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.