2. Use the Laplace transform to solve Y" – 2y = 2 y(0) = 0, y'(0)...
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
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)
1. (5 points) Use a Laplace transform to solve the initial value problem: y' + 2y + y = 21 +3, y(0) = 1,5 (0) = 0. 2. (5 points) Use a Laplace transform to solve the initial value problem: y + y = f(t), y(0) = 1, here f(0) = 2 sin(t) if 0 Str and f(0) = 0 otherwise.
B) Use the Laplace transform to solve the given initial-value problem: y' + 2y = 2cos(2+), with y(0) = 1
Use the Laplace transform to solve the following problem: y' + 2y – 38* ydx = 5 + 5x, y(0) = 2 This will require several expansions in partial fractions.
Use the Laplace transform to solve the initial value problem: y" - 3y' + 2y = 4t + ezt, y(0) = 1, y'(0) = -1
7. 49 + 2y(t) = 4t; y(0-3. Use Laplace transform to solve this problem.
(4 points) Use the Laplace transform to solve the following initial value problem: y" – 2y + 5y = 0 y(0) = 0, y'(0) = 8 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}| find the equation you get by taking the Laplace transform of the differential equation = 01 Now solve for Y(3) By completing the square in the denominator and inverting the transform, find g(t) =
use Laplace transform to solve! 10. 0 , y"(0) 1 . 0 , y,(0) sin(3t) , y(0) y" + 2y"-у,-2y
1) y'' -2y'+y=xE^x, y(0)=y'(0)=0 Solve the initial value problem using the Laplace transform. y" – 2y + y = xe*, y(0) = y'(0) =