## 2280 12:55pm Lectures Week 5 S2017

Last Modified: October 19, 2016, 03:16 MDT.    Today: September 24, 2018, 11:15 MDT.
```Topics
Sections 3.1 to 3.6
The textbook topics, definitions, examples and theoremsEdwards-Penney Ch 3, 3.1 to 3.4 (16.5 K, txt, 04 Jan 2015)Edwards-Penney Ch 3, 3.5 to 3.7 (17.7 K, txt, 02 Jan 2015)```

#### Monday-Tuesday: Nth Order Linear Differential Equations

```Second order and higher order differential Equations.
Detailed look at the second order case
REF. Theorems 1,2,3 in section 3.3 of Edwards-Penney.
Constant-coefficient SECOND ORDER homogeneous differential equations.
Characteristic equation is quadratic. Its factors determine the atoms.
The three possible cases
1. Two real roots
2. Two equal real roots
3. A complex conjugate pair of roots
The three possible kinds of solutions
1. y=c1 exp(r1 x) + c2 exp(r2 x)
If one root is zero, then the exponential equals 1.
2. y=c1 exp(r1 x) + c2 x exp(r1 x)
3. y= c1 exp(ax) cos(bx) + c2 exp(ax) sin(bx)
If in root a+ib, the real part a=0, then this is a pure
harmonic oscillation [because exp(ax)=1]
y= c1 cos(bx) + c2 sin(bx)
Sample equations:
y''=0, y''+2y'+y=0, y''-4y'+4y=0, y'' + 3y' + 2y=0,
x'' + x = 0, x'' + 2x' + 5x = 0, x'' + 8x' + 16x=0,
0.03I'' + 0.005 I = 0, Q'' + 100Q=0,

REVIEW OF SECTIONS 3.1, 3.2, 3.3  PROBLEM TYPES.
Example: Linear DE given by roots of the characteristic equation.
Example: Linear DE given by factors of the characteristic polynomial.
Example: Construct a linear DE of order 2 from a list of two atoms that
are known to be solutions.
Example: Construct a linear DE from characteristic equation roots.
Example: Construct a linear DE from its general solution.

COMPLEX ROOTS of the CHARACTERISTIC EQUATION
Solving a DE when the characteristic equation has complex roots.
Equations with both real roots and complex roots.
(D^4 + D^2)y = 0, (D+1)^2(D^2+4)^2 y = 0
An equation with 4 complex roots. How to find the 4 atoms.
(D^2+4)(D^2+16)y=0
SHORTCUT. One pair of complex conjugate roots identifies two Euler
solution atoms. Only one of the two complex roots is required to
construct the two atoms.

Nth order equations.
Solution space theorem for linear differential equations.
Superposition.
Independence and Wronskians. Independence of atoms.
Wronskian Test. Sampling Test.
Main theorem on constant-coefficient equations
THEOREM. Solutions are linear combinations of atoms.
Euler's substitution y=exp(rx).
Shortcut to finding the characteristic equation.
Euler's basic theorem:
y=exp(rx) is a solution <==> r is a root of the characteristic
equation.
Euler's multiplicity theorem:
y=x^n exp(rx) is a solution <==> r is a root of multiplicity
n+1 of the characteristic equation.
How to solve any constant-coefficient nth order homogeneous
differential equation.
1. Find the n roots of the characteristic equation.
2. Apply Euler's theorems to find n distinct solution atoms.
2a. Find the base atom for each distinct real root. Multiply
each base atom by powers 1,x,x^2, ... until the number of
atoms created equals the root multiplicity.
2b. Find the pair of base atoms for each conjugate pair of
complex roots. Multiply each base atom by powers 1,x,x^2,
... until the number of atoms created equals the root
multiplicity.
3. Report the general solution as a linear combination of the n atoms.

Constant coefficient equations with complex roots.
Applying Euler's theorems to solve a DE.
Examples of order 2,3,4. Exercises 3.1, 3.2, 3.3.
3.1-34:  y'' + 2y' - 15y = 0
3.1-36:  2y'' + 3y' = 0
3.1-38:  4y'' + 8y' + 3y = 0
3.1-40:  9y'' -12y' + 4y = 0
3.1-42:  35y'' - y' - 12y = 0
3.1-46:  Find char equation for y = c1 exp(10x) + c2 exp(100x)
3.1-48:  Find char equation for y = l.c. of atoms exp(r1 x), exp(r2 x)
where r1=1+sqrt(2) and r2=1-sqrt(2).
3.2-18:  Solve for c1,c2,c3 given initial conditions and general solution.
y(0)=1, y'(0)=0, y''(0)=0
y = c1 exp(x) + c2 exp(x) cos x + c3 exp(x) sin x.
3.2-22:  Solve for c1 and c2 given initial conditions y(0)=0, y'(0)=10
and y = y_p + y_h = -3 + c1 exp(2x) + c2 exp(-2x).
3.3-8:   y'' - 6y' + 13y = 0         (r-3)^2 +4 = 0
3.3-10:  5y'''' + 3y''' = 0          r^3(5r+3) = 0
3.3-16:  y'''' + 18y'' + 81 y = 0    (r^2+9)(r^2+9) = 0
dsolve(de,y(x)); # maple only
3.3-32:  Theory of equations and Euler's method. Char equation is
r^4 + r^3 - 3r^2 -5r -2 = 0. Use the rational root theorem
and long division to find the factorization (r+1)^3(r-2)=0.
solve(r^4 + r^3 - 3*r^2 -5*r -2 = 0,r);
The answer is a linear combination of 4 atoms, obtained from
the roots -1,-1,-1,2.
```

#### Tuesday: Damped and Undamped Motion. Section 3.4

```Lecture: Applications. Damped and undamped motion.
Last time: Theory of equations and 5.3-32.
Spring-mass equation,
Spring-mass DE derivation
Spring-mass equation mx''+cx'+kx=0 and its physical parameters.
Spring-mass system,
my'' + cy' + ky = 0
harmonic oscillation,
y'' + omega^2 y = 0,
Forced systems.
Forcing terms in mechanical systems. Speed bumps.
Harmonic oscillations: sine and cosine terms of frequency omega.
Damped and undamped equations. Phase-amplitude form.
Slides:
Shock-less auto.
Rolling wheel on a spring.
Swinging rod.
Mechanical watch.
Bike trailer.
Physical pendulum.Slides: Unforced vibrations 2008 (647.6 K, pdf, 27 Feb 2014)Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012) Slides: Forced damped vibrations (263.9 K, pdf, 10 Feb 2016) Slides: Forced vibrations and resonance (253.0 K, pdf, 08 Mar 2014)Slides: Resonance and undetermined coefficients, cafe door, pet door, phase-amplitude (178.0 K, pdf, 08 Mar 2014) Slides: Electrical circuits (112.8 K, pdf, 19 Feb 2016)

Slides on Section 3.4
Damped oscillations
overdamped, critically damped, underdamped,
pseudo-period  [Chapter 5]
phase-amplitude form of the solution [chapter 5]
Cafe door.
Pet door.
Undamped oscillations.
Harmonic oscillator. Slides: Unforced vibrations (647.6 K, pdf, 27 Feb 2014) Slides: phase-amplitude, cafe door, pet door, damping classification (136.0 K, pdf, 08 Mar 2014) Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012) Slides: Forced damped vibrations (263.9 K, pdf, 10 Feb 2016)```

#### Wednesday: Damped and Undamped Motion. Non-Homogeneous Equations.

```Partly solved 3.4-20
See the FAQ at the web site for answers and details.Text: FAQ for chapter 3 (31.7 K, txt, 11 Mar 2015)
The problem breaks into two distinct initial value problems:
(1)   2x'' + 16x' + 40x=0, x(0)=5, x'(0)=4
Characteristic equation  2(r^2+8r+20)=0.
Roots r=-4+2i,r=-4-2i.
Solution Atoms=e^{-4t}cos 2t, e^{-4t}sin 2t.
UNDERDAMPED.

(2)   2x'' + 0x' + 40x=0, x(0)=5, x'(0)=4
Characteristic equation 2(r^2+0+20)=0.
Roots r=sqrt(20)i,r=-sqrt(20)i.
The Euler solution atoms are
cos( sqrt(20)t), sin( sqrt(20)t).
UNDAMPED HARMONIC OSCILLATION.

Each system has general solution a linear combination of Euler
solution atoms. Evaluate the constants in the linear combination, in
each of the two cases, using the initial conditions x(0)=5, x'(0)=4.
There are two linear algebra problems to solve.

Answers: (1)  Coefficients 5, 12  for 2x'' + 16x' + 40x=0
Amplitude = sqrt(5^2 + 12^2) = 13
(2)  Coefficients 5, 2/sqrt(5) for 2x'' + 0x' + 40x=0
Amplitude = sqrt(5^2 + 4/5) = sqrt(129/5)
Write each solution in phase-amplitude form, a trig problem. See section
3.4 for specific instructions. The book's answers:
(1) tan(alpha) =12/5   (2) tan(alpha) = 5 sqrt(5)/2

Partly solved 3.4-34.
See the FAQ at the web site for answers and details.Text: FAQ for chapter 3 (31.7 K, txt, 11 Mar 2015)
The DE is 3.125 x'' + cx' + kx=0. The characteristic equation
is 3.125r^2 + cr + kr=0 which factors into 3.125(r-a-ib)(r-a+ib)=0
having complex roots a+ib, a-ib.

Problems 32, 33 find the numbers a, b from the given information.
This is an inverse problem, one in which experimental data is used
to discover the differential equation model. The book uses its own
notation for the symbols a,b: a ==> -p and b ==> omega1.

Because the two roots a+ib, a-ib determine the quadratic equation,
then c and k are known in terms of symbols a,b.

References:  Sections 3.4, 3.6. Forced oscillations.
Slides: Unforced vibrations 2008 (647.6 K, pdf, 27 Feb 2014) Slides: phase-amplitude, cafe door, pet door, damping classification (136.0 K, pdf, 08 Mar 2014) Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012) Slides: Forced damped vibrations (263.9 K, pdf, 10 Feb 2016) Slides: Forced vibrations and resonance (253.0 K, pdf, 08 Mar 2014) Slides: Undetermined coefficients, pure resonance, practical resonance (152.8 K, pdf, 03 Mar 2013) Slides: Electrical circuits (112.8 K, pdf, 19 Feb 2016)```

#### Friday: Undetermined Coefficients

```    PREVIEW: Undetermined Coefficients
THEOREM. Solution y_h(x) is a linear combination of atoms.
THEOREM. Solution y_p(x) is a linear combination of atoms.
THEOREM. (Superposition)  y = y_h + y_p

Which equations can be solved by undetermined coefficients.
1. Constant coefficients
2. Forcing term a linear combination of Euler solution atoms.

Intro to the basic trial solution method
Solution of x'' + 9x = 30 sin(2t)
x(t)=c1 cos 3t + c2 sin 3t + a cos 2t + b sin 2t
= sum of two harmonics, of frequencies 3 and 2
= BEATS example
xh(t) = c1 cos 3t + c2 sin 3t, the first harmonic
xp(t) = a cos 2t + b sin 2t, the second harmonic
TRIAL SOLUTION
The equation y'' + y = 1 + x should have a solution
yp = a + bx for some constants a, b. We find the constants
by substitution of yp into  y'' + y = 1 + x.
Ways to find the Euler solution atoms in y_p(x).
1. Use Laplace theory (a bit slow, even if you already know Laplace theory).
2. Use rules from undetermined coefficient theory. Faster.
Euler solution atoms in y_h(x):  Roots of the  characteristic equation.
```
```Slides: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 28 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
Extra Credit Maple Project: Tacoma narrows. Explore an alternative
explanation for what caused the bridge to fail, based on the hanging cables.
```

#### Mon-Wed-Fri: Undetermined Coefficients. Sections 4.1,3.5

```    REVIEW: Undetermined Coefficients
Which equations can be solved
THEOREM. Solution y_h(x) is a linear combination of atoms.
THEOREM. Solution y_p(x) is a linear combination of atoms.
THEOREM. (superposition)  y = y_h + y_pSlides: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 28 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
EXAMPLE. How to find a shortest expression for y_p(x) using
Details for x''(t)+x(t) = 1+t
the trial solution x(t)=A+Bt
BASIC METHOD. Given a trial solution with undetermined coefficients,
find a system of equations for d1, d2, ... and solve it.
Report y_p as the trial solution with substituted
THEORY. y = y_h + y_p, and each is a linear combination of atoms.

How to find the homogeneous solution y_h(x) from the characteristic equation.
How to determine the form of the shortest trial solution for y_p(x)
METHOD. A rule for finding y_p(x) from f(x) and the DE.
Finding a trial solution with fewest symbols.
Rule I. Assume the right side f(x) of the differential equation
is a linear combination of atoms. Make a list of all distinct atoms
that appear in the derivatives f(x), f'(x), f''(x), ... . Multiply
these k atoms by undetermined coefficients d_1, ... , d_k, then
add to define a trial solution y.

This rule FAILS if one or more of the k atoms is a solution of
the homogeneous differential equation.

Rule II. If Rule I FAILS, then break the k atoms into groups
with the same base atom. Cycle through the groups, replacing atoms
as follows. If the first atom in the group is a solution of the homogeneous
differential equation, then multiply all atoms in the group by factor x. Repeat
until the first atom is not a solution of the homogeneous differential equation.
Multiply the constructed k atoms by symbols d_1, ... , d_k and add to define trial solution y.

Explanation: The relation between the Rule I + II trial solution and
the book's table that uses the mystery factor x^s.
EXAMPLES.
y'' = x
y'' + y = x exp(x)
y'' - y = x exp(x)
y'' + y = cos(x)
y''' + y'' = 3x + 4 exp(-x)

THEOREM. Suppose a list of k atoms is generated from the
atoms in f(x), using Rule I. Then the shortest trial
solution has exactly k atoms.

EXAMPLE. How to find a shortest trial solution using
Rules I and II.

Details for x''(t)+x(t) = t^2 + cos(t), obtaining
the shortest trial solution
x(t)=d1+d2 t+d3 t^2+d4 t cos(t) + d5 t sin(t).
How to use dsolve() in maple to check the answer.
EXAMPLE. Suppose the DE has order n=4 and the homogeneous
equation has solution atoms cos(t), t cos(t), sin(t),
t sin(t). Assume f(t) = t^2 + cos(t). What is the
shortest trial solution?
EXAMPLE. Suppose the DE has order n=2 and the homogeneous
equation has solution atoms cos(t), sin(t). Assume
f(t) = t^2 + t cos(t).
What is the shortest trial solution?
EXAMPLE. Suppose the DE has order n=4 and the homogeneous
equation has solution atoms 1, t, cos(t), sin(t).
Assume
f(t) = t^2 + t cos(t).
What is the shortest trial solution?
```

#### Friday: Resonance, Section 3.6

```Undetermined coefficients
Examples continued from the previous lecture.
EXAMPLES.
y'' = x
y'' + y = x exp(x)
y'' - y = x exp(x)
y'' + y = cos(x)
y''' + y'' = 3x + 4 exp(-x)

Shortest trial solution.
Two Rules to find the shortest trial solution.
1. Compute the atoms in f(x). The number k of atoms found
is the number needed in the shortest trial solution.
2. Correct groups with the same base atom, by
multiplication by x until the group contains no atom
which is a solution of the homogeneous problem
[eliminate homogeneous DE conflicts].
The x^s mystery factor in the book's table. The number s is the
multiplicity of the root in the homogenous DE characteristic
equation, which constructed the base atom of the group.

Wine Glass Experiment
The lab table setup
Speaker.
Amplifier with volume knob.
Wine glass.
x(t)=deflection from equilibrium of the radial component of the
glass rim, represented in polar coordinates, orthogonal to
the speaker front.
mx'' + cx' + kx = F_0 cos(omega t)  The model of the wine glass
m,c,k are properties of the glass sample itself
omega = frequency generator knob adjustment
Theory of Practical Resonance
The equation is
mx''+cx'+kx=F_0 cos(omega t)
THEOREM. The limit of x_h(t) is zero at t=infinity
THEOREM. x_p(t) = C(omega) cos(omega t - phi)
C(omega) = F_0/Z, Z^2 = A^2+B^2, A and B are the
undetermined coefficient answers for trial solution
x(t) = A cos(omega t) + B sin(omega t).
THEOREM. The output x(t) = x_h(t) + x_p(t) is graphically
just x_p(t) = C(omega) cos(omega t - phi) for large t.
Therefore, x_p(t) is the OBSERVABLE output.
THEOREM. The amplitude C(omega) is maximized over all possible
input frequencies omega>0 by the single choice
omega = sqrt(k/m - c^2/(2m^2)).
DEFINITION. The practical resonance frequency is the number omega
defined by the above square root expression.

Slides: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 28 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)Slides: Resonance and undetermined coefficients (178.0 K, pdf, 08 Mar 2014)
Variation of Parameters and Undetermined Coefficients references
Slides: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 28 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
Systems of Differential Equations references
Manuscript: Systems of DE examples and theory (730.9 K, pdf, 09 Apr 2014) Slides: Laplace resolvent method (81.0 K, pdf, 14 Mar 2016) Slides: Laplace second order systems (273.7 K, pdf, 14 Mar 2016) Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 14 Mar 2016) Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008) Slides: Sliding plates example (105.8 K, pdf, 20 Aug 2008)
Oscillations. Mechanical and Electrical.
Slides: Electrical circuits (112.8 K, pdf, 19 Feb 2016)Slides: Forced damped vibrations (263.9 K, pdf, 10 Feb 2016)Slides: Forced vibrations and resonance (253.0 K, pdf, 08 Mar 2014)Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)Slides: Resonance and undetermined coefficients (178.0 K, pdf, 08 Mar 2014)Slides: Unforced vibrations 2008 (647.6 K, pdf, 27 Feb 2014)
```