A 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 TheorySome engineering functions unit step ramp sawtooth wave other periodic waves next MondayApplicationsHow to solve differential equations LRC Circuit 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: step(t)=1 for t>=0, step(t)=0 for t<0. Pulse: pulse(t,a,b)=step(t-a)-step(t-b) Ramp: ramp(t-a)=(t-a)step(t-a) L(step(t-a)) = (1/s) exp(-as) [for a >= 0 only]Piecewise defined periodic wavesSquare wave Triangular wave SawtoothIntegral TheoremL(int(g(x),x=0..t)) = s L(g(t)) Applications to computing ramp(t-a) L(ramp(t-a)) = (1/s^2) exp(-as) [for a >= 0 only]

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.