1.11. Additional Exercise (Practice)

Find the Laplace Transform of the following using the First Shifting Theorem:

1. 𝑒4𝑡 ⋅cos(3𝑡)

To find the Laplace Transform of the function 𝑓(𝑡) = 𝑒^{4𝑡} imes ext{cos}(3𝑡), we can apply the First Shifting Theorem. First, we note that the Laplace Transform of cos(3t) is known:

Since we have an exponential term e^(4t), we can let 'a' be 4. According to the theorem, we will shift 's' to 's - 4'. Therefore:

Think of it like tuning a radio. When you're trying to catch a specific station (the frequency of cos(3t)), sometimes you need to adjust your position slightly to account for interference, like the 4 from e^(4t). This adjustment represents shifting your frequency to get a clearer signal.

2. 𝑒−2𝑡 ⋅𝑡^2

To solve for the Laplace Transform of the function 𝑓(𝑡) = 𝑒^{-2𝑡} imes t^2, we first find the Laplace Transform of t². The Laplace Transform of t^n is:

So, for n = 2:

Next, applying the First Shifting Theorem with a = -2, we shift 's' to 's + 2':

Imagine you're pushing a toy car that slows down over time due to friction. The quadratic term represents the distance it covers, and the e^{-2t} is like the slowing force. We adjust our expectation of distance based on how quickly it slows down, which mirrors how we adjust 's' in the transform.

3. 𝑒5𝑡 ⋅sin(2𝑡)

To find the Laplace Transform of the function 𝑓(𝑡) = 𝑒^{5𝑡} imes ext{sin}(2𝑡), we start by finding the Laplace Transform of sin(2t):

Now, since we have an exponential e^ ext{5t}, and setting a = 5, we apply the First Shifting Theorem, which leads to shifting 's' to 's - 5':

Think of this process like adjusting the brightness on your smartphone screen. The sine function (sin(2t)) behaves predictively like the brightness levels, but when someone introduces a constant increase (e^{5t}), it's like someone adjusting the overall brightness. The transformation reflects this change in the 's' domain.

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