# MA2051 - Ordinary Differential EquationsSolutions to first sample examOriginally Given 1994 B Term

1. You are solving

(a)
The corresponding homogeneous equation is . The characteristic equation is r + 0.5 = 0, and so r=-0.5 and the homogeneous solution is .
(b)
Use the method of Undetermined Coefficients and look for a solution of the form . Plug this into the original equation and equate coefficients to obtain A = 10/37 and B = -60/37.
(c)
The general solution is . The initial condition gives you 32 = T(0) = C - 60/37 and so C = 32 + 60/37.
(d)
The answer is yes: simply choose . (You just have to make the initial data match the initial value for .)

2. (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)

3. 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.

(a)
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 (ounces/minute). The rate out due to treatment is (ounces/minute).

(b)
The balance equation is

Collect terms, divide by and take the limit as approaches zero to obtain the model:

(c)
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.

(d)
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 treatment process.

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.

4: BONUS 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 good enough!

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