5.6-2
==================================

The characteristic equation is r^2+4=0 with roots 2i,-2i and atom list
cos 2t, sin 2t. Then

                       xh=c1 cos 2t + c2 sin 2t.

A particular solution xp is found from trial solution x=d1 cos 3t + d2
sin 3t by the method of undetermined coefficients. Ans: d1 = 0, d2 = -1.

The initial conditions x(0)=x'(0)=0 are applied to the general solution

                   x = c1 cos 2t + c2 sin 2t - sin 3t

in order to evaluate c1 = 0, c2 = 3/2. Then

                       x = (3/2) sin 2t - sin 3t

is a sum of two harmonic oscillations of frequencies 2 and 3.


5.6-4
==================================

x'' + 25 x = 90 cos 4t, x(0)=0, x'(0)=90
This is a BEATS problem. For another example, and more details, see
5.6-2, above.
Expand the solution as x = x1 + x2, where x1 has frequency 5 and x2 has
frequency 4. This identifies x1 as c1 cos 5t + c2 sin 5t, a homogeneous
solution x_h, and x2 as d1 cos 4t + d2 sin 4t, a non-homogeneous
solution x_p. Superposition implies x = x_h + xp. Find x2 by
undetermined coefficients and then determine c1, c2 from the initial
conditions x(0)=0, x'(0)=90 [not x1(0)=0, x1'(0)=90, which is a common
error]. To repeat, find c1, c2 using x(0)=0, x'(0)=90, from x=c1 cos 5t
+ c2 sin 5t + d1 cos 4t + d2 sin 4t, where d1, d2 are the two values
found from undetermined coefficients.

5.6-8
==================================

 x'' + 3x' + 5x = -4 cos 5t
The book formulas can be used directly, because
 (-x)'' + 3(-x)' + 5(-x) = 4 cos 5t,
and then F_0 = 4, omega=5, m=1, c=3, k=5 
in section 5.6 of E&P. It follows that
  (1) -x(t) = A cos 5t + B sin 5t 
with A,B given by formula (19). Then  (1) implies
  (2)  x(t) = (-A) cos 5t + (-B) sin 5t.
The challenge left is to convert this expression to phase-amplitude form
   x(t) = C cos(5t - alpha)
with C=sqrt((-A)^2+(-B)^2) and tan(alpha)=(-B)/(-A). This is tricky only
because you must manually adjust the quadrant for alpha from the point
(-A,-B).

Read section 5.4, especially between equations (11)-(12), to find out
what to do. Beware of the advice in section 5.6, which suggests the
calculator answer or the calculator answer plus Pi [see (22) in section
5.6], because that advice applies only to the special problem being
studied in section 5.6.

5.6-10
==================================

The characteristic equation is r^2+3r+3=0. Because 10i, -10i are not roots
of this equation, then no fixup rule is required and the trial solution
for f(x)=8 cos 10t + 6 sin 10t is given by

                      x = d1 cos 10t + d2 sin 10t

A particular solution xp is found from this trial solution 
by the method of undetermined coefficients. Ans: 

                   d1 = -956/10309, d2 = - 342/10309.

Conversion to phase-amplitude form

                          x(t) =  C cos(10t-a)

is routinely done from the formulas

            C = sqrt(d1^2+d2^2), C cos a = d1, C sin a = d2,

with answers

      C = 10sqrt(61)/793, a = Pi + arctan(171/478) == 3.49 radians

5.6-16
==================================

The equation is x'' + 4x' + 5x = 10 cos(wt). The theory says that
practical resonance occurs exactly for input frequency w determined by
the equation [E&P problem 5.6-27]

                     w = sqrt((k/m) - (1/2)c^2/m^2)

Because m=1, c=4, k=5, then 

                             w = sqrt(5-9)

This value is NOT REAL, therefore practical resonance does not occur.
Contrast this result with 5.6-18, in which w=sqrt(600) and practical resonance does occur.

5.6-18
==================================

The equation is x'' + 10x' + 650x = 100 cos(wt). The theory says that
practical resonance occurs exactly for input frequency w determined by
the equation [E&P problem 5.6-27]

                     w = sqrt((k/m) - (1/2)c^2/m^2)

Because m=1, c=10, k=650, then 

                             w = sqrt(600)

This value is REAL, therefore practical resonance occurs.

======== end of problem notes 5.6 ========