TopicsThe textbook definitions and theorems

Edwards-Penney 2.4, 2.5, 2.6 (11.3 K, txt, 18 Dec 2013)

Edwards-Penney 3.1, 3.2, 3.3, 3.4 (16.5 K, txt, 04 Jan 2015)

Partial Fraction TheorySeparation of variables and partial fractions(Delayed from week 2) Exercise solution problem 2.1-8 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. Partial fraction details for 1/((u(u-13)) = A/u + B/(u-13)

Section 2.3: Newton's force and friction modelsIsaac Newton ascent and descent kinematic models. Free fall with no air resistance F=0. Linear air resistance models F=kx'. Non-linear air resistance models F=k|x'|^2.The tennis ball problem.Does it takelonger to rise or longer to fall?: Bolt shot Example 2.3-3 (1.0 K, txt, 16 Dec 2012)Text: Newton kinematics with air resistance. Projectiles. (134.8 K, pdf, 30 Jan 2018)SlidesA rocket from the earth to the moon: Jules Verne Problem (124.2 K, pdf, 01 Jan 2015)SlidesReading assignment:Proofs of 2.3 theorems in the textbook and derivation of details for the rise and fall equations with air resistance.

Problem notes for Chapter 2 (10.8 K, txt, 22 Dec 2014)Numerical Solution of y'=f(x,y)Two problems will be studied. First problem y' = -2xy, y(0)=2 Symbolic solution y = 2 exp(-x^2) This problem appears in the Week 3 homework. Second problem y' = (1/2)(y-1)^2, y(0)=2 Symbolic solution y = (x-4)/(x-2) Not assigned, only a lecture example.Numerical Solution of y'=F(x)Example: y'=2x+1, y(0)=1 Symbolic solution y=x^2 + x + 1. Dot table. Connect the dots graphic. The exact answers for y(x)=x^2+x+1 are (x,y) = [0., 1.], [.1, 1.11], [.2, 1.24], [.3, 1.39], [.4, 1.56], [.5, 1.75], [.6, 1.96], [.7, 2.19], [.8, 2.44], [.9, 2.71], [1.0, 3.00]Maple support for making a connect-the-dots graphic.Example: L:=[[0., 1.], [2,3], [3,-1], [4,4]]; plot(L);: connect-the-dots graphic (11.2 K, jpg, 11 Sep 2010)JPG ImageExample: Find y(2) when y'=x exp(x^3), y(0)=1. No symbolic solution! How to draw a graphic with no solution formula? Make the dot table by approximation of the integral of F(x). REFERENCE:: Numerical methods (138.0 K, pdf, 15 Feb 2017) RECTANGULAR RULE int(F(x),x=a..b) = F(a)(b-a) approximately for small intervals [a,b] Geometry and the Rectangular RuleSlidesExample: y'=2x+1, y(0)=1 Rectangular rule applied to y(1)=1+int(F(x),x=0..1) for y'=F(x), in the case F(x)=2x+1. Dot table steps for h=0.1, using the rule 10 times. Answers: (x,y) = [0, 1], [.1, 1.1], [.2, 1.22], [.3, 1.36], [.4, 1.52], [.5, 1.70], [.6, 1.90], [.7, 2.12], [.8, 2.36], [.9, 2.62], [1.0, 2.90] The correct answer y(2)=3.00 was approximated as 2.90.Where did [.2,1.22] come from?y(.2) = y(.1)+int(F(x),x=0.1 .. 0.2) [exactly] = y(.1)+(0.2-0.1)F(0.1) [approximately, RECT RULE] = y(.1)+0.1(2x+1) where x=0.1 = 1.1 + 0.1(1.2) [approx, from data [.1,1.1]] = 1.22Rect, Trap, Simp rules from calculusRECT Replace int(F(x),x=a..b) by rectangle area (b-a)F(a) TRAP Replace int(F(x),x=a..b) by trapezoid area (b-a)(F(a)+F(b))/2 SIMP Replace int(F(x),x=a..b) by quadratic area (b-a)(F(a)+4F(a/2+b/2)+F(b))/6 The Euler, Heun, RK4 rules from this course: how they relate to calculus rules RECT, TRAP, SIMP Numerical Integration Numerical Solutions of DE RECT Euler TRAP Heun [modified Euler] SIMP Runge-Kutta 4 [RK4]Example:y'=3x^2-1, y(0)=2 with solution y=x^3-x+2.Example:y'=2x+1, y(0)=1 with solution y=x^2+x+1. Dot tables, connect the dots graphic. How to draw a graphic without knowing the solution equation for y. What to do when int(F(x),x) has no formula? Key example y'=x exp(x^3), y(0)=2. Challenge: Can you integrate F(x) = x exp(x^3)? Making the dot table by approximation of the integral of F(x). Accuracy: Rect, Trap, Simp rules have 1,2,4 digits resp.Maple code for the RECT ruleApplied to the quadrature problem y'=2x+1, y(0)=1. # Quadrature Problem y'=F(x), y(x0)=y0. # Group 1, initialize. F:=x->2*x+1: x0:=0:y0:=1:h:=0.1:Dots:=[x0,y0]:n:=10: # Group 2, repeat n times. RECT rule. for i from 1 to n do Y:=y0+h*F(x0); x0:=x0+h:y0:=Y:Dots:=Dots,[x0,y0]; od: # Group 3, display dots and plot. Dots; plot([Dots]);Example 1, for your study:Problem: y'=x+1, y(0)=1 It has a dot table with x=0, 0.25, 0.5, 0.75, 1 and y= 1, 1.25, 1.5625, 1.9375, 2.375. The exact solution y = 0.5(1+(x+1)^2) has values y=1, 1.28125, 1.625, 2.03125, 2.5000. Determine how the dot table was constructed and identify which rule, either Rect, Trap, or Simp, was applied.Example 2, for your study:Problem: y'=x exp(x^3), y(0)=2 Find the value of y(2)=2+int(x*exp(x^3),x=0..1) to 4 digits. Elementary integration won't find the integral, it has to be done numerically. Choose a method and obtain 2.781197xxxx. MAPLE ANSWER CHECK F:=x->x*exp(x^3); int(F(x),x=0..1); # Re-prints the problem. No answer. evalf(%); # ANS=0.7811970311 by numerical integration.

Second lecture on numerical methodsStudy problems like y'=-2xy, which have the form y'=f(x,y). New algorithms are needed. Rect, Trap and Simp won't work, because of the variable y on the right.Euler, Heun, RK4 algorithmsComputer implementation in maple Geometric and algebraic ideas in the derivations. Numerical Integration Numerical Solutions of y'=f(x,y) RECT Euler TRAP Heun [modified Euler] SIMP Runge-Kutta 4 [RK4]Referencefor the ideas is this: Numerical methods (138.0 K, pdf, 15 Feb 2017)SlidesNumerical Solution of y'=f(x,y)Two problems will be studied. First problem y' = -2xy, y(0)=2 Symbolic solution y = 2 exp(-x^2) Second problem y' = (1/2)(y-1)^2, y(0)=2 Symbolic solution y = (x-4)/(x-2) MAPLE TUTOR for NUMERICAL METHODS # y'=-2xy, y(0)=2, by Euler, Heun, RK4 with(Student[NumericalAnalysis]): InitialValueProblemTutor(diff(y(x),x)=-2*x*y(x),y(0)=2,x=0.5); # The tutor compares exact and numerical solutions.ExamplesWeb references contain two kinds of examples. The first three are quadrature problems dy/dx=F(x). y'=3x^2-1, y(0)=2, solution y=x^3-x+2 y'=exp(x^2), y(0)=2, solution y=2+int(exp(t^2),t=0..x). y'=2x+1, y(0)=3 with solution y=x^2+x+3. The fourth is of the form dy/dx=f(x,y), which requires a non-quadrature algorithm like Euler, Heun, RK4. y'=1-x-y, y(0)=3, solution y=2-x+exp(-x). ==> Left off here on Friday, will continue on MondayWORKED EXAMPLEy'=1-x-y, y(0)=3, solution y=2-x+exp(-x). We will make a dot table by hand and also by machine. The handwritten work for the Euler Method and the Improved Euler Method [Heun's Method] are available:: Handwritten example y'=1-x-y, y(0)=3, Euler and Heun (427.8 K, jpg, 15 Dec 2012) The basic reference:Jpeg: Numerical methods (138.0 K, pdf, 15 Feb 2017)SlidesEULER METHODLet f(x,y)=1-x-y, the right side of the differential equation. Use step size h=0.2 from x=0 to x=0.4. The dot table has 3 rows. Table row 1: x0=0, y0=3 Taken from initial condition y(0)=3 Table row 2: x1=0.2, y1=2.6 Compute from x1=x0+h, y1=y0+hf(x0,y0)=3+0.2(1-0-3) Table row 3: x2=0.4, y2=2.24 Compute from x2=x1+h, y2=y1+hf(x1,y1)=2.6+0.2(1-0.2-2.6) MAPLE EULER: [0, 3], [.2, 2.6], [.4, 2.24]HEUN METHODLet f(x,y)=1-x-y, the right side of the differential equation. Use step size h=0.2 from x=0 to x=0.4. The dot table has 3 rows. Table row 1: x0=0, y0=3 Taken from initial condition y(0)=3 Table row 2: x1=0.2, y1=2.62 Compute from x1=x0+h, tmp=y0+hf(x0,y0)=2.6, y1=y0+h(f(x0,y0)+f(x1,tmp))/2=2.62 Table row 3: x2=0.4, y2=2.2724 Compute from x2=x1+h, tmp=y1+hf(x1,y1)=2.62+0.2*(1-0.2-2.62) y2=y1+h(f(x1,y1)+f(x2,tmp))/2=2.2724 MAPLE HEUN: [0, 3], [.2, 2.62], [.4, 2.2724]RK4 METHODLet f(x,y)=1-x-y, the right side of the differential equation. Use step size h=0.2 from x=0 to x=0.4. The dot table has 3 rows. The only Honorable way to solve RK4 problems is with a calculator or computer. A handwritten solution is not available (and won't be). Table row 1: x0=0, y0=3 Taken from initial condition y(0)=3 Table row 2: x1=0.2, y1=2.618733333 Compute from x1=x0+h, 5 lines of RK4 code Table row 3: x2=0.4, y2=2.270324271 Compute from x2=x1+h, 5 lines of RK4 code MAPLE RK4: [0, 3], [.2, 2.618733333], [.4, 2.270324271]EXACT SOLUTIONWe solve the linear differential equation by the integrating factor method to obtain y=2-x+exp(-x). # MAPLE evaluation of y=2-x+exp(-x) F:=x->2-x+exp(-x);[[j*0.2,F(j*0.2)] $j=0..2]; # Answer [[0., 3.], [0.2, 2.618730753], [0.4, 2.270320046]]COMPARISON GRAPHICThe three results for Euler, Heun, RK4 are compared to the exact solution y=2-x+exp(-x) in the: y'=1-x-y Compare Euler-Heun-RK4 (17.8 K, jpg, 09 Dec 2012) The comparison graphic was created with thisGRAPHIC: y'=1-x-y by Euler-Heun-RK4 (1.1 K, txt, 09 Dec 2012)MAPLE TEXT

EULER SOLUTION ATOMS and INDEPENDENCE.Def. atom=x^n(base atom) base atom = 1, exp(ax), cos(bx), sin(bx), exp(ax) cos(bx), exp(ax) sin(bx) for nonzero real a and positive real b THEOREM. Euler solution atoms are independent. EXAMPLE. Show 1, x^2, x^9 are independent by 3 methods. EXAMPLE. Independence of 1, x^2, x^(3/2) by the Wronskian test. Equation y''+10y'=0 has general solution y=c1 + c2 exp(-10x) The Euler solution atoms for this example are 1 and exp(-10x). Differential equations like this one have general solution a linear combination of atoms. THEOREM. An nth order scalar homogeneous linear differential equation has general solution a linear combination of Euler solution atoms.TOOLKIT for SOLVING LINEAR CONSTANT DIFFERENTIAL EQUATIONSPicard: Order n of a DE = dimension of the solution space. General solution = linear combination n independent atoms. Euler's theorem(s), an algorithm for finding solution atoms.Summary for Higher Order Differential Equations: Atoms, Euler's theorem, 7 examples (130.5 K, pdf, 25 Feb 2013)Slides: Base atom, atom, basis for linear DE (106.8 K, pdf, 07 Feb 2018) EXAMPLE. The equation y'' +10y'=0. How to solve y'' + 10y' = 0, chapter 1 methods. Idea: Let v=x'(t) to get a first order DE in v and a quadrature equation x'(t)=v(t). Solve the first order DE by the linear integrating factor method. Then insert the answer into x'(t)=v(t) and continue to solve for x(t) by quadrature. Vector space of functions: solution space of a differential equation. A basis for the solution space of y'' + 10y'=0 is {1,exp(-10x)} PICARDS THEOREM The theorem of Picard and Lindelof for y'=f(x,y) has an extension for systems of equations, which applies to scalar higher order linear differential equations. THEOREM [Picard] A homogeneous nth order linear differential equation with continuous coefficients has a general solution written as a linear combination of n independent solutions. This means that the solution space of the differential equation has dimension n.Slides