This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here are the solutions for the given Laplace transform problems. 3.1 L\(e^2t - 1)^2\ Step 1: Expand the expression (e^2t - 1)^2. (e^2t - 1)^2 = (e^2t)^2 - 2(e^2t)(1) + 1^2 = e^4t - 2e^2t + 1 Step 2: Apply the linearity property of the Laplace transform. L\e^4t - 2e^2t + 1\ = L\e^4t\ - 2L\e^2t\ + L\1\ Step 3: Use the standard Laplace transform formulas L\e^at\ = (1)/(s-a) and L\1\ = (1)/(s). L\e^4t\ = (1)/(s-4) L\e^2t\ = (1)/(s-2) L\1\ = (1)/(s) Step 4: Substitute these transforms back into the expression. L\(e^2t - 1)^2\ = (1)/(s-4) - (2)/(s-2) + (1)/(s) To combine into a single fraction: = (s(s-2) - 2s(s-4) + (s-4)(s-2))/(s(s-4)(s-2)) = (s^2-2s - 2s^2+8s + s^2-6s+8)/(s(s-4)(s-2)) = ((s^2 - 2s^2 + s^2) + (-2s + 8s - 6s) + 8)/(s(s-4)(s-2)) = (8)/(s(s-2)(s-4)) The final answer is (8)/(s(s-2)(s-4)). 3.2 L^-1\e^-3ss^3\ Step 1: Identify the form for the Second Shifting Theorem. The expression is of the form e^-asF(s), where a=3 and F(s) = (1)/(s^3). Step 2: Find the inverse Laplace transform of F(s) = (1)/(s^3). We know that L\t^n\ = (n!)/(s^n+1). For n=2, L\t^2\ = (2!)/(s^3) = (2)/(s^3). Therefore, f(t) = L^-1\(1)/(s^3)\ = (1)/(2)L^-1\(2)/(s^3)\ = (1)/(2)t^2 Step 3: Apply the Second Shifting Theorem, which states L^-1\e^-asF(s)\ = f(t-a)u(t-a). Here a=3 and f(t) = (1)/(2)t^2. So, f(t-3) = (1)/(2)(t-3)^2. L^-1\e^-3ss^3\ = (1)/(2)(t-3)^2 u(t-3) where u(t-3) is the Heaviside step function. The final answer is (1)/(2)(t-3)^2 u(t-3). 3.3 L^-1\(3(s+2))/(s^2+4s+13)\ Step 1: Complete the square in the denominator. s^2+4s+13 = (s^2+4s+4) + 9 = (s+2)^2 + 3^2 Step 2: Rewrite the expression using the completed square. (3(s+2))/(s^2+4s+13) = (3(s+2))/((s+2)^2 + 3^2) Step 3: Apply the First Shifting Theorem. The expression is of the form F(s-a), where a=-2. We know that L^-1\(s-a)/((s-a)^2+k^2)\ = e^at(kt). In this case, a=-2 and k=3. So, L^-1\(s+2)/((s+2)^2+3^2)\ = e^-2t(3t). Step 4: Include the constant factor of 3. L^-1\(3(s+2))/(s^2+4s+13)\ = 3 L^-1\(s+2)/((s+2)^2+3^2)\ = 3e^-2t(3t) The final answer is 3e^-2t(3t). That's 2 down. 3 left today — send the next one.