Math 293 Prelim 1 Solutions and regrades. 10/3/96
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1) REGRADE POLICY. If you think your prelim has been graded
improperly:
a) Don't put any marks on your prelim. Photocopy the
relevant page(s) if you need to make marks.
b) Write a note about why you think points should be
add or subtracted for a given problem and why.
Attach this note to your prelim with a paperclip
(put your name on the note).
c) turn this in to your TA at your section on
October 11 or 12 (or some time sooner).
2) The 1 page solutions to prelim 1 passed out in lecture
have an arithmetic error in problem 2b. The mistake happens at the
calculation of x1 and propagates to the end. Problem 2b should
read
x0 = 5
x1 = 5 + .5 * (2*5 + 1) = 10.5 (not 5.55)
x2 = 10.5 + .5 * (2*10.5 + 1) = 21.5 (not 11.6 )
^^^^
3) Rough reasoning for problem bonus problem 4.
Given y'=sqrt(y), y(0)=0, y(3)=1: any solutions?
Note that for any point with y>0 there exists a unique solution
through that point.
For y=0 the uniqueness criteria is not met and there is the
possibility of multiple solutions.
In fact y=( (xc)^2 )/4 (*) satisfies the ode for all c (found by
separation of variables).
Also y=0 satisfies the ODE (found by 'inspection').
Any solution that passes through the point (x,y) = (3,1) is
therefor unique until it hits the x axis.
y = ((x1)^2)/4 is such a solution but it is not unique
once it touches the x axis at (x,y) = (1,0) where is has a
slope of zero.
The function y=0 satisfies the ODE and goes through both
the point (0,0) and the point (1,0).
By pasting the solutions together, namely
_
y =  0 for 0 <= x <= 1 (**)
 ((x1)^2)/4 for 1 <= x <= 3

you get a curve that is continuous, has a continuous first
derivative, satisfies the ODE everywhere, and goes through the
desired points.
Are there more such solutions? No. In the range 1