1. ## differencial equations

1.A fruit fly population of 182 flies is in a closed container. The number of flies grows exponentially, reaching 340 in 18 days. Find the doubling time (time for the population to double) to the nearest tenth of a day.

19.9 days
20.0 days
18.0 days
2.0 days

2.One thousand dollars is invested at 5% continuous annual interest. This means the value of the investment will grow exponentially, with k equaling the decimal rate of interest. What will the value of the investment be after 7 1/2 years?
$1,375.00$375.00
$1,454.99$454.99

3.A population grows exponentially, doubling in 4 hours. How long, to the nearest tenth of an hour, will it take the population to triple?
8.0 hours
10.3 hours
6.3 hours
6.6 hours

4.The half-life of a newly discovered radioactive element is 30 seconds. To the nearest tenth of a second, how long will it take for a sample of 9 grams to decay to 0.72 grams?
109.3 seconds
42.5 seconds
30.8 seconds
99.7 seconds

5.
Find the particular solution to y dx = x dy given that y(8) = 2.

x = 4y
y = 4y
ln|x|=ln|y|+ln6
ln|x|=ln|y|+ln4

6.Which of the following is a first-order homogeneous differential equation?
dy/dx = (3x^3-2x^2y)/xy^2
dy/dx=(x^2y^3 - y^5)/(2xy^5)
dy/dx=(5xy-5y-5x)/y
none

2. Originally Posted by harry
1.A fruit fly population of 182 flies is in a closed container. The number of flies grows exponentially, reaching 340 in 18 days. Find the doubling time (time for the population to double) to the nearest tenth of a day.

19.9 days
20.0 days
18.0 days
2.0 days
The population t later is:

p(t) = p(0) e^{lambda t}

so:

340 = 182 e^{lambda 18}

18 lambda = ln(340/182) ~= 0.624939

lambda = 0.03471883.

For the doubling time t_2, we have:

364 = 182 e^{0.03471883 t_2}

so:

t_2 = ln(2)/0.03471883 ~=19.9646

So to the nearest 1/10th of a day the doubling time is 20 days.

RonL

3. Originally Posted by harry

4.The half-life of a newly discovered radioactive element is 30 seconds. To the nearest tenth of a second, how long will it take for a sample of 9 grams to decay to 0.72 grams?
109.3 seconds
42.5 seconds
30.8 seconds
99.7 seconds

Mass remaining after t seconds is:

M = M_0 2^(-t/30),

so for this problem we have:

0.72 = 9 2^(-t/30)

t = (-30) log_2(0.72/9) = (-1/30) ln(0.72/9)/ln(2) ~= 109.3

RonL

4. Originally Posted by harry

5.
Find the particular solution to y dx = x dy given that y(8) = 2.

x = 4y
y = 4y
ln|x|=ln|y|+ln6
ln|x|=ln|y|+ln4
The DE may be rewritten:

(1/x) dx = (1/y) dy

ln(|x|) = ln(|y|) + C,

Now as y(8) = 2, we have:

ln(8) = ln(2) + C,

so

C = ln(8) - ln(2) = ln(8/2) = ln(4)

hence:

ln(|x|) = ln(|y|) + ln(4).

Except as the DE has a singularity in dy/dx at x=0 and we cannot integrate
through a singularity this solution is only valid for x>0 (as the initial condition
is in this segment of the real line

RonL