# 2250-1 7:30am Lecture Record Week 2 S2010

Last Modified: January 24, 2010, 09:43 MST.    Today: July 15, 2018, 13:28 MDT.

## 19 Jan: Theory of Linear First Order Differential Equations. Section 1.5.

```Topics delayed from Week 1.
implicit solution ln|y|=2x+c for y'=2y
explicit solution y = C exp(2x) for y'=2y
Troubles with explicit solutions of y'= 3 sqrt(xy) [1.4-6].
Key Examples
Separable DE with no equilibrium solutions.
Separable DE with finitely many 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.
Examples for Midterm 1 problem 2:
y'=x+y, y'=x+y^2, y'=x^2+y^2
```
```Review and Drill Section 1.4
Variables separable method.
Discuss remaining exercises 1.4-6,12,18.
Problem Notes 1.4 at the web site.
Equilibrium solutions and how to find them.
```
```Lecture on Section 1.5
Theory of linear DE y'=-p(x)y+q(x).
Integrating factor W=e^Q(x), Q(x) = int( p(x),x)
(Wy)'/W, the fraction that replaces two-termed expression y'+py.
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)
Finding F and G in a separable equation y'=F(x)G(y)

Linear integrating factor method 1.5
Application to y'+2y=1 and y'+y=e^x.
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.
Main theorem on linear DE and explicit general solution.
```

## 20 Jan: Linear Applications. Section 1.5

Collect first package of dailies in class: 1.2, 1.3.
```Review and Drill
Variables Separable method
Equilibrium solutions from G(y)=0 and
Non-equilibrium solutions from G(y) nonzero.
```
```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 sybstitution 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 deal with 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.
```
```Partial fractions
How to solve y'=(1-y)y
Def: a partial fraction = constant / polynomial with one root
Theorem. A polynomial quotient p(x)/q(x) is a sum of partial
fractions, provided degree(p) < degree(q).
The possible partial fractions have denominator
dividing the denominator of q(x).
How to solve for partial fraction constants A,B,C,...
Clear the fractions
Substitute invented values for x to get a system of equations
for A,B,C,..., then solve the system.
Methods:
Sampling method [described above]
Method of atoms [multiply out, match powers]
Heaviside's coverup method
```
```General Verhulst DE
Solving y'=(a-by)y by a substitution
Let u=y/(a-by).
Then substitution into the DE gives u'=au
Solve u'=au to get u=u0 exp(ax).
Back-substitute u(x) into u=y/(a-by), then solve for y.
Solving y'=(a-by)y by partial fractions
Divide the DE by (a-by)y
Find the constants in the partial fractions on the left.
a y0
y(x) = --------------------------
b y0 + (a - b y0) exp(-ax)
where y0=y(0)=initial population size.
```
```Linear Differential Equation y'+p(x)y=q(x)
Section 1.5
Definition: Linear DE
Test: y'=f(x,y) is linear if and only if the partial
derivative f_y is independent of y.
Algorithm
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)dx))
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.
Two Methods for solving first order equations:
Linear integrating factor method,
Superposition + equilibrium solution for
constant-coefficient linear,

```

## 20-21 Jan: Murphy

Present problems 2, 3 of the midterm 1 sample [f2009 midterm 1 key].
Exam 1 date is 17 Feb 1-3pm in WEB 104 or 18 Feb 6:50am in JTB 140. Other exam times were pre-set by agreement at the start of the semester, on an individual basis. The plan was created to provide extra time to write the exam, which is designed for 50 min.
Sample Exam: Exam 1 key from F2009. See also S2009, Exam 1.
HTML: 2250 midterm exam samples S2010 (15.6 K, html, 16 May 2010)
Questions on textbook sections 1.3, 1.4.
Review and drill Ch1.

## 22 Jan: Autonomous systems and applications section 2.1

```Superposition Theory
Superposition for y'+p(x)y=0.
Superposition for y'+p(x)y=q(x)
A faster way to solve y'+2y=1
```
```Problem 1.5-34
The expected model is
x'=1/4-x/16,
x(0)=20,
using units of millions of cubic feet.
Model Derivation
Law:  x'=input rate - output rate.
Definition:  concentration == amt/volume.
Use of percentages
0.25% concentration means 0.25/100 concentration
```
```Drill Section 1.5
Three linear examples: y'+(1/x)y=1, y'+y=e^x, y'+2y=1.