Question

C.2. a. Find the inverse transform. C-12 b. Verify that your answer to part a is correct by finding the transform of your answer. C.3. a. Find the inverse transform. b. Verify that your answer to part a is correct by finding the transform of your answer. C.4. a. Find Eitse -2t) b. Verify that your answer to part a is correct by finding the inverse transform of your answer. a. Use the Laplace transform to solve the initial value problem: b. Verify that your answer to part a is correct by substitution into the original differential C.5. t4e3t equation. are aske

Answer 1

Lscr^-1 {squareroot 2/s^4} Because Lscr^-1 {1/s^n} = t^n - 1/(n)! Lscr^-1 {squareroot 2/s^4} = squareroot 2 t^3/3! = squareroot 2 /6 t^3 Now Lscr {squareroot 2/6 t^3} Because Lacr{t^n} = n!/s^n + 1 Because Lscr {squareroot 2/6 t^3} = squareroot 2/6 3!/t^4 = squareroot 2/t^4 Hence get value as in (a) part of 9 th Hence verified Lscr^-1 {3/s^3 - 2/s^2}^2 Lscr^-1 {9/s^6 + 4/s^4 - 12/s^5} {Because Lscr^-1 {1/s^n} = t^n -1/(n - 1)!} therefore Lscr^-1 {9/s^6 + 4/s^4 - 12/s^5} = 9 t^5/5! + 4 t^3/3! - 12 t^4/4! Lscr^-1 {(3/s^3 - 1/s^2)^2} = 3/40 t^5 + 2/3 t^3 - 1/2 t^4 Now Lscr {3/40 t^5 + 2/3 t^3 - 1/2 t^4} { Because Lscr {t^} = n!/s^n + 1} Lscr {3/40 t^5 + 2/3 t^3 - 1/2 t^4} = 3/40 5!/s^6 + 2/3 3!/s^4 - 1/2 4!/s^5 = 9/s^6 + 4/s^4 - 12/s^5 = (3/s^3 - 2/s^2)^2 Hence verified, we get the result as in part (a). \r\n Lscr{t^3 e^-2t} Using First slitting theorem
C.5

Answer (b)