Edwards-Penney, sections 5.4, 5.5, 5.6, 7.1 The textbook topics, definitions and theorems

Edwards-Penney 5.1, 5.2, 5.3, 5.4 (15.6 K, txt, 19 Dec 2013)

Edwards-Penney 5.5, 5.6 (15.5 K, txt, 19 Dec 2013)

PROBLEM SESSIONChapter 4 exercises. 4.7 Independence. Classify as indep/dep. Cite these results. THEOREM. Any set of distinct Euler solution atoms is independent. THEOREM. Wronskian determinant nonzero => functions in Wronskian row 1 are independent. THEOREM. Subsets of independent sets are independent. THEOREM. S1 is a finite list of k vectors. S2 is a finite list of k vectors (the same number). span(S1) = span(S2) If S2 is independent, then S1 is also independent. Proof: If S1 is dependent, then dim(span(S1)) < k., which violates dim(span(S1))=dim(span(S2))=k. 4.7-13: sin x, cos x [Euler solution atoms are independent] 4.7-14: exp(x), x exp(x) [Euler solution atoms are independent] 4.7-15: 1+x,1-x, 1-x^2 [This is S1. Verify S2 = 1,x,x^2 works in the theorem.] 4.7-16: 1+x, x+x^2, 1-x^2 [Dependent, v2+v3=v1] 4.7-17: cos 2x, sin^2 x, cos^2 x [Dependent, by trig identity cos 2x = cos^2 x - sin^2 x, or v1=v3-v2] 4.7-18: 2 cos x + 3 sin x, 4 cos x + 5 sin x [This is S1. What is S2?]

Second order and higher order differential Equations.Detailed look at the second order caseREF. Theorems 1,2,3 in section 5.3 of Edwards-Penney. Quadratic equations again. Constant-coefficient SECOND ORDER homogeneous differential equations. Characteristic equation is quadratic. Its factors determine the atoms. The three possible cases 1. Two real roots 2. Two equal real roots 3. A complex conjugate pair of roots The three possible kinds of solutions 1. y=c1 exp(r1 x) + c2 exp(r2 x) If one root is zero, then the exponential equals 1. 2. y=c1 exp(r1 x) + c2 x exp(r1 x) 3. y= c1 exp(ax) cos(bx) + c2 exp(ax) sin(bx) If in root a+ib, the real part a=0, then this is a pure harmonic oscillation [because exp(ax)=1] y= c1 cos(bx) + c2 sin(bx) Sample equations: y''=0, y''+2y'+y=0, y''-4y'+4y=0, y'' + 3y' + 2y=0, x'' + x = 0, x'' + 2x' + 5x = 0, x'' + 8x' + 16x=0, 0.03I'' + 0.005 I = 0, Q'' + 100Q=0, REVIEW OF SECTIONS 5.1, 5.2, 5.3 PROBLEM TYPES. Example: Linear DE given by roots of the characteristic equation. Example: Linear DE given by factors of the characteristic polynomial. Example: Construct a linear DE of order 2 from a list of two atoms that are known to be solutions. Example: Construct a linear DE from characteristic equation roots. Example: Construct a linear DE from its general solution. COMPLEX ROOTS of the CHARACTERISTIC EQUATION Solving a DE when the characteristic equation has complex roots. Equations with both real roots and complex roots. (D^4 + D^2)y = 0, (D+1)^2(D^2+4)^2 y = 0 An equation with 4 complex roots. How to find the 4 atoms. (D^2+4)(D^2+16)y=0 SHORTCUT. One pair of complex conjugate roots identifies two Euler solution atoms. Only one of the two complex roots is required to construct the two atoms.Slides on Section 5.4Damped oscillations overdamped, critically damped, underdamped, pseudo-period [Chapter 5] phase-amplitude form of the solution [chapter 5] Cafe door. Pet door. Undamped oscillations. Harmonic oscillator.: Unforced vibrations 2008 (647.6 K, pdf, 28 Feb 2014)Slides: phase-amplitude, cafe door, pet door, damping classification (136.0 K, pdf, 08 Mar 2014)Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)Slides: Forced damped vibrations (264.0 K, pdf, 08 Mar 2014)SlidesPartly solved 5.4-20See the FAQ at the web site for answers and details.: FAQ for section 5.4 (4.7 K, txt, 05 Dec 2012) The problem breaks into two distinct initial value problems: (1) 2x'' + 16x' + 40x=0, x(0)=5, x'(0)=4 Characteristic equation 2(r^2+8r+20)=0. Roots r=-4+2i,r=-4-2i. Solution Atoms=e^{-4t}cos 2t, e^{-4t}sin 2t. UNDERDAMPED. (2) 2x'' + 0x' + 40x=0, x(0)=5, x'(0)=4 Characteristic equation 2(r^2+0+20)=0. Roots r=sqrt(20)i,r=-sqrt(20)i. The Euler solution atoms are cos( sqrt(20)t), sin( sqrt(20)t). UNDAMPED HARMONIC OSCILLATION. Each system has general solution a linear combination of Euler solution atoms. Evaluate the constants in the linear combination, in each of the two cases, using the initial conditions x(0)=5, x'(0)=4. There are two linear algebra problems to solve. Answers: (1) Coefficients 5, 2 for 2x'' + 16x' + 40x=0 Amplitude = sqrt(5^2 + 12^2) = 13 (2) Coefficients 5, 2/sqrt(5) for 2x'' + 0x' + 40x=0 Amplitude = sqrt(5^2 + 4/5) = sqrt(129/5) Plots can be made from these answers directly. Write each solution in phase-amplitude form, a trig problem. See section 5.4 for specific instructions. The book's answers: (1) tan(alpha) = 5/12 (2) tan(alpha) = 5 sqrt(5)/2TextPartly solved 5.4-34.See the FAQ at the web site for answers and details.: FAQ for section 5.4 (4.7 K, txt, 05 Dec 2012) The DE is 3.125 x'' + cx' + kx=0. The characteristic equation is 3.125r^2 + cr + kr=0 which factors into 3.125(r-a-ib)(r-a+ib)=0 having complex roots a+ib, a-ib. Problems 32, 33 find the numbers a, b from the given information. This is an inverse problem, one in which experimental data is used to discover the differential equation model. The book uses its own notation for the symbols a,b: a ==> -p and b ==> omega1. Because the two roots a+ib, a-ib determine the quadratic equation, then c and k are known in terms of symbols a,b. References: Sections 5.4, 5.6. Forced oscillations.Text: Unforced vibrations 2008 (647.6 K, pdf, 28 Feb 2014)Slides: phase-amplitude, cafe door, pet door, damping classification (136.0 K, pdf, 08 Mar 2014)Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)Slides: Forced damped vibrations (264.0 K, pdf, 08 Mar 2014)Slides: Forced vibrations and resonance (253.0 K, pdf, 08 Mar 2014)Slides: Undetermined coefficients, pure resonance, practical resonance (152.8 K, pdf, 03 Mar 2013)Slides: Electrical circuits (112.9 K, pdf, 08 Mar 2014)SlidesPREVIEW: Undetermined CoefficientsTHEOREM. Solution y_h(x) is a linear combination of atoms. THEOREM. Solution y_p(x) is a linear combination of atoms. THEOREM. (Superposition) y = y_h + y_p Which equations can be solved by undetermined coefficients. 1. Constant coefficients 2. Forcing term a linear combination of Euler solution atoms. Intro to the basic trial solution method Solution of x'' + 9x = 30 sin(2t) x(t)=c1 cos 3t + c2 sin 3t + a cos 2t + b sin 2t = sum of two harmonics, of frequencies 3 and 2 = BEATS example xh(t) = c1 cos 3t + c2 sin 3t, the first harmonic xp(t) = a cos 2t + b sin 2t, the second harmonic TRIAL SOLUTION The equation y'' + y = 1 + x should have a solution yp = a + bx for some constants a, b. We find the constants by substitution of yp into y'' + y = 1 + x. Ways to find the Euler solution atoms in y_p(x). 1. Use Laplace theory (a bit slow, even if you already know Laplace theory). 2. Use rules from undetermined coefficient theory. Faster. Euler solution atoms in y_h(x): Roots of the characteristic equation.: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)SlidesExtra Credit Maple Project: Tacoma narrows. Explore an alternative explanation for what caused the bridge to fail, based on the hanging cables.## Wednesday: Undetermined Coefficients. Sections 7.1,5.5

REVIEW: Undetermined CoefficientsWhich equations can be solved THEOREM. Solution y_h(x) is a linear combination of atoms. THEOREM. Solution y_p(x) is a linear combination of atoms. THEOREM. (superposition) y = y_h + y_p: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012) EXAMPLE. How to find a shortest expression for y_p(x) using Details for x''(t)+x(t) = 1+t the trial solution x(t)=A+Bt the answer x_p(t)=1+t. BASIC METHOD. Given a trial solution with undetermined coefficients, find a system of equations for d1, d2, ... and solve it. Report y_p as the trial solution with substituted answers d1, d2, d3, ... THEORY. y = y_h + y_p, and each is a linear combination of atoms. How to find the homogeneous solution y_h(x) from the characteristic equation. How to determine the form of the shortest trial solution for y_p(x) METHOD. A rule for finding y_p(x) from f(x) and the DE. Finding a trial solution with fewest symbols.SlidesRule I. Assume the right side f(x) of the differential equation is a linear combination of atoms. Make a list of all distinct atoms that appear in the derivatives f(x), f'(x), f''(x), ... . Multiply these k atoms byundetermined coefficientsd_1, ... , d_k, then add to define atrial solution y. This ruleFAILSif one or more of the k atoms is a solution of the homogeneous differential equation.Rule II. If Rule IFAILS, then break the k atoms into groups with the samebase atom. Cycle through the groups, replacing atoms as follows. If the first atom in the group is a solution of the homogeneous differential equation, then multiply all atoms in the group by factor x. Repeat until the first atom is not a solution of the homogeneous differential equation. Multiply the constructed k atoms by symbols d_1, ... , d_k and add to define trial solution y.Explanation: The relation between the Rule I + II trial solution and the book's table that uses the mystery factor x^s. EXAMPLES. y'' = x y'' + y = x exp(x) y'' - y = x exp(x) y'' + y = cos(x) y''' + y'' = 3x + 4 exp(-x) THEOREM. Suppose a list of k atoms is generated from the atoms in f(x), using Rule I. Then the shortest trial solution has exactly k atoms. EXAMPLE. How to find a shortest trial solution using Rules I and II. Details for x''(t)+x(t) = t^2 + cos(t), obtaining the shortest trial solution x(t)=d1+d2 t+d3 t^2+d4 t cos(t) + d5 t sin(t). How to use dsolve() in maple to check the answer. EXAMPLE. Suppose the DE has order n=4 and the homogeneous equation has solution atoms cos(t), t cos(t), sin(t), t sin(t). Assume f(t) = t^2 + cos(t). What is the shortest trial solution? EXAMPLE. Suppose the DE has order n=2 and the homogeneous equation has solution atoms cos(t), sin(t). Assume f(t) = t^2 + t cos(t). What is the shortest trial solution? EXAMPLE. Suppose the DE has order n=4 and the homogeneous equation has solution atoms 1, t, cos(t), sin(t). Assume f(t) = t^2 + t cos(t). What is the shortest trial solution?## Friday: Resonance, Section 5.6

Undetermined coefficientsExamples continued from the previous lecture. EXAMPLES. y'' = x y'' + y = x exp(x) y'' - y = x exp(x) y'' + y = cos(x) y''' + y'' = 3x + 4 exp(-x) Shortest trial solution. Two Rules to find the shortest trial solution. 1. Compute the atoms in f(x). The number k of atoms found is the number needed in the shortest trial solution. 2. Correct groups with the same base atom, by multiplication by x until the group contains no atom which is a solution of the homogeneous problem [eliminate homogeneous DE conflicts].The x^s mystery factorin the book's table. The number s is the multiplicity of the root in the homogenous DE characteristic equation, which constructed the base atom of the group.Reference: Edwards-Penney, Differential Equations and Boundary Value Problems, 4th edition, section 3.7 [math 2280 textbook]. Extra pages supplied by Pearson with bookstore copies of the 2250 textbook. Also available as a xerox copy in case your book came from elsewhere. Check-out the 2280 book in the math center or the Math Library. All editions of the book have identical 3.7 and 7.6 sections.Wine Glass ExperimentThe lab table setup Speaker. Frequency generator with adjustment knob. Amplifier with volume knob. Wine glass. x(t)=deflection from equilibrium of the radial component of the glass rim, represented in polar coordinates, orthogonal to the speaker front. mx'' + cx' + kx = F_0 cos(omega t) The model of the wine glass m,c,k are properties of the glass sample itself F_0 = volume knob adjustment omega = frequency generator knob adjustmentTheory of Practical ResonanceThe equation is mx''+cx'+kx=F_0 cos(omega t) THEOREM. The limit of x_h(t) is zero at t=infinity THEOREM. x_p(t) = C(omega) cos(omega t - phi) C(omega) = F_0/Z, Z^2 = A^2+B^2, A and B are the undetermined coefficient answers for trial solution x(t) = A cos(omega t) + B sin(omega t). THEOREM. The output x(t) = x_h(t) + x_p(t) is graphically just x_p(t) = C(omega) cos(omega t - phi) for large t. Therefore, x_p(t) is the OBSERVABLE output. THEOREM. The amplitude C(omega) is maximized over all possible input frequencies omega>0 by the single choice omega = sqrt(k/m - c^2/(2m^2)). DEFINITION. Thepractical resonance frequencyis the number omega defined by the above square root expression. Projection: glass-breaking video. Wine glass experiment. Tacoma narrows.: Wine glass breakage (QuickTime MOV) (96.8 K, mov, 21 Mar 2013)Video: Wine glass experiment (12mb mpg, 2min) (12493.8 K, mpg, 01 Apr 2008)Video: Tacoma Narrows Bridge Nov 7, 1940 (18mb mpg, 4min) (18185.8 K, mpg, 01 Apr 2008)Video: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)Slides: Resonance and undetermined coefficients (178.0 K, pdf, 08 Mar 2014)SlidesVariation of Parameters and Undetermined Coefficients references: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)SlidesSystems of Differential Equations references: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)Manuscript: Laplace resolvent method (88.1 K, pdf, 04 Mar 2012)Slides: Laplace second order systems (288.1 K, pdf, 04 Mar 2012)Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 10 Apr 2014)Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)SlidesOscillations. Mechanical and Electrical.: Electrical circuits (112.9 K, pdf, 08 Mar 2014)Slides: Forced damped vibrations (264.0 K, pdf, 08 Mar 2014)Slides: Forced vibrations and resonance (253.0 K, pdf, 08 Mar 2014)Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)Slides: Resonance and undetermined coefficients (178.0 K, pdf, 08 Mar 2014)Slides: Unforced vibrations 2008 (647.6 K, pdf, 28 Feb 2014)Slides