## 2250 8:05am Lectures Week 8 S2015

Last Modified: March 08, 2015, 18:54 MDT.    Today: December 16, 2018, 17:32 MST.
``` Edwards-Penney, sections 5.1, 5.2, 5.3, 5.4
The textbook topics, definitions and theoremsEdwards-Penney 5.1, 5.2, 5.3, 5.4 (15.6 K, txt, 18 Dec 2013)```

### Week 8: Sections 5.1, 5.2, 5.3, 5.4

#### Monday: Linear Differential Equations of order n

```PARTIAL FRACTION THEORY. MAPLE ASSIST.
top:=x-1; bottom:=(x+1)*(x^2+1);
convert(top/bottom,parfrac,x);
top:=x-1; bottom:=(x+1)^2*(x^2+1)^2;
convert(top/bottom,parfrac,x);

PARTIAL FRACTION THEORY. MUPAD ASSIST.
top:=x-1; bottom:=(x+1)*(x^2+1);
partfrac(top/bottom,x)

PARTIAL FRACTION THEORY. MATLAB CALL BUILT-IN MUPAD.
syntax: y = evalin(symengine,'MuPAD_Expression')
y = evalin(symengine,'top:=x-1;bottom:=(x+1)*(x^2+1);partfrac(top/bottom,x)')

PROBLEM 4.7-26. Done last time.
How to solve y''+10y'=0 for general solution y=c1 + c2 exp(-10x)
TOOLKIT for SOLVING LINEAR CONSTANT DIFFERENTIAL EQUATIONS
Picard: Order n of a DE = dimension of the solution space.
General solution = linear combination n independent atoms.
Euler's theorem(s), an algorithm for finding solution atoms.

Linear DE Slides. Slides: Picard-Lindelof, linear nth order DE, superposition (181.5 K, pdf, 03 Mar 2012)  Slides: How to solve linear DE or any order (153.3 K, pdf, 24 Feb 2014)  Slides: Atoms, Euler's theorem, 7 examples (130.5 K, pdf, 25 Feb 2013)
EXAMPLE. The equation y'' +10y'=0. Review.
How to solve y'' + 10y' = 0, chapter 1 methods. Midterm 1 problem 1(d).
Idea: Let v=x'(t) to get a first order DE in v and a quadrature
equation x'(t)=v(t). Solve the first order DE by the linear
integrating factor method. Then insert the answer into
x'(t)=v(t) and continue to solve for x(t) by quadrature.
Vector space of functions: solution space of a differential
equation.
A basis for the solution space of y'' + 10y'=0 is {1,exp(-10x)}

PICARDS THEOREM
The theorem of Picard and Lindelof for y'=f(x,y) has an extension
for systems of equations, which applies to scalar higher order
linear differential equations.

THEOREM [Picard]
A homogeneous nth order linear differential equation with
continuous coefficients has a general solution written as a linear
combination of n independent solutions. This means that the
solution space of the differential equation has dimension n.

PICARD FAILURE.
Although Picard's structure theorem does not provide an algorithm
for construction of independent solutions, the theorems of Euler
do that. Combined, there is an easy path to finding a basis for
the solution space of an nth order linear differential equation.

EULER'S THEOREM for CONSTRUCTING a SOLUTION BASIS
It says y=exp(rx) is a solution of ay'' + by' + cy = 0 <==>
r is a root of the characteristic equation ar^2+br+c=0.

REAL EXPONENTIALS: If the root r is real, then the exponential is a
real solution.

THEOREM.
A real root r=a (positive, negative or zero) produces one Euler
solution atom exp(ax).

COMPLEX EXPONENTIALS: If a nonreal root r=a+ib occurs, a complex
number, then there is a conjugate root a-ib. The pair of roots
produce two real solutions from EULER'S FORMULA (a trig topic):
exp(i theta) = cos(theta) + i sin(theta)
Details to obtain the two solutions will be delayed. The answer is

THEOREM.
A conjugate root pair a+ib,a-ib produces two independent Euler
solution atoms exp(ax) cos(bx), exp(ax) sin(bx).

HIGHER MULTIPLICITY
For roots of the characteristic equation of multiplicity
greater than one, there is a correction to the answer obtained in
the two theorems above:
Multiply the answers from the theorems by powers of x until
the number of Euler solution atoms produced equals the
multiplicity.
EXAMPLE: If r=3,3,3,3,3 (multiplicity 5), then multiply exp(3x) by
1, x, x^2, x^3, x^4 to obtain 5 Euler solution atoms.
EXAMPLE: If r=5+3i,5+3i (multiplicity 2), then there are roots
r=5-3i,5-3i, making 4 roots. Multiply the two Euler atoms
exp(5x)cos(3x), exp(4x)sin(3x) by 1, x to obtain 4 Euler
solution atoms.

SHORTCUT: The characteristic equation can be synthetically formed
from the differential equation ay''+by'+cy=0 by the formal
replacement y ==> 1, y' ==> r, y'' ==> r^2.

EULER'S SOLUTION ATOMS
Leonhard Euler described a complete solution to finding n
independent solutions in the special case when the coefficients are
constant. The Euler solutions are called atoms in these
lectures.

THE TERM ATOM.
The term atom abbreviates Euler solution atom of a
linear differential equation. The main theorem says that the
answer to a homogeneous constant coefficient linear differential
equation of higher order is a linear combinations of atoms.

EULER SOLUTION ATOMS for LINEAR DIFFERENTIAL EQUATIONS
DEF. Base atoms are 1, cos(b x), sin(b x),
exp(a x), exp(ax)cos(bx), exp(ax)sin(bx).
DEF: atom = x^n (base atom) for n=0,1,2,...

THEOREM. Euler solution atoms are independent.

THEOREM.
Solutions of constant-coefficient homogeneous differential
equations are linear combinations of a complete set of Euler
solution atoms.

EULER SOLUTION ATOMS and INDEPENDENCE.
EXAMPLE. Show 1, x^2, x^9 are independent [atom theorem]
EXAMPLE. Show 1, x^2, x^(3/2) are independent [Wronskian test]

EXAMPLE.
The equation y''+10y'=0 has characteristic equation r^2+10r=0 with
roots r=0, r=-10. Then Euler's theorem says exp(0x) and exp(-10x)
are solutions. By vector space dimension theory, the Euler solution
atoms 1, exp(-10x) are a basis for the solution space of the
differential equation. Then the general solution is
y = a linear combination of the Euler solution atoms
y = c1 (1) + c2 (exp(-10x)).
Summary for Higher Order Differential Equations
Slides: Atoms, Euler's theorem, 7 examples (130.5 K, pdf, 25 Feb 2013)Slides: Base atom, atom, basis for linear DE (117.6 K, pdf, 03 Mar 2012)
PICARD'S THEOREM.  It says that nth order equations have a solution
space of dimension n.
EULER'S THEOREM. It says y=exp(rx) is a solution of ay''+by'+cy=0 <==> r is
a root of the characteristic equation ar^2+br+c=0.
Shortcut: The characteristic equation can be synthetically formed
from the differential equation ay''+by'+cy=0 by the
formal replacement
y ==> 1, y' ==> r, y'' ==> r^2.
EXAMPLE. The equation y''+10y'=0 has characteristic equation r^2+10r=0
with roots r=0, r=-10. Then Euler's theorem says exp(0x) and
exp(-10x) are solutions. By vector space dimension theory,
{1, exp(-10x)} is a basis for the solution space of the
differential equation. Then the general solution is
y = c1 (1) + c2 (exp(-10x)).

Survey of topics for this week.
Linear DE Slides.
Slides: Picard-Lindelof, linear nth order DE, superposition (181.5 K, pdf, 03 Mar 2012)  Slides: How to solve linear DE or any order (153.3 K, pdf, 24 Feb 2014)  Slides: Atoms, Euler's theorem, 7 examples (130.5 K, pdf, 25 Feb 2013)```
```Theory of Higher Order Constant Equations:
Homogeneous and non-homogeneous structure.
Superposition.
Picard's Theorem.
Solutions form a vector space of dimension n
Dimension of the solution set = n = order of highest derivative
Euler Solution Atoms.
Definition of atom.
Independence of atoms.
Euler's theorem.
Real roots
Non-real roots [complex roots].
How to deal with conjugate pairs of factors (r-a-ib), (r-a+ib).
The Euler formula exp(i theta)=cos(theta) + i sin(theta).
How to solve homogeneous equations:
Use Euler's theorem to find a list of n distinct solution atoms.
Examples:   y''=0, y''+3y'+2y=0, y''+y'=0, y'''+y'=0.

Second order equations.
Homogeneous equation.
Harmonic oscillator example y'' + y=0.
Picard-Lindelof theorem.
Dimension of the solution space.
Structure of solutions.
Non-homogeneous equation. Forcing term.
Nth order equations.
Solution space theorem for linear differential equations.
Superposition.
Independence and Wronskians. Independence of atoms.
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.
```

#### Mon-Wed: Sections 5.1, 5.2, 5.3

```REVIEW 5.1, 5.2, 5.3
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 5.1, 5.2, 5.3.
5.1-34:  y'' + 2y' - 15y = 0
5.1-36:  2y'' + 3y' = 0
5.1-38:  4y'' + 8y' + 3y = 0
5.1-40:  9y'' -12y' + 4y = 0
5.1-42:  35y'' - y' - 12y = 0
5.1-46:  Find char equation for y = c1 exp(10x) + c2 exp(100x)
5.1-48:  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).
5.2-18:  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.
5.2-22:  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).
5.3-8:   y'' - 6y' + 13y = 0         (r-3)^2 +4 = 0
5.3-10:  5y'''' + 3y''' = 0          r^3(5r+3) = 0
5.3-16:  y'''' + 18y'' + 81 y = 0    (r^2+9)(r^2+9) = 0
Check all answers with Maple, using this example:
de:=diff(y(x),x,x,x,x)+18*diff(y(x),x,x)+81*y(x) = 0;
dsolve(de,y(x));
5.3-32:  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.
Check the root answer in Maple, using the code
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.
```

#### Wed-Fri: Sections 5.2, 5.3

```Continued: Constant coefficient equations with complex roots.
Applying Euler's theorems to solve a DE.
Examples of order 2,3,4. Exercises 5.1, 5.2, 5.3.
5.1-48:  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).
5.2-18:  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.
5.2-22:  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).
5.3-8:   y'' - 6y' + 13y = 0         (r-3)^2 +4 = 0
5.3-10:  5y'''' + 3y''' = 0          r^3(5r+3) = 0
5.3-16:  y'''' + 18y'' + 81 y = 0    (r^2+9)(r^2+9) = 0
Check all answers with Maple, using this example:
de:=diff(y(x),x,x,x,x)+18*diff(y(x),x,x)+81*y(x) = 0;
dsolve(de,y(x));
5.3-32:  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.
Check the root answer in Maple, using the code
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.
Quadratic equations. Cubic equations. Theory of equations.
Inverse FOIL, complete the square, quadratic formula.Slides: Theory of equations, quadratics. (78.1 K, pdf, 07 Dec 2014)
```

#### Friday: Damped and Undamped Motion. Section 5.4

```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.
LRC Circuit equation
LRC derivation
Electrical-Mechanical analogy
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 (264.0 K, pdf, 08 Mar 2014) 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 (112.9 K, pdf, 08 Mar 2014)
```