Problem 17 - Differential Equations and Linear Algebra

Question

Use the linearity of \(L\) and the formulas derived in this section to determine \(L[f]\). $$f(t)=7 e^{-2 t}+1$$

Step-by-Step Solution

Answer

The Laplace Transform of the given function \(f(t) = 7e^{-2t} + 1\) is \(L[f(t)] = \frac{7}{s+2} + \frac{1}{s}\).

Step 1: Use linearity property of Laplace Transform

The linearity property of Laplace Transform states that if \(c_1\) and \(c_2\) are constants and \(f(t)\) and \(g(t)\) are functions of time, then the Laplace Transform of their linear combination is given by: \[L[c_1 f(t) + c_2 g(t)] = c_1 L[f(t)] + c_2 L[g(t)]\] Here, the given function is \(f(t) = 7e^{-2t} + 1\), which can be considered as a linear combination of two functions: \(h(t) = 7e^{-2t}\) and \(k(t) = 1\), with constants \(c_1 = 1\) and \(c_2 = 7\). So, we have: \[L[f(t)] = L[7e^{-2t} + 1] = L[7h(t) + k(t)] = 7L[h(t)] + L[k(t)]\]

Step 2: Find Laplace Transform of 7h(t)

Now, we need to find the Laplace Transform of \(7h(t) = 7e^{-2t}\). Recall the formula for the Laplace Transform of an exponential function: \[L[e^{at}] = \frac{1}{s-a},\quad (\text{for } a \in \mathbb{R} \text{ and } \text{Re}(s) > \text{Re}(a))\] In our case, \(a = -2\). Thus, the Laplace Transform of \(7h(t) = 7e^{-2t}\) is: \[L[7e^{-2t}] = 7L[e^{-2t}] = \frac{7}{s-(-2)} = \frac{7}{s+2}\]

Step 3: Find Laplace Transform of k(t)

The function \(k(t) = 1\) is a constant function. The Laplace Transform of a constant function is given by: \[L[c] = \frac{c}{s}, \quad (\text{for } c \in \mathbb{C} \text{ and } \text{Re}(s) > 0)\] So, the Laplace Transform of \(k(t) = 1\) is: \[L[1] = \frac{1}{s}\]

Step 4: Combine the results using linearity property

Now, we will combine the results obtained in Step 2 and Step 3 using the linearity property: \[L[f(t)] = 7L[h(t)] + L[k(t)] = \frac{7}{s+2} + \frac{1}{s}\] Therefore, the Laplace Transform of the given function \(f(t) = 7e^{-2t} + 1\) is: \[L[f(t)] = \frac{7}{s+2} + \frac{1}{s}\]

Key Concepts

linearity of Laplace Transform

The linearity property of the Laplace Transform is a fundamental principle that simplifies the process of transforming functions. It tells us how to handle the Laplace Transform of a combination of functions. If you have two functions, say \( f(t) \) and \( g(t) \), and they are multiplied by constants \( c_1 \) and \( c_2 \) respectively, the Laplace Transform of their combination is the same as the sum of the Laplace Transforms of the individual functions, each multiplied by their respective constants. In mathematical terms, this is expressed as:

• \( L[c_1 f(t) + c_2 g(t)] = c_1 L[f(t)] + c_2 L[g(t)] \)

This property illustrates the flexibility of Laplace Transforms in operations involving sums and differences. In our specific example, \( f(t) = 7e^{-2t} + 1 \), this property allows us to break it down into two separate, simpler problems: finding the transform of \( 7e^{-2t} \) and \( 1 \), and then combining the results.

exponential functions

Exponential functions are a type of mathematical function that often appear in various scientific fields, especially when modeling growth or decay processes. When it comes to the Laplace Transform, exponential functions have a neat formula associated with them. The Laplace Transform of an exponential function \( e^{at} \) is given by:

• \( L[e^{at}] = \frac{1}{s-a} \)
• This formula is valid when the real part of \( s \) is greater than the real part of \( a \).

For the function \( 7e^{-2t} \) from our exercise, \( a = -2 \). By plugging \( a \) into the formula, we find:

• \( L[7e^{-2t}] = 7 \cdot L[e^{-2t}] = 7 \cdot \frac{1}{s-(-2)} = \frac{7}{s+2} \)

This result is a direct application of both the linearity of Laplace Transforms and the known formula for exponential functions, illustrating how powerful these properties are when combined.

constant functions

A constant function is the simplest kind of function where the output value is the same for any input. In the realm of Laplace Transforms, dealing with constant functions is quite straightforward. The Laplace Transform of a constant function \( c \) is:

• \( L[c] = \frac{c}{s} \)
• This formula applies provided that the real part of \( s \) is greater than zero.

In our example from the exercise, we encounter the constant function \( 1 \). Using the above formula, we quickly find that:

• \( L[1] = \frac{1}{s} \)

This simplicity highlights why the Laplace Transform is such a favored tool in engineering and physics: it turns constant functions, which reflect steady-state behavior, into easily manageable algebraic terms that fit neatly into more complex calculations.