13.1.7. Solved Examples

Example 1: Find ℒ−1{ \frac{1}{s(s+1)} }

Solution: We know, • ℒ−1{ \frac{1}{s} } = 1 • ℒ−1{ \frac{1}{s+1} } = e^{-t}

So,

ℒ^{-1}{ \frac{1}{s(s+1)} } = 1*e^{-t} = \int_0^t 1 \cdot e^{-(t-\tau)} d\tau

=t\int_0^t e^{-(t-\tau)} d\tau = e^{-t}\int_0^t e^{\tau} d\tau = e^{-t}(e^t - 1) = 1 - e^{-t}

In this example, we want to find the inverse Laplace transform of the function 1/(s(s+1)). The solution begins by recognizing that we can break apart the inverse transform into simpler components: the first part corresponds to 1/s, which has an inverse transform of 1, and the second part corresponds to 1/(s+1), which has an inverse transform of exp(-t). We then utilize the property of convolution, which tells us that the inverse Laplace transform of the product of two transforms is the convolution of their respective time-domain functions. By computing the integral for the convolution, we arrive at the final result, simplifying the expression to 1 - exp(-t).

Think of finding the inverse Laplace transform like breaking down a complex recipe into simpler parts. Instead of trying to prepare a complicated dish in one go, you might first gather all the ingredients (the parts of the Laplace transform), and then follow step-by-step instructions for each component before combining them at the end to create the final meal (the final function). This process of breaking down the recipe helps ensure that each part is correctly executed.

Example 2: Find the inverse Laplace transform of \frac{1}{s^2(s+2)}

Solution: Let 𝐹(s) = \frac{1}{s^2}, 𝐺(s) = \frac{1}{s+2}.

We know: • ℒ^{-1}{ \frac{1}{s^2} } = t • ℒ^{-1}{ \frac{1}{s+2} } = e^{-2t}

So by convolution:

f* g = \int_0^t \tau \cdot e^{-2(t - \tau)} d\tau = e^{-2t}\int_0^t \tau \cdot e^{2\tau} d\tau.

Use integration by parts: Let • u = \tau ⇒ du = d\tau • dv = e^{2\tau} d\tau ⇒ v = \frac{1}{2} e^{2\tau}.

Then, we have: e^{-2t}[\tau \cdot \frac{1}{2} e^{2\tau}|_0^t - \int_0^t \frac{1}{2} e^{2\tau} d\tau] = e^{-2t}[\frac{1}{2}(te^{2t} - 0) - \frac{1}{2}(\frac{1}{2} e^{2t} - 1)]

= (1 - e^{-2t}) + \frac{1}{4}

In the second example, we find the inverse Laplace transform of the function 1/(s^2(s+2)). We first define F(s) and G(s), which represent two simpler functions. Each of these functions has a known inverse Laplace transform (t and e^(-2t), respectively). Using the convolution theorem, we can compute the convolution of these two functions to get the final inverse transform. The integral involves integration by parts, allowing us to manage the computation effectively, resulting in a clear form that combines exponential decay with a linear term.

Imagine trying to calculate the total distance traveled on a road trip that consists two different segments: one part of the trip has a steady speed (like the function \frac{1}{s^2}), and the other part involves driving a little slower due to traffic (like the function \frac{1}{s+2}). Just as you would calculate the distance for each segment separately before adding them up to arrive at the total distance traveled, here we similarly handle each part and combine effects through the convolution process, allowing us to achieve an understanding of the total outcome.

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