# 2280 12:55pm Lectures Week 2 S2018

Last Modified: January 22, 2018, 12:45 MST.    Today: December 14, 2018, 00:27 MST.
```Topics
Sections 1.5, 2.1, 2.2, 2.3
The textbook topics, definitions, examples and theoremsEdwards-Penney 1.2, 1.3, 1.4, 1.5 (14.9 K, txt, 17 Jan 2018)Edwards-Penney 2.1, 2.2, 2.3 (15.8 K, txt, 17 Dec 2014)PDF: Week 2 Examples (91.9 K, pdf, 17 Oct 2016)```

### Tue: Theory and Examples for Separable Equations, sections 1.4, 2.1

```Partial Fractions
Start topic of partial fractions, to be applied again in 2.1-2.2.
To be studied: Heaviside's method. Sampling method [a Fail-safe
method]. The method of atoms.
References on partial fractionsSlides: Partial Fraction Theory (148.6 K, pdf, 14 Dec 2014)Manuscript: Heaviside coverup partial fraction method (290.2 K, pdf, 07 Jan 2014)Manuscript: Heaviside's method and Laplace theory (352.3 K, pdf, 07 Jan 2014)
Definition: A partial fraction is a constant divided by a polynomial
with exactly one root, that is, c/(x-r)^k. The root can be real or complex.
Definition of separable DE.
Examples: 1.4-6,12,18.
Complete answers and methods for 1.4-6 and 1.4-18 appear in link TEXT: Problem notes 1.4 (15.9 K, txt, 07 Jan 2018)
Some separability tests.
References for separable DE.Slides: Separable DE method. Tests I, II, III. Equilibrium solutions (148.9 K, pdf, 12 Jan 2018)Manuscript: Method of quadrature (258.8 K, pdf, 13 Jan 2016)Manuscript: Separable Equations (314.8 K, pdf, 06 Jan 2014)Text: How to do a maple answer check for y'=y+2x (0.3 K, txt, 06 Jan 2014)Transparencies: Section 1.4 and 1.5 Exercises (465.0 K, pdf, 26 Aug 2003)
Theory of separable equations section 1.4.
Separable equations depend on partial fraction theory, reading below.
Separation test:
Define F(x)=f(x,y0)/f(x0,y0),
G(y)=f(x0,y),
then FG=f if and only if y'=f(x,y) is separable.
Non-Separable Test
TEST I. f_x/f depends on y ==> y'=f(x,y) not separable
TEST II. f_y/f depends on x ==> y'=f(x,y) not separable
Review: Basic theory of y'=F(x)G(y):
y(x) = H^(-1)( C1 + int(F)),
H(u)=int(1/G,u0..u).
Solutions y=constant are called equilibrium solutions.
Find them using G(c)=0.
Non-equilibrium solutions arise from y'/G(y)=F(x) and a
Implicit and explicit solutions.
Discussion of answer checks for implicit solutions and also
explicit solutions.
Exercise 1.4-6: Trouble with explicit solutions of y'= 3 sqrt(xy)
Separable DE with no equilibrium solutions.
Separable DE with infinitely many equilibrium solutions.
The list of answers to a separable DE.
Influence of an initial condition to extract just one solution
formula from the list of solutions.
Key Examples for Midterm 1 problem 2:
y'=x+y, y'=x+y^2, y'=x^2+y^2
Example 1: Show that y'=x+y is not separable using TEST I or II
TEST I. f_x/f depends on y ==> y'=f(x,y) not separable
TEST II. f_y/f depends on x ==> y'=f(x,y) not separable
Example 2: Find the factorization f=F(x)G(y) for y'=f(x,y), given
(1) f(x,y)=2xy+4y+3x+6
(2) f(x,y)=(1-x^2+y^2-x^2y^2)/x^2
Answers: (1) F=x+2, G=2y+3; (2) F=(1-x^2)/x^2, G=1+y^2
Idea: F(x)=f(x,y)/G(y) implies F(x)=f(x,0)/G(0) for y=0, which
implies \$F(x)=f(x,0)\$ divided by some constant c. Backsub
implies G(y)=f(x,y)/F(x)= c f(x,y)/f(x,0). Cleverly select
y=y0 instead of y=0 for a general method.
implicit solution ln|y|=2x+c for y'=2y
explicit solution y = C exp(2x) for y'=2y
Answer check for y'= 3 sqrt(xy) [1.4-6].
Answer checks for midterm examples y'=x+y, y'=x+y^2, y'=x^2+y^2
```

### Wednesday-Friday: Solving Linear DE. Section 1.5.

```Lecture on Section 1.5
We will study linear DE y'=-p(x)y+q(x).
Classification of y'=f(x,y)
quadrature [Q], separable [S], linear [L].
Venn diagram of classes Q, S, L.
Examples of various types.
Test for linear (f_y indep of y)
Test for not separable (f_y/f depends on x ==> not sep)
Section 1.4. Separable DE. Review and Drill, as time allows.
Variables separable method.
Finding F and G in a separable equation y'=F(x)G(y)
Equilibrium solutions from G(y)=0 and
Non-equilibrium solutions from G(y) nonzero.
Method of Quadrature: When to use it.
Discuss remaining exercises 1.4-6,12,18.
Problem Notes Ch2Text: Problem notes Chapter 2 (10.8 K, txt, 22 Dec 2014)
Detailed derivations for 1.4-6
y' = 3 sqrt(-x) sqrt(-y)  on quadrant 3, x<0, y<0
y' = 3 sqrt(x) sqrt(y)  on quadrant 1, x>0, y>0
Equilibrium solution
Found by substitution of y=c into the DE y'=3 sqrt(xy)
Ans: y=0 is an equilibrium solution
Non-equilibrium solution
Found from y'=F(x)G(y) by division by G(y),
followed by the method of quadrature.
y = ( x^(3/2)+c)^2
y = - ((-x)^(3/2)+c)^2
List of 3 solutions cannot be reduced in number
How to test separable and non-separable equations
Theorem. If f_y/f depends on x, then y'=f(x,y) is not separable
Theorem. If f_x/f depends on y, then y'=f(x,y) is not separable
Theorem. If y'=f(x,y) is separable, then f(x,y)=F(x)G(y) is
the separation, where F and G are defined by the formulas
F(x) = f(x,y0)/f(x0,y0)
G(y) = f(x0,y).
The invented point (x0,y0) may be chosen conveniently,
subject to f(x0,y0) nonzero.
Section 1.5. Linear integrating factor method
```
```Linear Differential Equation y'+p(x)y=q(x)
Section 1.5
Definition: Linear DE y'+p(x)y=q(x)
Test: y'=f(x,y) is linear if and only if the partial
derivative f_y is independent of y.
Examples:
Testing linear DE y'=f(x,y) by f_y independent of y.
Classifying linear equations and non-linear equations.
Picard's theorem implies a linear DE has a unique solution.
We don't check for equilibrium solutions or exceptions.
THEOREM. A linear DE has an explicit general solution.
Def. Integrating factor W=exp(Q(x)), where Q(x) = int(p(x),x)
THEOREM. The integrating factor fraction (Wy)'/W replaces the
two-termed expression y'+py.
Application Examples: y'+2y=1 and y'+y=e^x.
ALGORITHM.
How to solve a linear differential equation
Test the DE for linear
Identify p(x), q(x) in the standard form y'+py=q.
Determine an integrating factor W(x)=exp(int(p(x),x))
Replace y'+py in the standard form y'+py=q by the quotient
(Wy)' / W
and then clear fractions to get the quadrature equation
(Wy)' = qW
Solve by the method of quadrature.
Divide by W to find an explicit solution y(x).
Three linear examples: y'+(1/x)y=1, y'+y=e^x, y'+2y=1.