Math 2250-1
Wed Oct 24
5.4 experiments; then begin 5.5
,
Let's conduct two (undamped) experiments, using a pendulum configuration in one, and an oscillating
mass-spring configuration in the other. The differential equation for an undamped swinging pendulum
turns out to be identical to the mass-spring equation, once we "linearize":
Pendulum: measurements and prediction (we'll check these numbers).
> restart :
Digits d 4 :
> L d 1.526;
g d 9.806;
g
wd
; # radians per second
L
f d evalf w / 2$ Pi ; # cycles per second
T d 1 / f; # seconds per cycle
Experiment:
Mass-spring:
compute Hooke's constant:
> 98.7 K 83.4; #displacement from extra 50g
.05$9.806
> kd
; # solve k$x=m$g for k.
.153
> m d .1; # mass for experiment is 100g
k
wd
; # predicted angular frequency
m
w
f d evalf
; # predicted frequency
2$Pi
1
T d ; # predicted period
f
Experiment:
We neglected the KEspring , which is small but could be adding intertia to the system and slowing down the
oscillations. We can account for this:
Improved mass-spring model
Normalize TE = KE C PE = 0 for mass hanging in equilibrium position, at rest. Then for system in
motion,
KE C PE = KEmass C KEspring C PEwork .
x
PEwork =
k s ds =
0
1 2
kx ,
2
KEmass =
1
m x# t
2
2
,
KEspring = ????
How to model KEspring ? Spring is at rest at top (where it's attached to bar), moving with velocity x# t at
bottom (where it's attached to mass). Assume it's moving with velocity µ x# t at location which is
fraction µ of the way from the top to the mass. Then we can compute KEspring as an integral with respect
to µ , as the fraction varies 0 % µ % 1 :
1
KEspring =
0
1
= mspring x# t
2
1
µ x# t
2
1
2
2
2
µ dµ=
0
mspring dµ
1
m
x# t
6 spring
2
.
Thus
TE =
1
2
mC
1
m
3 spring
x# t
2
C
1 2 1
k x = M x# t
2
2
2
C
1 2
kx ,
2
where
M= mC
1
m
3 spring
Dt TE = 0 0
M x# t x## t C k x t x# t = 0 .
x# t M x##C k x = 0 .
Since x# t = 0 only at isolated tKvalues, we deduce that the corrected equation of motion is
M x##C k x = 0
with
w0 =
k
=
M
k
1
m C mspring
3
Does this lead to a better comparison between model and experiment?
> ms d .011; # spring has mass 11g
1
M d m C $ms; # "effective mass"
3
k
; # predicted angular frequency
M
w
f d evalf
; # predicted frequency
2$Pi
1
T d ; # predicted period
f
> wd
>
.
Section 5.5: Finding yP for non-homogeneous linear differential equations
L y =f
(so that you can use the general solution y = yP C yH to solve initial value problems).
There are two methods we will use:
, The method of undetermined coefficients uses guessing algorithms, and works for constant coefficient
linear differential equations and certain classes of functions f x . The method seems magic, but actually
relies on vector space theory. This is the main focus of section 5.5.
, The method of variation of parameters is more general, and yields an integral formula for a particular
solution yP , assuming you are already in possession of a basis for the homogeneous solution space. It
works for any linear differential equation and any (continuous) function f , but the formulas can get
computationally messy especially for differential equations of order n O 2 . We'll focus on the case n = 2.
The easiest way to explain the method of undetermined coefficients is with examples.
Exercise 1) Find a particular solution yP x for the differential equation
L y d y##C 4 y#K5 y = 10 x C 3 .
Hint: try yP x = A x C B (because L A x C B = c x C d ).
Exercise 2) Use your work in 1 and your expertise with homogeneous linear differential equations to find
the general solution to
y##C 4 y#K5 y = 10 x C 3
Exercise 3) Find a particular solution to
L y = y##C 4 y#K5 y = 14 e2 x .
Hint: try yP = C e2 x since L C e2 x = d e2 x .
Exercise 4a) Use superposition and your work from the previous exercises to find the general solution to
L y = y##C 4 y#K5 y = 14 e2 x K 20 x K 6 .
4b) Solve (or at least set up the problem to solve) the initial value problem
y##C 4 y#K5 y = 14 e2 x K 20 x K 6
y 0 =4
y# 0 =K4 .
4c) Check your answer with technology.
Exercise 5) Find a particular solution to
L y d y##C 4 y#K5 y = 2 cos 3 x .
Hint: To solve L y = f we hope that f is in some finite dimensional subspace V that is preserved by L, i.e.
L : V/V .
, In Exercise 1 V = span 1, x and so we guessed yP = A C B x.
, In Exercise 3 V = span e2 x and so we guessed yP = C e2 x.
, What's the smallest subspace V we can take in the current exercise? Can you see why
V = span cos 3 x and a guess of yP = C cos 3 x won't work?
All of the previous exercises relied on:
Method of undetermined coefficients (base case): If f is in a finite dimensional subspace V with the
property that L preserves V, and if the only function g 2 V for which L g = 0 is g = 0, then there is a
unique yP 2 V with L yP = f .
proof: This is a special case of the following fact: if V and W are vector spaces with the same finite
dimension, and if L : V/W is a linear transformation between them, then the equation L v = w has
unique solutions v 2 V for each w 2 W if and only if the only solution v 2 V to L v = 0 is v = 0 . We
already know this fact is true for matrix transformations L x = An # n x with L : =n /=n (because each
requirement is equivalent to saying that the reduced row echelon form of A is the identity), but the result is
really more general then that special case. We will either discuss it in class, or you will think about it in
next week's homework assignment.
Exercise 6) Use the method of undetermined coefficients to guess the form for a particular solution yP x
for a constant coefficient differential equation
L y d y n C an K 1 y n K 1 C... C a1 y#C a0 y = f
(assuming the only such solution which would solve the homogeneous DE is the zero solution):
6a) L y = x3 C 6 x K 5
6b) L y = 4 e2 xsin 3 x
6c) L y = x cos 2 x
BUT LOOK OUT
Exercise 7a) Find a particular solution to
y##C 4 y#K5 y = 4 ex .
Hint: since yH = c1 ex C c2 eK5 x , a guess of yP = A ex will not work (and span ex does not satisfy the
vector space conditions for undetermined coefficients). Instead try
yP = A x ex
and factor L = D2 C 4 D K 5 I = D C 5 I + D K I .
7b) check work with technology
The vector space theorem alluded to above for L : V/W, combined with our understanding of how to
factor constant coefficient differential operators (as in this week's homework) leads to the more general
method of undetermined coefficients algorithm, for right hand sides which can be written as sums of
functions having the indicated forms. See the discussion in pages 341-346 of the text, and the table on
page 346, reproduced here. If L has a factor D K r I s then the guesses you would have made in the
"base case" need to be multiplied by xs , as in Exercise 7.