MA2051 - Ordinary Differential Equations
Solutions to first sample exam
Originally Given 1994 B Term
You are solving
The corresponding homogeneous equation is
The characteristic equation is r + 0.5 = 0, and so r=-0.5
and the homogeneous solution is .
Use the method of Undetermined Coefficients and look for a solution of
. Plug this into the original
equation and equate coefficients to obtain
A = 10/37 and B = -60/37.
The general solution is .
The initial condition gives you 32 = T(0) = C - 60/37 and
so C = 32 + 60/37.
The answer is yes: simply choose .
(You just have to make the initial data match the initial
value for .)
(a) Separate and integrate:
The initial data determines the constant to be 26. (The solution is
the top half of an ellipse.)
(b) Separate and integrate again:
Use the initial data and simplify to obtain
y = 1/(12-4t)
Let A(t) denote the amount of TCE in the tank at time t (in ounces)
and let C(t) denote the concentration of TCE in the tank at time t
(in ounces per gallon). Then C(t) = A(t)/V = A(t)/10,000.
TCE enters the tank at rate (ounces/minute).
There are two ways that TCE ``leaves'' the tank: in the water flowing
out and through treatment.
The rate out due to flow is
The rate out due to treatment is (ounces/minute).
The balance equation is
Collect terms, divide by and take the limit as
approaches zero to obtain the model:
If you are going to have a constant solution, then
and the model tells you that this happens when 1000 - 0.17A(t) = 0.
Thus the amount of TCE in the tank approaches the constant level
ounces and so the concentration
approaches the constant level
(ounces per gallon).
This is the concentration of TCE in the water flowing out of the tank.
There are (at least) three ways to reduce the concentration of TCE in
the water leaving the tank: reduce the flow rate,
increase the size of the tank, increase the efficiency of the
If you go back to the model, and let denote the flow rate in
(as well as out), let V denote the volume, and r denote the
treatment rate, the model becomes
You can solve for the constant solution as before, and you find
that the concentration approaches the constant level
In other words, the treatment tank multiplies the in-flow concentration by
the fraction . You want to make this fraction small
and you can do that by (1) decreasing , or (2) increasing V,
or (3) increasing r.
If you want to stay in business, you need
One way to attain this is to
keep V=10,000, r = 0.12 and decrease the flow rate . If you
do this you find that the maximum admissible flow rate is about 30.77
gallons per minute. (Slow flow.)
Another way to do it is to keep the flow rate equal to 500 gallons/minute,
but increase the volume of the tank. If you want to do this, you need a tank
with volume 162,500 gallons. (Big tank.)
It is interesting to note that you cannot satisfy the government
regulations even if the treatment is 100% efficient; r=1 isn't
© 1997 by MA2051 Webmasters
All rights Reserved. File last modified on January 31, 1997