Question

1 point) Use the Laplace transform to solve the following initial value problem: y" - 9y' + 18y-0, y(0) -3, y' (0) 3 (1) First, using Y for the Laplace transform of y(t), i.e., Y-C00), find the equation you get by taking the Laplace transform of the differential equation to obtain (2) Next solve for Y (3) Now write the above answer in its partial fraction form, Y- (NOTE: the order that you enter your answers matter so you must order your terms so that the first corresponds to a and the second s- a s-b to b, where a < b. Also note, for example that -2< 1) 1 (4) Finally apply the inverse Laplace transform to find y(t) y(t)

Answer 1

yPrime - 9yprime + 18y = 0: y(0) = 3, yprime(0) = 3 Taking Laplace transform both sides: L{yPrime} - 9L {yprime} + 18L {y} = 0 So, s^2Y(s) -s, y(0) - yprime(0) - 9(sY(s) - y(0))+ 18Y (s) = 0 Apply initial conditions: So, s^2.Y(s) - s.(3) - 3 - 9(s.Y(s) - 3) + 18Y (s) = 0 So, s^2.Y(s) - s.(3) - 3 - 9s.Y(s) + 27 + 18Y (s) = 0 So, s^2.Y(s) - 9s.Y (s) + 18Y (s) = 3s - 24 So, Y(s)(s^2 - 9s + 18) = 3s - 24 So, Y(s) = 3s - 24/s^2 - 9s + 18 Take partial fraction Y(s) = 3s - 24/s^2 - 9s + 18 = 3s - 24/(s - 3) (s - 6) = A/s - 3 + B/s - 6