Math 2250-1
Mon Oct 29
5.5-5.6 Finding yP for non-homogeneous linear differential equations, and applications to forced
mechanical systems.
On Friday we talked about the method of undetermined coefficients for finding particular solutions yP for
L y =f
(as a step in finding the general solution y = yP C yH to solve initial value problems). This method works
if
(i) (base case) There is a finite-dimensional subspace V , with f 2 V, L : V/V, and such that the only
solution v 2 V to L v = 0 is v = 0. In this case there is a unique yP 2 V with L yP = f .
Or if
(ii) (xs Kcorrected case): There is a finite dimensional W, with f 2 W, and a subspace V of the same
dimension as W, and so that L : V/W, and such that the only solution v 2 V to L v = 0 is v = 0. In this
case there is a unique yP 2 V with L yP = f .
Both cases above rely on this linear algebra theorem, whose proof is in Friday's notes:
Theorem:
, Let V and W be vector spaces with the same dimension n !N.
, Let L : V/W be a linear transformation, (i.e. L u C v = L u C L v and L c u = c L u holds
c u, v 2 V, c 2 =.)
, Then the equation problem L v = f has a unique solution v 2 V for each f 2 W if and only if the
only solution to L v = 0 is v = 0 .
Exercise 1) What form should the undetermined coefficients guess take in the following examples?
a) y = y x to be a particular solution to
y##C 4 y#C 4 y = 3e x
p r = r2 C 4 r C 4 = r C 2 2
L = D2 C 4 D C 4 I = D C 2 I + D C 2 I .
b) y = y x to be a particular solution to
y##C 4 y#C 4 y = 3e K2 x
Variation of Parameters: This is an alternate method for finding particular solutions. Its advantage is that
it always provides one, even for non-homogeneous problems in which the right-hand side doesn't fit into a
nice finite dimensional subspace preserved by L. The formula for the particular solutions can be somewhat
messy to work with, however, once you start computing.
Here's the formula: Let y1 x , y2 x ,...yn x be a basis of solutions to the homogeneous DE
L y d y n C pn K 1 x y n K 1 C...C p1 x y#C p0 x y = 0 .
Then yp x = u1 x y1 x C u2 x y2 x C ... C un x yn x is a particular solution to
L y =f
provided the coefficient functions (aka "varying parameters") u1 x , u2 x ,...un x have derivatives
satisfying the matrix equation
u1 #
u2 #
:
= W y1, y2,..., yn
un #
where W y1 , y2 ,..., yn
K1
0
0
:
f
is the Wronskian matrix.
Here's how to check this fact when n = 2: Write
yp = y = u1 y1 C u2 y2 .
Thus
y#= u1 y1 #C u2 y2 #C u1 #y1 C u2 #y2 .
Set
u1 #y1 C u2 #y2 = 0.
Then
y##= u1 y1 ##C u2 y2 ##C u1 #y1 #C u2 #y2 # .
Set
u1 #y1 #C u2 #y2 # = f .
Notice that the two "set" equations are equivalent to the matrix equation
y1 y2
u1 #
0
=
y1 # y2 #
u2 #
f
which is equivalent to the n = 2 version of the claimed condition for yp . Under these conditions we
compute
p0 y = u1 y1 C u2 y2
C p1 y#= u1 y1 #C u2 y2 #
C 1 y##= u1 y1 ##C u2 y2 ##C f
L y = u1 L y1 C u2 L y2 C f
L y = 0C0Cf = f
A
Exercise 2a) Refind a particular solution for the problem in Exercise 1b) using variation of parameters:
y x to solve
y##C 4 y#C 4 y = 3e K2 x
2b) What is the general solution to the DE above?
Section 5.6: forced oscillations in mechanical (and electrical) systems. We will continue to discuss section
5.6 on Tuesday and Wednesday.
Overview for solutions x t to
m x##C c x#C k x = F0 cos w t
based on section 5.5 undetermined coefficients algorithms. We'll fill in the details today and tomorrow.
,
undamped c = 0 :
In this case the complementary homogeneous differential equation for x t is
m x##C k x = 0
k
x##C
x=0
m
2
x##C w0 x = 0
which has simple harmonic motion solutions xH t = C cos w0 t K a . So for the nonhomongeneous DE:
k
, w s w0 d
0 xP = A cos w t C B sin w t
m
0 x = xP C xH = C cos w t K a C C0 cos w0 t K a 0 .
,
w s w0 but w z w0 , C z C0 Beating!
,
w = w0
0 xP = t A cos w0 t C B sin w0 t
0 x = xP C xH = C t cos w t K a C C0 cos w0 t K a 0 .
(Resonance!)
,
.
damped c O 0 :
in all cases xP = A cos w t C B sin w t = C cos w t K a because w s w0
x = xP C xH = C cos w t K a C eKp t C1 cos w1 t .
,
underdamped:
,
critically-damped: x = xP C xH = C cos w t K a C eKp t c1 t C c2 .
Kr t
Kr t
over-damped:
x = xP C xH = C cos w t K a C c1 e 1 C c2 e 2 .
in all three cases above, xH t /0 exponentially and is called the transient solution xtr t .
xP t as above is called the steady periodic solution xsp t .
F0
, if c is small enough and w z w0 then the amplitude C of xsp t can be large relative to
, and
m
the system can exhibit practical resonance. This can be an important phenomenon in electrical circuits,
where amplifying signals is important.
,
,
Exercise 3a) Solve the initial value problem for x t :
x##C 9 x = 80 cos 5 t
x 0 =0
x# 0 = 0 .
3b) This superposition of two sinusoidal functions is periodic because there is a common multiple of their
(shortest) periods. What is this (common) period?
3c) Compare your solution and reasoning with the display at the bottom of this page.
> with plots :
> plot1 d plot K5$cos 5$t , t = 0 ..10, color = green, style = point :
plot2 d plot 5$cos 3$t , t = 0 ..10, color = blue, style = point :
plot3 d plot K5$cos 5$t C 5$cos 3$t , t = 0 ..10, color = black :
display plot1, plot2, plot3 , title ='superposition' ;
superposition
8
6
4
2
0
K2
K4
K6
K8
2
4
6
t
8
10
In general:
There is an interesting beating phenomenon for w z w0 (but still with w s w0 ). This is explained
analytically via trig identities, and is familiar to musicians in the context of superposed sound waves:
cos a K b K cos a C b = cos a cos b C sin a sin b
K cos a cos b K sin a sin b
= 2 sin a sin b .
1
1
Set a =
w C w0 t, b =
w K w0 t in the identity above, to rewrite the first term in x t as a
2
2
product rather than a difference:
F
xt =
0
2
2
m w K w0
2 sin
1
2 w C w0 t sin
v
1
0
w
K
w
t
C
x
cos
w
t
C
sin w0t
0
0
2
0
w0
.
In this product of sinusoidal functions, the first one has angular frequency and period close to the original
angular frequencies and periods of the original sum. But the second sinusoidal function has small angular
frequency and long period, given by
1
4p
angular frequency:
w K w0 ,
period:
.
2
w K w0
We will call half the period the beating period, as explained by the next exercise:
2 F0
2p
beating period:
,
beating amplitude:
2
2
w K w0
m w Kw
.
0
Exercise 4a) Use one of the formulas on the previous page to write down the IVP solution x t to
x##C 9 x = 80 cos 3.1 t
x 0 =0
x# 0 = 0 .
4b) Compute the beating period and amplitude. Compare to the graph shown below.
> plot 262.3$sin 3.05$t sin .05$t , t = 0 ..100, color = black, title = `beating` ;
beating
200
100
0
K100
K200
20
40
60
t
80
100
(You can also get this solution by letting w/w0 in the beating formula.)
Exercise 5a) Solve the IVP
x##C 9 x = 80 cos 3 t
x 0 =0
x# 0 = 0 .
5b) Compare the solution graph below with the beating graph in exercise 4.
> plot
40
t$sin 3$ t , t = 0 ..40, color = black, title = `resonance` ;
3
resonance
400
200
0
K200
K400
10
20
t
30
40