Challenge Problem 7
Due Tues, November 27th, 2012
1. Consider a matrix
The Determinant, which we denote D, is a11 a12 a12 a21 , and the Trace, denoted T , is
a11 + a22 .
1. Write the eigenvalues of A in terms of T and D. Hin
Due Tuesday, November 13th, 2012
Here is a method for getting real solution more eciently by using complex numbers.
Besides ei = cos + i sin , youll have to remember how to invert complex numbers, that
1 a ib
Challenge Problem 5 Due Thursday, October 25th, 2012
1. An alternative way to solve (linear) ODEs. Assume that the solution to the ODE
L[y ] = g (t) is analytic at the point t = 0. An analytic function is one which is equal to
its power series e
Challenge Problem 4
Due Tues, October 9th, 2012
1. Using the following scaold, sketch a proof of the existence and uniqueness theorem
for rst order IVPs:
Suppose f (t, y ) is a continuous function on an interval I containing t0 . Then
Challenge Problem 3
Due Tues, September 18th, 2012
1. Suppose a open-topped hemispherical tank of radius 10 feet full of water has a closed,
one foot radius circular drain at the bottom. By Torricellis Law, water ows out of the
hole with the sam
Challenge Problem 2
Due Tuesday, September 11th, 2012
1. Typically, the rst thing people will do in using a DE to model some part of the Real
WorldTM is introduce dimensionless variables, which reduces the number of parambers in
the DE. Here, we
Challenge Problem 1 Due Tuesday, September 4th, 2012
1. A dog standing at point (c, 0) in the plane spots a rabbit standing at the origin. The
rabbit bolts along the line y = ax (a < 0), traveling at a constant velocity VR . The dog
Thursday, November 29th, 2012
1. Compute from the denition L [e3t+1 ](s).
est e3t+1 dt
= e lim
t (s 3)
if s > 3.
2. Using the facts that L [cos(kt)](s) =
Thursday, November 8th, 2012
1. Find the general solution to the ODE.
Begin by nding the eigenvalues. That is, solve
0 = ( 8)( + 4) (6)(6)
= 2 4 + 4
= ( 2)2
So, the matrix has one eigenvalue, = 2, of mu
Thursday, October 18th, 2012
1. Find a particular solution to the non-homogeneous ODE:
y 2y + 5y = t + 1 et
We are going to apply the method of undetermined coecients. First, it is vital to be
sure that our naive guess doesnt involv
Thursday, October 4th, 2012
1. Solve the IVP:
y 2y + 5 = 0
y (0) = 3
y (0) = 5
This is a homogeneous linear ODE, so we guess that the solution is of the form u = ert .
Plugging this in, we obtain the characteristic equation
0 = r2 2r
Thursday, September 6th, 2012
1. Without solving, nd and classify the stability of all equilibrium solutions to
= y 3 4y 2 + 4y
The right hand side of the equation factors as f (y ) = y (y 2)2 , which has zeroes at
y = 0 and y
Thursday, September 6th, 2012
1. Solve the IVP.
3t2 y = t2
y (0) = 1
This is a linear rst order ODE, so we can solve it by guessing an integrating factor. We
know the proper guess is er(t) , where r(t) = 3t2 dt = t3 + C , and w
Thursday, August 30th, 2012
1. Consider the ODE y = ty . What is the largest rectangle R on which this ODE
satises the conditions of the the E/U theorem for the IVP y (1) = 2?
The function f (y, t) = ty is continuous on its do
Thursday, August 23rd, 2012
1. Check that u(t) = Aet sin(t) is a solution of the ODE y 2y cot(t) = 0.
To solve this problem, we plug u(t) in for y in the given ODE.
= Ae sin(t) + Ae cos(t) + Ae cos(t) Ae sin(t)