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MA2051--C97SOLUTIONS FOR TEST I 1. You are solving


The corresponding homogeneous equation is tex2html_wrap_inline51 . The characteristic equation is r +3 = 0, so r=-3 and the homogeneous solution is tex2html_wrap_inline57 .
Use the method of Undetermined Coefficients and look for a solution of the form tex2html_wrap_inline59 . Plug this into the original equation and equate coefficients to obtain A = 7, so tex2html_wrap_inline63 .
The general solution is tex2html_wrap_inline65 . The initial condition gives you 5 = y(0) = C + 7 and so C = -2.
The answer is yes: simply choose tex2html_wrap_inline71 (and remember that solutions are unique).

2. Let A(t) denote the amount of sugar (in Tbsp) in the tank at time t and let C(t) denote the concentration of sugar in the tank at time t (in tablespoons per gallon). Then C(t) = A(t)/V = A(t)/1000.

Sugar enters the tank at rate tex2html_wrap_inline83 (Tbsp/min).

Sugar leaves the tank at the rate tex2html_wrap_inline85 (Tbsp/minute). The assumption that the tank is ``well-mixed'' implies that the concentration of sugar is uniform throughout the tank.

The balance equation is Net change between t and tex2html_wrap_inline89

= (rate in) tex2html_wrap_inline93 (rate out) tex2html_wrap_inline95 ,



Collect terms, divide by tex2html_wrap_inline95 and take the limit as tex2html_wrap_inline95 approaches zero to obtain the model:


The equation has a steady state at tex2html_wrap_inline105 , so the concentration will be close to C = 22/3 = 7.333 if you wait long enough.

3. Just separate and integrate:


gives you


when you plug in the initial data.

The disaster occurs as t approaches 5/k, when the population becomes larger than the number of electrons in the universe.

4. The differential equation tells you that The steady states are y=+6 and y=+9, where tex2html_wrap_inline121 .

tex2html_wrap_inline123 when (or where) y(t) < 6 or y(t)> 9

tex2html_wrap_inline129 when 6 < y(t) < 9

Your graph starting at y=0 should increase, be concave down, and have a horizontal asymptote at y=6. The graph starting at y=12 should be increasing and concave up. The graph starting at y=8.9 should decrease, be concave down initially but switch to concave up (at y=7.5), and have a horizontal asymptote at y=6. Bonus Problem Define Q(t) = 1/P(t) and differentiate it:


but this simplifies to


This is linear for Q and the solution is


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Laura Jean Cooper
Thu Feb 6 17:34:47 EST 1997