MA2051 - Ordinary Differential Equations
Solutions to Second Sample Exam

Originally Given 1996 D Term

(a) Just integrate:

and use the initial data:


integrate and simplify to obtain

Use the initial data to find C:

2. You are solving

The corresponding homogeneous equation is . The characteristic equation is , and so and the homogeneous solution is .
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 = 2 and B = -4.
The general solution is . The initial condition gives you and so C = 36.
The answer is yes: simply choose . (You just have to make the initial data match the initial value for .)

3. Let denote the amountof carbon monoxide in the lecture hall at time t (in cubic meters) and let denote the concentrationof carbon monoxide in the hall at time t (in cubic meters per cubic meter). Then .

CO enters the hall at rate (/minute).

CO leaves the hall at the rate (/minute).

The balance equation is Net change in the amount between t and = (rate in) (rate out) ,


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

To turn this into an equation for the concentration, replace by everywhere that you see it.

4. Sketch the graph of . The differential equation tells you

when (or where)



The last condition tells you that the inside temperature can have a max or a min only when the the solution curve crosses the graph of . It also tells you that cannot follow exactly (because would have to be zero everywhere along the solution).

For the initial data given starts at 0 and stays strictly below until the curves cross (at some time after when is strictly less than 10). This shows that the amplitude of th inside temperature is smaller than the amplitude of the outside temperature.


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