The sample exam problems are meant to give you an idea of the
typed and difficulty of the exam problems. Studying them exclusively
is definitely NOT a good idea--you should be doing many problems. The
Chapter Review problems at the end of chapter 1 are a good place to
start.
Reading solutions does you almost nothing--you only get better at problems
by doing them. Several people have asked for solutions to the sample
problems. It is much better to do the problems yourselves, compare answers,
discuss, ask me and/or Eric if you are completely stuck. In particular,
if you are asking "is this right?" a great deal, then you should
do a more basic review--the nice thing about math is if you do
the right steps, you get the right answer (not like life).
Also, it is a very good idea to think about "related" problems.
These won't be the problems on the exam, but what similar questions
could I ask?
With that disclaimer, here are some suggestions on the sample
exam problems.
- 1. We did this one completely in class Tuesday.
- 2. We have done many of these (and there are many more in the
text). For part (a), you can say that all solutions "come together"
as t gets large to the particular solution which is periodic
(you should be able to
say why.) You can say the period of the particular solution is
2pi/5 (and say why). You can give a bound on the amplitude of
the particular solution (in this case 2/3--Why? (look at slope field)).
Remark: In ENG you will say all solutions approach the "steady state"
solution. The steady state solution is what we call the particular solution.
"Steady state" in this context does NOT mean constant--it just means all
solutions approach regardless of initial value.
For part (b) you just calculate--you can check your own answer by
differentiating.
For part(c) you can now give a better estimate of amplitude of the
"steady state" (we will give ways to compute the exact amplitude
later).
- 3. This is a trick question--the solution with y(0)=0 is an equilibrium
so Euler's method works perfectly...However, this is a good point to think
about what I could ask. What if y(0)=1? Then note that Euler's method
follows tangent lines, but the solutions curves are increasing (dy/dt>0)
so concave up (y^3 +ty^5 is increasing in y and t). So the tangent lines
are below the actual solution...(if in doubt, draw the Euler approximation
and the actual solution carefully). What if $y(0)=-1$?
- 4.(a) Remember, sources the arrows point out, sinks, they point in.
(b) Remember, arrow up means f(y)>0. Arrow down means f(y)<0.
(c) You don't know how positive or how negative f(y) is...
- 5.(a) Don't try to find the general solution--we have no tools
for this. However, you can look for very simple solutions (equilibrium
solutions)--Are there any equilibrium solutions?
(b) The only tool we have is the uniqueness theorem, so use that
(we can't say more because the equation is not autonomous).
- 6. (a) This is separable. You can check your own solution by
differentiating.
(b) Solve for the constant so that y(0)=1...
(c) Do bad things happen at y=0?
- (a) S(t)=sugar in oz, t=time minutes (don't forget to say this),
dS/dt = rate in (oz/min) - rate out (oz/min) . Rate in is 5 times 0.3,
the rate out is 0.3 times S/5 (Why--check units).
(b) Find the general solution. The equation is both separable and linear,
so take your pick of methods--you can check your own answer (and should!)
by differenitating.
(c) Here the volume changes--Volume =5-0.2t because more is going out
than coming in... so the rate out is now .5*S/(5-0.2t)
(d) If you have looked at section 1.9, give this a try...if not, don't
worry about it.
- 8. Draw the graph of y^2 -y^4/4 . Adding or subracting a constant
moves the graph up and down--as you slide the graph up and down, draw
the phase line in each case (move to the right as the value of a increases.)
(a=0 has 3 equilibria and a=0 is a bifurcation value). You can
compute the other bifurcation values by computing the maximum values
of y^2-y^4/4-- there is one more bifurcation value of a and it is
negative.
<9> Have fun with this one...