To find the inverse Laplace transform of the given expression $\frac{2}{(p+1)} \cdot \frac{1}{p}$, we can split the expression and compute each part using standard Laplace transform properties. Here's the step-by-step solution:

Expression:

$\mathcal{L}^{-1} \left( \frac{2}{(p+1)} \cdot \frac{1}{p} \right)$

We can split the product into two simpler terms using convolution:

$\mathcal{L}^{-1} \left( \frac{2}{p} \right) * \mathcal{L}^{-1} \left( \frac{1}{p+1} \right)$

Step 1: Find the inverse Laplace of each term

1. Inverse Laplace of $\frac{2}{p}$: $\mathcal{L}^{-1} \left( \frac{2}{p} \right) = 2$ (This is a standard result as $\mathcal{L}(1) = \frac{1}{p}$).

2. Inverse Laplace of $\frac{1}{p+1}$: $\mathcal{L}^{-1} \left( \frac{1}{p+1} \right) = e^{-t}$ (This is another standard result as $\mathcal{L}(e^{-t}) = \frac{1}{p+1}$).

Step 2: Apply Convolution

The result is given by the convolution of the two functions:

$f(t) = 2 * e^{-t}$

The convolution of $a$ and $b$, denoted $(a * b)(t)$, is:

$(2 * e^{-t})(t) = \int_0^t 2 \cdot e^{-(t-\tau)} \, d\tau$

Now, solving this:

= 2 \cdot (1 - e^{-t})$$ ### Final Result: $$\mathcal{L}^{-1} \left( \frac{2}{(p+1)} \cdot \frac{1}{p} \right) = 2(1 - e^{-t})$$ ### Conclusion: The inverse Laplace transform of the given expression is: $$2(1 - e^{-t})$$ --- Would you like more details on any specific step or have any further questions? Here are 5 related questions you might find interesting: 1. How do you compute the convolution of two functions in general? 2. What is the Laplace transform of $$e^{at}$$? 3. How does the Laplace transform simplify solving differential equations? 4. Can you explain the connection between Laplace and Fourier transforms? 5. How is the initial value theorem applied in Laplace transforms? **Tip**: The convolution theorem is very useful for handling products of Laplace transforms and simplifies inverse Laplace calculations!