Problem 18 - Differential Equations and Linear Algebra

Question

Use the linearity of \(L\) and the formulas derived in this section to determine \(L[f]\). $$f(t)=3 t^{2}-5 \cos 2 t+\sin 3 t$$

Step-by-Step Solution

Answer

The short answer is: \[L[f(t)] = \frac{6}{s^3} - \frac{5s}{s^2+4} + \frac{3}{s^2+9} \]

Step 1: Find Laplace transform of \(g_1(t) = 3t^2 \)

Using the standard Laplace transform formula: \[L[t^n] = \frac{n!}{s^{n+1}}\] We can find the Laplace transform of the function \(g_1(t) = 3t^2\): \[L[g_1(t)] = L[3t^2] = 3L[t^2] = \frac{3 \cdot 2!}{s^{2+1}} = \frac{6}{s^3}\]

Step 2: Find Laplace transform of \(g_2(t) = -5\cos(2t) \)

Using the standard Laplace transform formula for cosine functions: \[L[\cos(at)] = \frac{s}{s^2+a^2}\] We can find the Laplace transform of the function \(g_2(t) = -5\cos(2t)\): \[L[g_2(t)] = L[-5\cos(2t)] = -5L[\cos(2t)] = -5 \cdot \frac{s}{s^2+(2)^2} = \frac{-5s}{s^2+4}\]

Step 3: Find Laplace transform of \(g_3(t) = \sin(3t) \)

Using the standard Laplace transform formula for sine functions: \[L[\sin(at)] = \frac{a}{s^2+a^2}\] We can find the Laplace transform of the function \(g_3(t) = \sin(3t)\): \[L[g_3(t)] = L[\sin(3t)] = \frac{3}{s^2+(3)^2} = \frac{3}{s^2+9}\]

Step 4: Combine Laplace transforms using linearity property

Now that we have the Laplace transforms for all three functions, let's combine them using the linearity property of the Laplace transform. Since \(f(t) = g_1(t) + g_2(t) + g_3(t)\), we have: \[L[f(t)] = L[g_1(t)] + L[g_2(t)] + L[g_3(t)] = \frac{6}{s^3} + \frac{-5s}{s^2+4} + \frac{3}{s^2+9}\] So, the Laplace transform of the given function is: \[L[f(t)] = \frac{6}{s^3} - \frac{5s}{s^2+4} + \frac{3}{s^2+9}\]