Laplace transform of the unit step function
Given equation is
Taking Laplace transform of both sides, we get
6 (5) Solve the differential equation using a Laplace Transform: y 3y' +2y t y(0) 0, y'(0) 2
10. Use the Laplace transform to solve y" - 3y' +2y f(t), y(0)-0,'(0) 0, where (t)-(0 for 0 st < 4; for t 2 4 No credit will be given for any other method. (10 marks)
2. Use the Laplace Transform to solve the initial value problem y"-3y'+2y=h(t), y(O)=0, y'(0)=0, where h (t) = { 0,0<t<4 2, t>4
Use the Laplace transform to solve the initial value problem: y" - 3y' + 2y = 4t + ezt, y(0) = 1, y'(0) = -1
use the Laplace transform to solue the initial value problem y - 3y +2y = h (t) Y(o)=0 yo(o)=0 where h(t): 0,0 Lt L4 t> 4 { 2
Consider the initial value problem y" +3y' +2y = (t-1)+r(t), y(0) = y(0) = 0, where 8(t-1) is Dirac's delta function and S4 if 0<t<1 r(t) 8 if t > 1 (a) Represent r(t) using unit step functions. (b) Find the Laplace Transform of 8(t-1)+r(t). (c) Solve the above initial value problem. {
Find the solution of the given IVP y" + 3y' + 2y = uz(t); y(0) = 0, y'(0) = 1 a. y = et-e-t + uz(t) [] + e-(6+2) +22(6+2) b. y = ef +e-t+uz(t)ſ - e-(6-2) + şe-26-2)] + uz(t) - e-(1-2) 3e=2(-2)] e + C. y = e-t-e-27 d. None of these
Find the solution of the given IVP y" + 3y' + 2y = Uz(t); y(0) = 0, y'(0) = 1 + e-(t+2) e-2(t+2) + e 2 a. y=et-e-t + uz(t) [+ b. y=et +e-+ + uz(t) [ – e-(6-2) + že=2(t-2)] c. y = e-t-e-2t + uz(t) (2) - e-(4-2) + že=2(t-2)] + d. None of these
Laplace transform of the unit step function
y" + 4y = ſi, if 0 <t<, y(0) = 0, y'(0) = 0. 10, if a St<oo.'
Express the forcing function in terms of unit step function, then use the laplace transform to solve the initial value problem: y"+y={1,4 0<=t<2, t>=2 y(0)=2 y'(0)=1