University:
North Central College
Course:
MATH 152 | Calculus II
Academic year:

2024
Views:

77

Pages:

4
Author:

nesikhiyawl0

Question 1 Find the inverse Laplace transform of G(s) = 1/(s² - 4s + 5) Answer Step 1 Calculation Function is given as 1/(s² - 4s + 5) to find laplace inverse transform 1/(s² - 4s + 5) = 1/(s² - 4s + 4 + 1) = 1/((s - 2)² + 1) Step 2 Applying inverse laplace transform rule: L{F(s)} = f(t) then L⁻¹{F(s – a)} = e^(at) f(t) SO, L⁻¹{1/((s-2)²+1)} = e^(2t)L⁻¹{1/(s²+1)} L⁻¹{1/(s²+1)} = sin(t) L⁻¹{1/((s-2)²+1)} = e^(2t) sin t SO, L⁻¹{1/(s²-4s+5)} = e^(2t)sin(t) Question 2 The Laplace transform of L{d³x(t)/dt³} is given by Select one: sX(s) - x(0) s³X(s) – s²x(0) – sx'(0) – x"(0) s³X(s) – s²x(0) – sx'(0) – x(0) s²X(s) – sx(0) - x'(0) Answer Step 1 L{d³x(t)/dt³} L[x'(t)] = ∫(0 to ∞) e^(-st) . x'(t) dt = [e^(-st) . x(t)](0 to ∞) - ∫(0 to ∞) (-se^(-st))x(t)dt = -x(0) + s∫(0 to ∞) e^(-st)x(t)dt (e^(-st)x(t) = 0 when t = ∞) = -x(0) + sL{x(t)} L{x'(t)} = sL{x(t)} - x(0) L{x''(t)} = sL{x'(t)} - x'(0) = s(sX(s) - x(0)) - x'(0) = s²X(s) - sx(0) - x'(0) L{x'''(t)} = sL{x''(t)} - x''(0) = s(s²X(s) - sx(0) - x'(0)) - x''(0) = s³X(s) - s²x(0) - sx'(0) - x''(0) L{d³x(t)/dt³} = s³X(s) – s²x(0) – sx'(0) – x''(0) Question 3 Use the table of laplace transform to find the inverse Laplace transform of F(s) = (2 + s(s + 1))/(s(s² - s - 6)) Answer Step 1 solution: We need to find the inverse laplace transform of F(s) = (2 + s(s + 1))/(s(s² - s - 6)) Step 2 For this F(s) = (2 + s(s + 1))/(s(s² - s - 6)) F(s) = (2 + s² + s)/(s(s² - s - 6)) = (s²+s+2)/(s(s²-s-6)) = (s²+s+2)/(s(s-3)(s+2)) F(s) = (2/(s(s-3)(s+2))) + ((s+1)/((s-3)(s+2))) using partial fraction (2/(s(s-3)(s+2))) = A/s + B/(s-3) + C/(s+2) 2 = A(s-3)(s+2) + Bs(s+2) + Cs(s-3) s=0 => 2 = -6A => A = -1/3 s=3 => 2 = 15B => B = 2/15 s=-2 => 2 = 10C => C = 1/5 (s+1)/((s-3)(s+2)) = A/(s-3) + B/(s+2) s+1 = A(s+2) + B(s-3) s = -2 => -1 = -5B => B = 1/5 s = 3 => 4 = 5A => A = 4/5 F(s) = (-1/3)/s + (2/15)/(s-3) + (1/5)/(s+2) + (4/5)/(s-3) + (1/5)/(s+2) F(s) = (-1/3)/s + (14/15)/(s-3) + (2/5)/(s+2) Taking inverse laplace transform is f(t) = L⁻¹{F(s)} = -1/3 L⁻¹{1/s} + 14/15 L⁻¹{1/(s-3)} + 2/5 L⁻¹{1/(s+2)} f(t) = -1/3 + 14/15 e^(3t) + 2/5 e^(-2t) Question 4 Using only the Laplace transform table, obtain the Laplace transforms of the following function: cosh(4t) - 4t where "cosh" stands for hyperbolic cosine and cosh x = (e^x+e^(-x))/2 Answer f(t) = cosh(4t) - 4t f(t) = (e^(4t) + e^(-4t))/2 - 4t By applying Laplace transform: F(s) = 1/2(1/(s-4) + 1/(s+4)) - 4/s² F(s) = s/(s²-16) - 4/s² Question 5 find the Laplace transform of f (t). f(t) = t²cos 2t Answer Given f(t) = t² cos 2t Since L(cos(at)) = s/(s² + a²) L(cos(2t)) = s/(s² + 2²) = s/(s² + 4) Using property, L{t^n*g(t)} = (-1)^n d^n/ds^n L{g(t)} L(t² cos 2t) = (-1)² d²/ds² (s/(s² + 4)) L(t² cos 2t) = d/ds (d/ds(s/(s² + 4))) Using quotient Rule, d/ds(u/v) = (v du/ds - u dv/ds)/v² d/ds (s/(s² + 4)) = ((s²+4)1 - s2s) / (s²+4)² = (s²+4 - 2s²)/(s²+4)² = (4-s²)/(s²+4)² L(t² cos 2t) = d/ds( (4 - s²)/(s² + 4)² ) = ((s²+4)²(-2s) - (4-s²) * 2(s²+4)*2s )/(s²+4)⁴ = (s²+4)(-2s) - (4-s²)(4s) / (s²+4)³ = (-2s³ - 8s -16s + 4s³)/(s²+4)³ = (2s³-24s)/(s²+4)³ = 2s(s² - 12)/(s²+4)³