Grades on the final exam

The answers to the final exam are up. The curve for the final is as follows:
55+ A
40-55 B
25-40 C
15-25 D
0-15 F

Solutions to those practice problems

1. Let A be a square matrix with A*A = Identity. Let M be a square matrix, such that AMA = M. Let v be an eigenvector of M with eigenvalue e.

a) Show that Av is one too.

ANSWER.
We want to show MAv = eAv.
We don't know much about MA, but we do about AMA:
AMAv = Mv = ev.
That's ev on the right, but we want eAv or Aev. So left multiply by A:
AAMAv = Aev = eAv.
Now we use A*A = identity to cancel the AA at the beginning.

b) Assuming that, show that v+Av, v-Av are too.

ANSWER. This is easier.
M(v+Av) = Mv + MAv = ev + eAv = e(v+Av)
Same thing with minuses.

2. Find the exponential of t times the matrix M

[2 0 3 9 0]
[0 2 1 6 0]
[0 0 2 0 0]
[0 0 0 2 0]
[0 0 0 0 3]

Hint: it would be easier if you subtracted 2 * Identity... how do you make use of that?

ANSWER.
exp(A+B) = exp(A)exp(B) if A and B commute. For example if A = 2t * Identity.
Breaking this up that way, and looking at exp(B) = 1 + B + B^2/2! + B^3/3! + B^4/4!

exp(Bt) = [please ignore the periods, only there to keep Blogger from collapsing the line]

[1 0 0 0 0] . [0 0 3 9 0] .. [0 0 0 0 0] ....... [0 0 0 0 0]
[0 1 0 0 0] . [0 0 1 6 0] .. [0 0 0 0 0] ....... [0 0 0 0 0]
[0 0 1 0 0] + [0 0 0 0 0]t + [0 0 0 0 0]t^2/2! + [0 0 0 0 0]t^3/3! + ...
[0 0 0 1 0] . [0 0 0 0 0] .. [0 0 0 0 0] ....... [0 0 0 0 0]
[0 0 0 0 1] . [0 0 0 0 1] .. [0 0 0 0 1] ....... [0 0 0 0 1]

Higher powers only contribute in the lower right corner, giving exactly
the exponential series for exp(t). So the total is

1	0	3t	9t	0
0	1	t	6t	0
0	0	0	1	0
0	0	0	0	exp(t)

Q. Solve the ODE system y' = My for arbitrary initial conditions.

ANSWER. y = exp(tM)c. So if c =
[f]
[g]
[h]
[j]
[k],

then exp(tM)y = exp(2t) times
[f + 3th + 9tj]
[g + th + 6tj]
[h]
[j]
[exp(t)k].

3. Let y' = f(t,y).

a) What is the general form of f, if this equation is separable and linear homogeneous?

ANSWER. If it's linear homogeneous, then f(t,y) = g(t)y. In which case it's
automatically separable.

b) What is the general form of f, if this equation is separable and linear but not necessarily homogeneous?

ANSWER. If it's linear inhomogeneous, then f(t,y) = g(t)y + h(t) = g * (y + h/g).
To be separable, that h(t)/g(t) should only depend on y, not on t. So it doesn't depend on y, nor on t; it must be some constant C. Hence f(t,y) = g(t) * (y+C).


4. Let c_0 = 0, c_{n+1} = c_n + 2n + 1. Find c_0,c_1,c_2,c_3,c_4,c_5 and guess a general formula for c_n. Then determine the radius of convergence of the power series \sum_n c_n x^n.

ANSWER.
c_0..5 = 0,1,4,9,16,25. By the way, one way to prove that c_n = n^2 (though you weren't asked to) is to think of the numbers being added on like this:

1

12 22

123 223 333

1234 2234 3334 4444

...

or if you weren't taking a test, you could look it up here.

Anyway, the ratio test or the root test tell you that the radius of convergence is 1.

5. Assume that
f(x) y dx + (x+y) dy = 0
is exact. Determine the possibilities for f, and solve this ODE. (You should have two constants in your answer: one because you don't know f precisely, and the other from solving the ODE.)

ANSWER. Oops, actually you do know f precisely; I was thinking of a slightly harder problem. Probably a good one for the final.

Anyway, exactness says partial/partial_y (f(x) y) = partial/partial_x (x+y),
or f(x) = 1.

Now we do the integrals to determine F(x,y) such that y dx + (x+y) dy = dF:

	integral_{t=0}^x P(t,0) dt	+	integral_{s=0}^y Q(x,s) ds
=	integral_{t=0}^x 0 dt	+	integral_{s=0}^y (x+s) ds
=	0	+	(xs + 1/2 s^2) from s=0 to y

= xy + 1/2 y^2.

So the solutions are given by xy + 1/2 y^2 = C.