![Find L[e3t + e^-3t] and L^-1[1/(s(s+1))].](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1775230904843-4365ba063ade82eb.png&w=3840&q=75)
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 to the problems: 6. Find L[e^3t + e^-3t] Step 1: Use the linearity property of Laplace transforms. L[e^3t + e^-3t] = L[e^3t] + L[e^-3t] Step 2: Apply the standard Laplace transform formula L[e^at] = (1)/(s-a). L[e^3t] = (1)/(s-3) L[e^-3t] = (1)/(s-(-3)) = (1)/(s+3) Step 3: Combine the terms and simplify. L[e^3t + e^-3t] = (1)/(s-3) + (1)/(s+3) L[e^3t + e^-3t] = ((s+3) + (s-3))/((s-3)(s+3)) L[e^3t + e^-3t] = (2s)/(s^2 - 3^2) L[e^3t + e^-3t] = (2s)/(s^2 - 9) --- 7. Find L^-1[(1)/(s(s+1))] Step 1: Use partial fraction decomposition for (1)/(s(s+1)). Let (1)/(s(s+1)) = (A)/(s) + (B)/(s+1). Multiplying by s(s+1) gives 1 = A(s+1) + Bs. Setting s=0 1 = A(1) A=1. Setting s=-1 1 = B(-1) B=-1. So, (1)/(s(s+1)) = (1)/(s) - (1)/(s+1). Step 2: Apply the inverse Laplace transform to each term. L^-1[(1)/(s(s+1))] = L^-1[(1)/(s) - (1)/(s+1)] L^-1[(1)/(s(s+1))] = L^-1[(1)/(s)] - L^-1[(1)/(s+1)] Step 3: Use the standard inverse Laplace transform formulas L^-1[(1)/(s)] = 1 and L^-1[(1)/(s+a)] = e^-at. L^-1[(1)/(s(s+1))] = 1 - e^-t L^-1[(1)/(s(s+1))] = 1 - e^-t --- 8. Evaluate L^-1[((s^2+1)/(s(s+1)))] Step 1: Let F(s) = ((s^2+1)/(s(s+1))). Use logarithm properties to expand F(s). F(s) = (s^2+1) - (s) - (s+1)