Google Search in:

2280 12:55pm Lectures Week 2 S2018

Last Modified: January 22, 2018, 12:45 MST.    Today: December 14, 2018, 00:27 MST.
  Sections 1.5, 2.1, 2.2, 2.3
  The textbook topics, definitions, examples and theorems
Edwards-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)

Monday: Martin Luther King Day, no class

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 fractions
Slides: 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. Read the first slide link below, Tests I, II, III. 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 quadrature step. 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. Answer Checks and Key Examples. Discussion of answer checks 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 quadrature (f_y=0)
    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 Ch2
Text: 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. Applied to quadrant 1 y = ( x^(3/2)+c)^2 Applied to quadrant 3 y = - ((-x)^(3/2)+c)^2 List of 3 solutions cannot be reduced in number Graphic shows threaded solutions: quadrants 2,4 empty 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.
      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.
     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.
   Classification: separable, quadrature, linear.
  Superposition Theory
    Superposition for y'+p(x)y=0.
    Superposition for y'+p(x)y=q(x)
References for linear DE:
Slides: Linear integrating factor method (126.0 K, pdf, 15 Dec 2014)
Manuscript: Applications of linear DE, brine tanks, home heating and cooling (484.2 K, pdf, 16 Jan 2014)
Manuscript: 1st Order Linear DE part I. Integrating Factor Method, Superposition (303.6 K, pdf, 16 Jan 2014)
Manuscript: 1st Order Linear DE part II. Variation of Parameters, Undetermined Coefficients (238.7 K, pdf, 16 Jan 2014)
Manuscript: Kinetics, Newton's Models (343.8 K, pdf, 16 Jan 2014)
Transparencies: Linear integrating factor method, exercises 1.5-3,5,11,33. Brine mixing (375.0 K, pdf, 29 Jan 2006)
Text: How to do a maple answer check for y'=y+2x (0.3 K, txt, 06 Jan 2014)