Exam 3 ReviewPatrick will do exam 3 review this week on Thursday, JTB 140 7:30 to 9:25am. Exam 3 part I: April 12 The problems are 1,2 on the sample exam below, which matches 3,4 on the S2010 exam key. See the Fall 2010 exam 3 solution key for another sample exam, in which the problem numbering is the same as your exam. Exam 3 part II: April 19 The Thursday exam is at the usual 7:30am time in JTB 140. The problems are 3,4,5 on the sample exam below, which matches 1,2,5 on the S2010 exam key.

More Laplace ExamplesContinuing 10.3, 10.5 examples from last lecture.Review of the Laplace resolvent method for 2x2 systemsProblem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au. x' = 2x + y, y' = 3y ==> u'=Au Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au. u'=Au where A=<2,0|1,3> and u(0)=<1,2> The general vector-matrix DE Model u'=Au BASIC RESULT: Laplace of u(t) = Resolvent matrix times vector u(0) DEFINITION: Resolvent matrix = inverse(sI - A) Solve the systems by ch1 methods for x(t), y(t): x' = 2x, x(0)=100, y' = 3y, y(0)=50. Answer: x = 100 exp(2t), y = 50 exp(3t) x' = 2x+y, x(0)=1, y' = 3y, y(0)=2. Answer: y(t) = 2 exp(3t) and x(t) is the solution of the linear integrating factor problem x'(t)=2x(t)+2 exp(3t). Review of problem 10.2-16, a 3x3 system that can be solved with the resolvent equation shortcut. RULE: Use Cramer's rule or matrix inversion to solve the resolvent equation for the vector of components L(x), L(y), L(z). Any linear algebra problem Bu=c where B contains symbols should be solved this way, unless B is triangular.Variation of parametersThe second order formula. Application to y''=1+x Application to y''+y=sec(x) [see also slides] How to calculate y_p(x) from the five parameters y1(x) y2(x) W(x) = y1(x)y2'(x)-y1'(x)y2(x) A(x) = coefficient in the DE of y'' f(x) = input or forcing term, the RHS of the DE See (33) in section 5.5.

Undetermined CoefficientsWhich equations can be solved Intro to the basic trial solution method Laplace solution of y'' + y = 1+x [use x''(t)+x(t) = 1+t, x(0)=x'(0)=0] How to find the atoms in y_p(x). How to find the atoms in y_h(x) 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 EXAMPLE. How to find a shortest expression for y_p(x) using Laplace's method. Details for x''(t)+x(t) = 1+t, to obtain the trial solution x(t)=A+Bt and 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, ... METHODS to FIND the SOLUTION. Laplace solution of y''(t) + y(t) = f(t) when f(t)=linear combination of atoms We get L(y) = polynomial fraction. By partial fractions and the backward table, L(y)=L(g(t)), where g(t)=linear combination of atoms! This proves the general result: 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 1. Laplace theory. It works, but it is slow. Central to this method is dividing L(f(t)) by the characteristic polynomial, then writing this polynomial fraction as L(g(t)). The homogeneous solution atoms are shrunk from g(t) to obtain the trial solution. METHOD 2. A rule for finding y_p(x) from f(x) and the DE. We get rid of Laplace theory's t-variable and use x. Finding a trial solution with fewest symbols.Rule 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?

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.

Transform TerminologyConvolution theorem and x'' + 4x = cos(t), x(0)=x'(0)=0. Input Output Transfer FunctionCircuits EPbvp3.7: Electrical resonance. Derivation from mechanical problems 5.6. THEOREM: omega = 1/sqrt(LC). Impedance, reactance. Steady-state current amplitude Transfer function. Input and output equation.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.

Topics from linear systems:Brine tank models. Recirculating brine tanks. Pond pollution. Home heating. Earthquakes. Railway cars. All are 2x2 or 3x3 or nxn system applications that can be solved by Laplace methods.Systems of two differential equationsSolving a system from Chapter 1 methods The Laplace resolvent method for systems. Cramer's Rule, Matrix inversion methods. EXAMPLE: Solving a 2x2 dynamical system using Laplace's resolvent method. Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]). EXAMPLE: Problem 10.2-16, This problem is a 3x3 system for x(t), y(t), z(t) solved by Laplace theory methods. The resolvent formula (sI - A) L(u(t)) = u(0) with u(t) the fixed 3-vector with components x(t), y(t), z(t), amounts to a shortcut to obtain the equations for L(x(t)), L(y(t)), L(z(t)). After the shortcut is applied, in which Cramer's Rule is the method of choice, to find the formulas, there is no further shortcut: we have to find x(t), for example, by partial fractions and the backward table, followed by Lerch's theorem.

Systems of two differential equationsThe Laplace resolvent method for systems. Solving the resolvent equation for L(x), L(y). Cramer's Rule Matrix inversion Elimination Example: Solving a 2x2 dynamical system Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]). Dynamical system scalar form is x' = 2x + y, y' = 3y, x(0)=1, y(0)=2. Chapter 1: Linear integrating factor method. Laplace resolvent method The shortcut equations Solving for L(x), L(y) Backward table and Lerch's theorem Answers: x(t) = - e^{2t} + 2 e^{3t}, y(t) = 2 e^{3t}. Chapter 1+5 Shortcut Method [appears in Edwards-Penney] Solve w'+p(t)w=0 as w = constant / integrating factor. Then y(t) = 2 exp(3t) Stuff y(t) into the first DE to get the linear DE x' - 2x = 6 exp(3t) Superposition: x(t)=x_h(t)+x_p(t), x_h(t)=c exp(2t), x_p(t) = d1 exp(t) = 2 exp(3t) by undetermined coeff. Then x(t)= - exp(2t) + 2 exp(3t).