HW #2 Solutions1
2.3.6:
We are given that the force exerted by the spring is
F = F (x) = kx x3 ,
where k and are positive constants and x is the displacement from equilibrium.
It is reasonable to assume the force is an odd function of x since the spring s

HW #7 Solutions1
6.9.1: Observe that in this case the masses at site 1 and site N are connected
with a spring. (Visualize the system of N particles to be on a circle.) Thus there
are no end eects as in the case of the lectures. The only change in the matr

HW #3 Solutions1
3.5.1:
1. By Eulers formula
ei4 = cos 4 + i sin 4.
(1)
By the addition theorem and Eulers formula
4
ei4 = ei
= (cos + i sin )4 .
Expanding the last expression, and using the fact that i2 = 1 we obtain
ei4 = cos4 + 4i cos3 sin 6 cos2 sin2

HW #4 Solutions1
3.5.6:
The DE given is a 2nd order linear inhomogeneous equation. We rst
nd a particular solution. Based upon our experience with forced oscillations,
we guess a particular solution of the form
yp (t) = A cos(t)
Substituting this into the

HW #6 Solutions1
5.5.6:
We are again asked to analyze the DE
dx
= Ax
dt
where now A is a 2 2 matrix given by (5.16). The eigenvalues of A are the
roots of the characteristic polynomial of A. A simple calculation shows we must
nd the roots to
2 4 + 3 + 2 =

HW #5 Solutions1
5.5.1:
The rst step is to write the harmonic oscillator equation
2
x + 0 x = 0
as a rst order system. Let
x
x
X=
Then the rst order matrix equation is
dX
= AX
dx
where
0
2
0
A=
1
0
.
Now we compute the eigenvalues and eigenvectors of A. T

HW #8 Solutions1
6.9.8:
Assuming a solution of the form
uj (t) = eit vj , vj independent of t,
one simply substitutes this into the equation
mj
d2 uj
+ k (uj 1 + 2uj uj +1 ) = 0 for j = 1, 2, 3, . . . , 2N,
dt2
(1)
to obtain the equation for the vj
mj 2 v