ch5

Chapter 5: Applications The Laplace Transform to Differential Equations Chapter 5.1: ODE with constant coefficients : ƒ We wish to solve the equation: dny d n −1 y dy + a0 y = f (t ) an n + an −1 n −1 + + a1 dt dt dt to the initial conditions (IC) : Chapter (1)  y(0) = y0  '  y (0) = y1    y(n −1) (0) = y  n −1 [2] ƒ General Procedure: i Taking the Laplace Transform of both sides Equation in s-space ii Determine Y(s) iii Using Inverse Laplace Transform of finding the original function y(t) Recall: Chapter L { y (n ) (t)} = s n Y ( s ) − s n −1.y(0) − − y (n −1) (0) [3] ™Example1: Solve ODE by Laplace Transform Solve y ''− y = t y (0) = 1; y '(0) = with Step1: Laplace Transform s Y − sy (0) − y '(0) − Y = 1/s 2 Step2: Solving in s-space and partial fraction expansion gives: Y(s) = s −1 + s2 −1 − s2 1 Step3: we obtain the solution −1 y(t) = L {Y(s)} = e + sinh t − t Chapter t [4] 5.2: System of ODEs : ƒ We wish to solve the state variable equation:  dx  dt = Ax + By   dy = Cx + Dy  dt ƒ Applying : to the initial conditions (IC) :  x(0) = x0   y(0) = y0 L {dx / dt} = sX ( s ) − x to solve X(s) & Y(s) ƒ Find x(t) and y(t) using Inverse Laplace Transform Chapter [5] ™Example: System of ODEs  y1' = − 100 y1 + 100 y2 + Solve  ' 8  y2 = 100 y1 + 100 y2  y1 (0) = with   y2 (0) = 150 ( s + 0.08)Y1 − 0.02Y2 = 6s   −0.08Y1 + ( s − 0.08)Y2 = 150 Y1 = 100 s Y2 = 100 s Chapter − 62.5 s + 0.12 − 37.5 s + 0.04 + 125 s + 0.12 − 75 s + 0.04 [6] 5.3: Applications to Mechanics : Chapter [7]