Find the Laplace transform of (e2t - 1)2 and the inverse Laplace transform of e−3ss3.

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ScanSolve AI Tutor · Accepted 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) + 1² = e^(4t) − 2e^(2t) + 1 Step 2: Apply the linearity property of the Laplace transform. Le^(4t) − 2e^(2t) + 1 = Le^(4t) − 2Le^(2t) + L1; Step 3: Use the standard Laplace transform formulas Le^(at) = (1)/(s−a) and L1 = 1/s. Le^(4t) = (1)/(s−4); Le^(2t) = (1)/(s−2); L1 = 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)); = (s²−2s − 2s²+8s + s²−6s+8)/(s(s−4)(s−2)); = ((s² − 2s² + s²) + (−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^(−3s))/(s³) Step 1: Identify the form for the Second Shifting Theorem. The expression is of the form e^(−as)F(s), where a=3 and F(s) = (1)/(s³). Step 2: Find the inverse Laplace transform of F(s) = (1)/(s³). We know that Ltⁿ = (n!)/(sⁿ⁺¹). For n=2, Lt² = (2!)/(s³) = (2)/(s³). Therefore, f(t) = L^−1(1)/(s³) = 1/2L^−1(2)/(s³) = 1/2t²; Step 3: Apply the Second Shifting Theorem, which states L^−1e^(−as)F(s) = f(t−a)u(t−a). Here a=3 and f(t) = 1/2t². So, f(t−3) = 1/2(t−3)². L^−1(e^(−3s))/(s³) = 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))/(s²+4s+13) Step 1: Complete the square in the denominator. s²+4s+13 = (s²+4s+4) + 9 = (s+2)² + 3² Step 2: Rewrite the expression using the completed square. (3(s+2))/(s²+4s+13) = (3(s+2))/((s+2)² + 3²) 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+k²) = e^(at)cos(kt). In this case, a=−2 and k=3. So, L^−1(s+2)/((s+2)²+3²) = e^(−2t)cos(3t). Step 4: Include the constant factor of 3. L^−1(3(s+2))/(s²+4s+13) = 3 L^−1(s+2)/((s+2)²+3²); = 3e^(−2t)cos(3t) The final answer is 3e^(−2t)cos(3t).

Find the Laplace transform of (e2t - 1)2 and the inverse Laplace transform of e-3ss3.

Answers

Step-by-step solution

Part 3.1· 4 steps

Expand the expression $(e^{2t} - 1)^2$ .

Apply the linearity property of the Laplace transform.

Use the standard Laplace transform formulas $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ and $\mathcal{L}\{1\} = \frac{1}{s}$ .

Substitute these transforms back into the expression.

Part 3.2· 3 steps
$\frac{1}{2}(t-3)^2 u(t-3)$

Identify the form for the Second Shifting Theorem.

Find the inverse Laplace transform of $F(s) = \frac{1}{s^3}$ .

Apply the Second Shifting Theorem, which states $\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a)$ .

Part 3.3· 4 steps
$3e^{-2t}\cos(3t)$

Complete the square in the denominator.

Rewrite the expression using the completed square.

Apply the First Shifting Theorem.

Include the constant factor of 3.

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