Partly solved 5.4-34.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. See also the web site FAQ for more details.Partly solved 5.4-20The 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. Solution Atoms=cos( sqrt(20)t), sin( sqrt(20)t). Each system has general solution a linear combination of the 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)/2Laplace TheoryA brief Laplace table. Forward table. Backward table. Extensions of the Table. Laplace rules. Linearity. The s-differentiation theorem (d/ds)L(f(t))=L((-t)f(t)). Shift theorem. Parts theorem. Finding Laplace integrals using Laplace calculus. Solving differential equations by Laplace's method.Basic Theorems of Laplace TheoryLerch's theorem Linearity. The s-differentiation theorem (d/ds)L(f(t))=L((-t)f(t)). Shift theorem L(exp(at)f(t)) = L(f(t))|s->(s-a) Parts theorem L(y')=sL(y)-y(0)

Solving y' = -1, y(0)=2 Solving y''+y=0, y(0)=0, y'(0)=1 Solving y''+y=1, y(0)=y'(0)=0 Solving y''+y=cos(t), y(0)=y'(0)=0 Computing Laplace integrals Solving an equation L(y(t))=expression in s for y(t) Dealing with complex roots and quadratic factors Partial fraction methods Using the s-differentiation theorem Using the shifting theorem Harmonic oscillator y''+a^2 y=0

Basic Theorems of Laplace TheoryPeriodic function theorem Proof Some engineering functions unit step ramp sawtooth wave other periodic waves next Monday Convolution theorem application L(cos t)L(sin t) = L(0.5 t sin(t))ApplicationsHow to solve differential equations LRC Circuit Second shifting rule Specialized models. Pure Resonance x''+x=cos(t) Solution explosion, unbounded solution x=(1/2) t sin t. Resonance examples: Soldiers marching in cadence, Tacoma narrows bridge, Wine Glass Experiment. Theodore Von Karman and vortex shedding. Cable model of the Tacoma bridge, year 2000. Resonance explanations. Beats x''+x=cos(2t) Graphics for beats [x=sin(10 t)sin(t/2)], slowly-oscillating envelope, rapidly oscillating harmonic with time-varying amplitude.

Piecewise FunctionsUnit Step Pulse RampSecond shifting Theoremse^{-as}L(f(t))=L(f(t-a)step(t-a)) L(g(t)step(t-a))=e^{-as}L(g(t+a))Piecewise defined periodic wavesSquare wave Triangular wave Sawtooth Rectified sine Half-wave rectified sine Parabolic wavePeriodic function theoremLaplace of the square wave, tanh function.Laplace Resolvent Method.--> This method is considered a shortcut for systems by Laplace's method. --> It is also a convenient way to solve systems with maple. Consider problem 10.2-16 x'=x+z, y'=x+y, z'=-2x-z, x(0)=1, y(0)=0, z(0)=0 Write this as a matrix differential equation u'=Bu, u(0)=u0 Then u:=vector([x,y,z]); B:=matrix([[1,0,1],[1,1,0],[-2,0,-1]]); u0:=vector([1,0,0]); If we think of the matrix differential equation as a scalar equation, then its Laplace model is -u(0) + s L(u(t)) = BL(u(t)) or equivalently sL(u(t)) - B L(u(t)) = u0 Write s = sI where I is the 3x3 identity matrix. Then the Laplace model is (aI - B) L(u(t)) = u0 which is called the Resolvent Equation. TheRESOLVENTis the inverse of the matrix multiplier on the left: Resolvent == inverse(sI - B) It is so-named because the vector of Laplace answers is vector([L(x),L(y),L(z)]) = L(u(t)) = inverse(sI - B) u0 Briefly, Laplace of u(t) = RESOLVENT x u(0)Intro to the Laplace resolvent method for 2x2 systemsRemarks on problem 10.2-16, a 3x3 system that can be solved with the resolvent equation shortcut. Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au. Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au. The general vector-matrix DE Model u'=Au Laplace of u(t) = Resolvent x u(0) Resolvent = 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). 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.