MA2051 - Ordinary Differential Equations
Solutions to first sample exam

Originally Given 1994 B Term

1. You are solving


The corresponding homogeneous equation is tex2html_wrap_inline59 . The characteristic equation is r + 0.5 = 0, and so r=-0.5 and the homogeneous solution is tex2html_wrap_inline65 .
Use the method of Undetermined Coefficients and look for a solution of the form tex2html_wrap_inline67 . Plug this into the original equation and equate coefficients to obtain A = 10/37 and B = -60/37.
The general solution is tex2html_wrap_inline73 . The initial condition gives you 32 = T(0) = C - 60/37 and so C = 32 + 60/37.
The answer is yes: simply choose tex2html_wrap_inline79 . (You just have to make the initial data match the initial value for tex2html_wrap_inline81 .)

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.

TCE enters the tank at rate tex2html_wrap_inline99 (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 tex2html_wrap_inline101 (ounces/minute). The rate out due to treatment is tex2html_wrap_inline103 (ounces/minute).

The balance equation is


Collect terms, divide by tex2html_wrap_inline107 and take the limit as tex2html_wrap_inline107 approaches zero to obtain the model:


If you are going to have a constant solution, then tex2html_wrap_inline113 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 tex2html_wrap_inline117 ounces and so the concentration approaches the constant level tex2html_wrap_inline119 (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 treatment process.

If you go back to the model, and let tex2html_wrap_inline121 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 tex2html_wrap_inline131 . You want to make this fraction small and you can do that by (1) decreasing tex2html_wrap_inline121 , 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 tex2html_wrap_inline121 . 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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