of 3%. Two years later, an equal sum is invested at an annual rate of 5%. Determine after how many years the second investment starts being more profitable than the first one."

The accepted answer is t=3 years.

My work:

1.
P(t) = P_0 * e^((beta)-(delta)*t) (This equation is given in the book)
P_1 = 2*e^((1/(t+1) - 1/10)*t)
P_2 = 1*e^((2/(t+2) - 1/10)*t)
I am confused. What you show as the "equation ... given in the book" would be correct if "beta" and "delta" were constants- but "beta" is not constant in this problem. A complete solution to this problem, as given, would involve solving a "separable differential equation".

( http://www.wolframalpha.com/input/?i=2*e^%28%281%2F%28t%2B1%29+-+1%2F10%29*t%29+%3D+1*e^%28%282%2F%28t%2B2%29+-+1%2F10%29*t%29 )

P_1 = P_2 should be when one population surpasses another, since P_2 is growing faster than P_1 this should be the critical point.

However, I have set this up incorrectly.

3.
Let k = half life. Half life is decreasing so constant k is negative.
Let t = elapsed time.
Let N_0 = the number of atoms at t=0
Let N(t) = the number of atoms total.

Given equations:
1/2 * N_0 = N_0 * e^(-k*t)
N(t) = N_0 * e^(-k*t)
Personally, for "half life" problems I prefer "If the half life is T then N(t)= N_0*(1/2)^{t/T}". That is, t/T counts the number of "half lives" and the amount halves for each of those.
That can be converted to e^{-kt} by using "x= e^ln(x)" as (1/2)^{t/T}= e^ln((1/2)^{t/T}= e^(t/T)ln(1/2) so that "-k= ln(1/2)/T" (since 1/2< 1, ln(1/2) is negative).

My work:
N(4) = 6000
6000 = N_0 * e^(-8*4)
N_0 = 6000/(e^(-8*4))

This is not equal to 6000 * sqrt(2). My issue appears to be in my setup.
Your error is in thinking that k is the half life. It is not that easy. Using that set up, knowing that the half life is 8, you know that for given N(0),
N(8)= N(0)e^{-k(8)}= (1/2)N_0 (i.e. half as much)
Dividing both sides by N_0, e^{-8k}= 1/2. Taking the ln of both sides, -8k= ln(1/2)= -0.6931 (to four decimal places) so k= -0.6931/8= -.0866.
Solve 6000= N_0e^{-.0866(4)}

Or, using my formula, with half life T= 8, 6000= N_0(1/2)^{4/8}= N_0(1/2)^(1/2) so that N_0= 6000/(1/2)^(1/2)

5.
Let S_0 be the starting amount in the account.
Let k be the interest rate.
Let t be the time elapsed.
Let S(t) be the current amount in the account.

S(t) = s_0*e^(k*t) (given equation)

S(t)_1 = s_0*e^(.03*2 + .03*t) (We're starting with two years elapsed)
S(t)_2 = s_0*e^(.05*t)

Setting S(t)_1 = S(t)_2,

e^(.03*2 + .03*t) = e^(.05*t), t = 3.
Again, your error is in thinking that the "k" in the formula is the interest rate. With interest rate "r", an initial amount "S" will have increased to S+ rS= (1+ r)S in one year. So we would have to have Se^{k(1)}= Se^k= (1+ r)S so that e^k= 1+ r. From that k= ln(1+ r), not r. If r= .03, k= ln(1.03)= 0.02956 while if r= .05, k= ln(1.05)= 0.04879.