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2250-1 7:30am Lecture Record Week 2 S2012

Last Modified: January 26, 2012, 19:51 MST.    Today: October 21, 2017, 13:41 MDT.

Week 2, Jan 16 to 20: Sections 1.4, 1.5, 2.1, 2.2.

17 Jan: 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
     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.
  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.
  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].
  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).
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)
    Finding F and G in a separable equation y'=F(x)G(y)
  

18 Jan: 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
  Method of Quadrature
   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 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 showing threaded solutions: quadrants 2,4 empty

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.
Linear integrating factor method 1.5
   Def. Integrating factor W=e^Q(x), Q(x) = int( p(x),x)
  (Wy)'/W, the fraction that replaces two-termed expression y'+py.
  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.
    References for linear DE:
    Slides: Linear integrating factor method (129.8 K, pdf, 03 Mar 2012)
    Transparencies: Linear DE method, 1.5-3,5,11,33. Brine mixing (375.0 K, pdf, 29 Jan 2006)
    Manuscript: Applications of linear DE (374.2 K, pdf, 28 Jul 2009)
    Manuscript: Linear DE part I. Integrating Factor Method (152.7 K, pdf, 07 Aug 2009)
    Manuscript: Linear DE part II. Variation of Parameters, Undetermined Coefficients (134.1 K, pdf, 07 Aug 2009)
    Text: How to do a maple answer check for y'=y+2x (0.2 K, txt, 27 Jan 2005)
    Slides: Variation of Parameters. Integrating factor method (24.6 K, pdf, 23 Jan 2007)
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.
   classification: separable, quadrature, linear.
   Two Methods for solving first order equations:
     Linear integrating factor method,
     Superposition + equilibrium solution for
         constant-coefficient linear,
    References for linear applications
    Manuscript: Applications of linear DE (374.2 K, pdf, 28 Jul 2009)
    Slides: Brink tanks (95.3 K, pdf, 04 Mar 2012)
    Slides: Home heating (109.8 K, pdf, 04 Mar 2012)

19 Jan: Patrick B.

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 F2010. See also S2010, Exam 1.

HTML: 2250 midterm exam samples S2012 (18.9 K, html, 02 May 2012)

20 Jan: Solving Linear DE. Start Ch 2.

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
Drill Section 1.5
   Three linear examples: y'+(1/x)y=1, y'+y=e^x, y'+2y=1.
   classification: separable, quadrature, linear.
   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.
Problem 1.5-34
    The expected model is
      x'=1/4-x/16,
      x(0)=20,
    using units of millions of cubic feet.
  The answer is x(t)=4+16 exp(-t/16).
  Model Derivation
    Law:  x'=input rate - output rate.
    Definition:  concentration == amt/volume.
    Use of percentages
       0.25% concentration means 0.25/100 concentration
Introduction to Ch 2 topics
  Autonoumus DE y'=f(y)
  Solution of the Verhulst DE y'=(a-by)y
  Numerical solutions of DE. No exercises, but:
     Maple lab 3
     Maple lab 4
Midterm 1 sample exam is the F2010 exam, found at the course web site.
HTML: 2250 midterm exam samples S2012 (18.9 K, html, 02 May 2012)