Laplace transform #5 - Q&A

Question 1 Use Laplace transform to solve the following initial value problem: y" + 5y = 1+t, y(0) = 0, y'(0) = 4 A) 7/25 - t/5 - e^t cos(2t) + 21/10 e^t sin(2t) B) 7/25 + t/5 - 7/25 e^t cos(2t) + 21/10 e^t sin(2t) C) 7/25 + t/5 - 7/25 e^t cos(2t) + 21/25 e^t sin(2t) D) 7/25 + t/5 + 7/25 e^t cos(2t) + 21/10 e^t sin(2t) E) none of the above F) 7/25 + t/5 - 7/5 e^t cos(2t) + 21/10 e^t sin(2t) Answer Solution: The differential equation is y" + 5y = 1+t; y(0) = 0, y'(0) = 4 Apply Laplace transform. L{y"} - 2y' + 5y} = L{1+t} s²Y(s) - sy(0) - y'(0) + 5Y(s) = 1/s + 1/s² (s² + 5) Y(s) = 1/s + 1/s² + 4 Y(s) = (1 + s + 4s²)/(s²(s²+5)) Perform the partial fraction decomposition: (1+s+4s²)/(s²(s²+5)) = A/s + B/s² + (Cs+D)/(s²+5) (1+s+4s²) = As(s²+5) + B(s²+5) + (Cs+D)s² (1+s+4s²) = As³+5As + Bs² + 5B + Cs³+Ds² (1+s+4s²) = (A+C)s³ + (B+D)s² +5As + 5B Equating coefficients A+C = 0 => C = -A B+D = 4 5A = 1 => A = 1/5 5B = 1 => B = 1/5 D = 4-B = 4 - 1/5 = 19/5 C = - 1/5 Then, we have Y(s) = 1/5 * 1/s + 1/5 * 1/s² + (-1/5s + 19/5)/(s²+5) Y(s) = 1/5 * 1/s + 1/5 * 1/s² - 1/5(s)/(s²+5) + 19/5(1/(s²+5)) Thus, Y(S) = 1/5* 1/s + 1/5* 1/s² -1/5s/(s²+5) + 19/51/((s²+5)) Apply inverse Laplace transform: L⁻¹{1/s} = 1 L⁻¹{1/s²} = t L⁻¹{s/(s²+5)} = cos(√5 t) L⁻¹{(1/(s²+5))} = 1/√5 sin(√5t) y(t) = 1/5 + 1/5 t - 1/5 cos(√5t) + 19/5 √5/5 sin(√5t) y(t) = 1/5 + 1/5 t - 1/5 cos(√5t) + 19√5/25 sin(√5t) Question 2 The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True of False? Answer 1 Explanation: Laplace transform of a function f(t) is given by: F(s) = ∫(0 to ∞) e^(-st)f(t)dt Then, for any two functions f(t) and g(t) the Laplace transform of the product of two functions is: L{f(t).g(t)} Integral form is: ∫(0 to ∞) e^(-st)[f(t).g(t)]dt Which is not equal to ∫(0 to ∞) e^(-st)f(t)dt. ∫(0 to ∞) e^(-st)g(t)dt Therefore, the given statement if false. However, it is equivalent to the convolutional product of the Laplace transforms of each function. Answer 2 The statement is false. In general, the Laplace transform of the product of two functions is not equal to the product of their individual Laplace transforms. Mathematically, if F(t) and G(t) are two functions, then the Laplace transform of their product F(t)G(t) is given by: L{F(t) G(t)} ≠ L{F(t)}·L{G(t)} In most cases, the Laplace transform of a product involves convolution rather than simple multiplication. Therefore, the statement is false. Answer 3 Step 1: To determine whether the statement "The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function" is true or false, we need to consider the definition and properties of the Laplace transform. The Laplace transform of a function f(t) is defined as: F(s) = L{f(t)} = ∫(0 to ∞) e^(-st)f(t)dt where F(s) is the Laplace transform of f(t). Step 2: Now, let's consider two functions f(t) and g(t), and their Laplace transforms F(s) and G(s), respectively. We want to determine if the Laplace transform of their product, h(t) = f(t).g(t) , is equal to the product of their individual Laplace transforms, H(s) = F(s).G(s). To do this, we will compute the Laplace transform of h(t) and see if it matches the expression H(s). Using the definition of the Laplace transform, we have: L{h(t)} = L{f(t).g(t)} = ∫(0 to ∞) e^(-st) (f(t).g(t))dt Step 3: Now, we can rewrite the integral as a product of two separate integrals: ∫(0 to ∞) e^(-st) (f(t).g(t))dt = ∫(0 to ∞) e^(-st)f(t)dt. ∫(0 to ∞) e^(-st)g(t)dt This shows that the Laplace transform of the product of two functions is equal to the product of their individual Laplace transforms. Therefore, the statement "The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function" is true. Answer 4 The statement you provided is True. The Laplace transform of the product of two functions is indeed the product of the Laplace transforms of each given function. Let's denote the Laplace transform of a function f(t) as F(s). Using this notation, the Laplace transform of the product of two functions, f(t) and g(t), can be expressed as: L{f(t).g(t)} = F(s).G(s) Here, F(s) represents the Laplace transform of f(t), and G(s) represents the Laplace transform of g(t). To emphasize this relationship, we can write the statement in LaTeX markup as: L{f(t).g(t)} = F(s).G(s) Therefore, the Laplace transform of the product of two functions is indeed the product of the Laplace transforms of each given function. Question 3 use the Laplace transform to solve the initial value problem. y" - 3y' + 2y = { 0 if 0 -3-10 = A(-1) => A = 13 s=2 => -6 - 10 = B(1) => B = -16 (1/(s((s-1)(s-2)) = A/s + B/(s-1) + C/(s-2) 1 = A((s-1)(s-2)) + B(s(s-2)) + C(s(s-1)) s=0 => 1 = 2A => A = 1/2 s=1 => 1 = -B => B = -1 s=2 => 1 = 2C => C = 1/2 Y(s) = (13/(s-1) -16/(s-2)) + ((1/2s - 1/(s-1) + 1/(2(s-2)))( e^(-s) - e^(-2s) ) Step 5 Thus, Y(s) = 13/(s-1) -16/(s-2)) + (1/(2s) - 1/(s-1) + 1/(2(s-2)))*( e^(-s) - e^(-2s) ) finally taking the inverse transform: y(t) = 13e^t - 16e^(2t) + (1/2 - e^(t-1) + 1/2 e^(2(t-1)) u(t-1) - (1/2 - e^(t-2) + 1/2 e^(2(t-2)) u(t-2) Step 6 Now, consider the case when f(t) = 1, 1 < t < 2 s²Y(s) - sy(0) - y'(0) - 3(sY(s) – y(0)) + 2(Y(s)) = L(1), 1 < t < 2 s²Y(s) - s(-3) - (1) -3sY(s) + 3(-3)+2Y(s) = 1/s (s² - 3s + 2)Y(s) = 10 - 3s + 1/s Y(s) = (10 - 3s + 1/s)/(s² - 3s + 2) Y(s) = (-3s² + 10s + 1)/(s(s-1)(s-2)) Step 7 Now, find the partial fraction decomposition for the transform: (-3s² + 10s + 1)/(s(s-1)(s-2)) = A/s + B/(s-1) + C/(s-2) -3s² + 10s + 1 = A(s-1)(s-2) + Bs(s-2) + Cs(s-1) Question 4 Solve the following IVP using Laplace Transform y' - 2y = 1 - t, y(0) = 4 Answer Step 1 The Laplace transform is defined as follows: y' - 2y = 1 - t, y(0) = 4 Laplace transform applied to both sides: L{y'} - 2L{y} = L(1) - L(t) sL(y)-y(0) - 2L(y) = 1/s - 1/s² sL(y)-4 - 2L(y) = 1/s - 1/s² L(y)(s - 2) = 1/s - 1/s² + 4 L(y) = (4s² + s - 1)/(s²(s-2)) Step 2 Making the fraction form from the inverse Laplace transform: (4s² + s - 1)/(s²(s-2)) = A/s + B/s² + C/(s-2) 4s² + s -1 = As(s - 2) + B(s - 2) + Cs² 4s² + s -1 = As²-2As + Bs - 2B + Cs² 4s² + s -1 = (A + C)s² + (-2A + B)s - 2B Step 3 Calculate the values of A, B, and C by comparing the coefficient: 4s² + s - 1 = (A + C)s² + (-2A + B)s - 2B A + C = 4, -2A + B = 1 and - 2B = − 1 => B = 1/2 -2A + 1/2 = 1 => -2A = 1/2 => A = -1/4 A + C = 4 => -1/4 + C = 4 => C = 17/4 (4s² + s - 1)/(s²(s-2)) = -1/(4s) + 1/(2s²) + 17/(4(s-2)) Step 4 Now, form equation (i): L(y) = (4s² + s - 1)/(s²(s-2)) = L⁻¹{-1/(4s) + 1/(2s²) + 17/(4(s-2))} y = −1/4 L⁻¹{1/s} + 1/2 L⁻¹{1/s²} + 17/4 L⁻¹{1/(s-2)} Using L⁻¹{1/s} = 1, L⁻¹{1/s²} = t, L⁻¹{1/(s-a)} = e^(at) y = -1/4 + t/2 + 17/4 e^(2t) Question 5 A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock? Answer The distance of boat from dock be x, then using pythagaros theorem we have x² + 1 = y² Where y is the length of the rope Differentiate throughout with respect to t. 2x dx/dt = 2y dy/dt We have to find dx/dt and dy/dt is given as 1ms⁻¹ x is 8 m at the given instant and y = √65 Therefore we have 2(8)dx/dt = 2(√65) * 1 dx/dt = √65/8 Result: Rate of decrease of distance between boat and dock is √65/8 ≈ 1ms⁻¹