Math 282
Quiz
4 Name: Solutions
October 15, 2009
1. Consider the vector field to the right, in which x and y represent the populations of 2 species (in 1000’s).
a) Explain: What happens to each of the populations in the short run? In the long run?
In the short run, both
populations are decreasing, with y decreasing more rapidly than x.
In the long run, x
eventually rebounds and starts to grow again, why y heads to extinction.
(For more detail on x’s growth, see the note below, in part (b) )
b) Sketch a graph of x vs. time based on the vector field
(Note: The graph was intended to show that the
solution curve stops at (10,0), in which case x would start at 6, drop to a
little below 3, then rise to 10 before leveling off. Since it was not very clear that the graph
did not keep going, I accepted drawings that started as I just described, but
did not have x leveling off, but rather had it keep growing).

2. Use two steps of Euler’s method with step size 0.1 to approximate Y(.2) for the system
x'=x+y, y’=3y, x(0) = 3, y(0) = 2. Show all work.
|
t |
x |
y |
x' |
|
|
y' |
|
|
0 |
3 |
2 |
5 |
.1 |
.5 |
6 |
.6 |
|
.1 |
3.5 |
2.6 |
6.1 |
.1 |
.61 |
7.8 |
.78 |
|
.2 |
4.11 |
3.38 |
|
|
|
|
|
3. Solve the IVP from #2.
The 2nd
equation ha solution y=2e3t.
The 1st equation then becomes x’=x+2e3t, or x’-x
=2e3t .
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