# 2250-4 12:25pm Lecture Record Week 2 F2010

Last Modified: September 06, 2010, 09:09 MDT.    Today: July 23, 2018, 03:58 MDT.

## Aug 30 and 1 Sep: Michal Kordy

```Maple lab 1: quadratics, partial derivatives.
Present problem 1 from the midterm 1 sample [S2010 midterm 1 key].
Exam 1 date is in the syllabus and also the online due dates page.
Questions on textbook sections 1.3, 1.4.
Review and drill Ch1.
Sample Exam: Exam 1 key from S2010. See also F2009, Exam 1.
```

HTML: 2250 midterm exam samples S2010 (0.0 K, html, 31 Dec 1969)

## 31 Aug: Theory of Linear First Order Differential Equations. Section 1.5.

``` Theory of separable equations section 1.4.
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
f_x/f depends on y ==> y'=f(x,y) not separable
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.
Troubles with explicit solutions of y'= 3 sqrt(xy) [1.4-6].
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.
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
(partial derivative tests).
Example 2: Find the factorization f=F(x)G(y) for y'=f(x,y),
given
(1) f(x,y)=2xy+4y+3x+6 [ans: F=x+2, G=2y+3].
(2) f(x,y)=(1-x^2+y^2-x^2y^2)/x^2 [ans: F=(1-x^2)/x^2, G=1+y^2].
```
```Answer Checks and Key Examples.
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
```
```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.
```

## 31 Aug: Linear Applications. Section 1.5

```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.
```
```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 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.
```
```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,

```

## 02 Sep: 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.
Methods for solving first order equations:
Linear integrating factor method,
Superposition + equilibrium solution for
constant-coefficient linear DE
Drill: worksheet distributed in class, for the example
y' + 2y = 6. Solved in class y'+3y=6, y'+y=e^x, and several
homogeneous equations like y'+3y=0, y'+2y=0. Solved for
equilibrium solutions in more complicated examples like
2y' + Pi y = e^2.
```

## 02 Sep: Autonomous Differential Equations and Phase Diagrams. Sections 2.1, 2.2

```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
```

html: Problem notes F2010 (4.6 K, html, 26 Nov 2010)
```Examples and Applications
Growth-Decay model y'=ky and its algebraic model y=y(0)exp(kx).
Pharmokinetics of drug transport [ibuprofen]
Pollution models.
Three lake pollution model [Erie, Huron, Ontario].
Brine tanks.
One-tank model.
Two-tank and three-tank models.
Recycled brine tanks and limits of chapter 1 methods.
Linear cascades and how to solve them.
Method 1: Linear integrating factor method.
Method 2: Superposition and equilibrium solutions for
constant-coefficient y'+py=q. Uses the shortcut for
homogeneous DE y'+py=0.
Separation of variables
The equation y'=7y(y-13), y(0)=17
F(x) = 7, G(y) = y(y-13)
Separated form y'/G(y) = F(x)
Answer check using the Verhulst solution
P(t) = aP_0/(bP_0 + (a-b P_0)exp(-at))
Separation of variables details.
Review of partial fractions.
Partial fraction details for 1/((u(u-13)) = A/u + B/(u-13)
```
```  Partial fractions.
DEFINITION: partial fraction=constant/polynomial with exactly one root
THEOREM: P(x)/q(x) = a sum of partial fractions
Finding the coefficients.
Method of sampling
clear fractions, substitute samples, solve for A,B, ...
Method of atoms
clear fractions, multiply out and match powers, solve for A,B,...
Heaviside's cover-up method
partially clear fraction, substitute root, find one constant
Separation of variable solutions with partial fractions.
Exercise solutions to the problems due in 2.1.
```
Midterm 1 sample exam is the S2010 exam, found at the course web site.
HTML: 2250 midterm exam samples F2010 (17.7 K, html, 18 Dec 2010)