Math 44 – Differential Equations

Sunday, February4, 2007

Mini-quiz - SOLUTIONS

1. You are given a differential equation . Assume that f(t, y) is defined and continuous for all pairs .

Further, assume:

f ( t, y ) = +1 when y = t

f ( t, y ) = +1when y = t + 4

f ( t, y ) > +1 when y is strictly between t and t + 4

f ( t, y ) < +1when y is less than t or more than t + 4

a. Sketch a direction field for this equation. (Obviously, there is a limit to how

precise you can be.) (4 pts)

Here’s one possibility:

b. Find two different solutions to the equation (say y1(t) and y2(t) ) that are each

valid for every t. (4 pts)

y1(t) = t and y2 = t+4.

c. From the existence theorem, there must be a solution y3(t) on some interval with

y3(0) = 2. Is there a solution with y3(0) = 2 that is valid for all t ?(2 pts)

Yes; there is no way that a solution in the band can reach the edge of D, so there is no barrier to extending it over all of R.

d. Assuming that the solution y3(t) (from part c) is at least defined for 0 ≤ t ≤ 2,

say as much as you can about the value of y3(2). (4 pts)

y3(2) > 4 …because

y3(2) ≤ 6 ...well, one might suppose that y3(2) < 6 because

y3 can’t cross y2 due to uniqueness. That was what I was

looking for. But the problem doesn’t give enough information

to guarantee uniqueness, so y3 could actually merge with y2 and

y3(2) could actually be equal to 6. But y3(2) couldn’t be above 6,

because then some point on the curve above the band would

have to have slope greater than 1 (by the mean value theorem)

and that’s impossible. (3 pts for 2 < y3(2) < 6)

2. Consider the differential equation .It has a unique solution y with y(0) = 0.

(1)(a) Assuming that y(2) exists, estimate y(2) as accurately as you can.(Well, within 0.25 is surely close enough.)

What a horrible problem! I had intended to ask for y(1) but changed it to y(2) without enough checking. In fact, as t gets close to 2, the slope increases very rapidly, and no reasonable step size (with Euler’s method or any other method I know) gives a reasonable estimate. Full credit for any correct application of Euler’s method with step size less than 0.5. Here’s the result with step size 0.25:

t / y / y-prime / incr. to y
0.25 / 0 / 0 / 0 / 0
0.25 / 0.25 / 0 / 0.0625 / 0.015625
0.25 / 0.5 / 0.015625 / 0.250244 / 0.062561
0.25 / 0.75 / 0.078186 / 0.568613 / 0.142153
0.25 / 1 / 0.220339 / 1.048549 / 0.262137
0.25 / 1.25 / 0.482477 / 1.795284 / 0.448821
0.25 / 1.5 / 0.931298 / 3.117315 / 0.779329
0.25 / 1.75 / 1.710626 / 5.988743 / 1.497186
0.25 / 2 / 3.207812

But actually y(2) = 317.722…, and y(t) approaches infinity when t is about 2.00315. I got that with Mathematica, but even then it wasn’t easy; maybe we’ll try it in class. Lessons:

(1) It’s not so easy to tell whether a solution blows up.

(Euler’s method doesn’t help with this; it never blows up.)

(2) Mathematica is useful.

(b) How sure are you that y(2) even exists? (Or, might it be possible that y(t) grows

to infinity before t reaches 2 ?) (1 pt automatically)

Well, you probably shouldn’t have been sure. It’s a bad question, so 1/1 automatically; plus one bonus point if you were definitely not sure that y(2) is finite.

(end)