Question

Use the linearity property (7) along with the transforms found in Example 2, $$ \mathcal{L}\left\\{e^{a t}\right\\}=\frac{1}{s-a}, \quad s>a \quad \text { and } \quad \mathcal{L}\\{t\\}=\frac{1}{s^{2}}, \quad s>0, $$ to calculate the Laplace transform \(R(s)=\mathcal{L}\\{r(t)\\}\) of the given function \(r(t)\). For what values \(s\) does the Laplace transform exist? $$ r(t)=5 e^{-7 t}+t+2 e^{2 t} $$

Short answer

Answer: The Laplace transform of the given function is R(s) = (5/(s+7)) + (1/s^2) + (2/(s-2)). The Laplace transform exists for values of s > 2.

Step-by-step solution

Identify linearity property of Laplace Transform

The linearity property of the Laplace transform states that the Laplace transform of the sum of functions is equal to the sum of Laplace transforms of individual functions, i.e., $$ \mathcal{L}\\{a f(t) + b g(t)\\} = a \mathcal{L}\\{f(t)\\} + b\mathcal{L}\\{g(t)\\} $$ where a and b are constants.

Apply linearity property to the given function

The given function is $$ r(t) = 5 e^{-7 t} + t + 2 e^{2t} $$ So, the Laplace transform of r(t) can be found by applying the linearity property: $$ R(s) = \mathcal{L}\\{r(t)\\} = 5 \mathcal{L}\\{e^{-7t}\\} + \mathcal{L}\\{t\\} + 2 \mathcal{L}\\{e^{2t}\\} $$

Use the given Laplace transforms of individual functions

We have the Laplace transforms of individual functions: $$ \mathcal{L}\\{e^{at}\\} = \frac{1}{s-a}, \quad s>a \quad \text{and} \quad \mathcal{L}\\{t\\} = \frac{1}{s^2}, \quad s>0 $$ Applying these transforms to the function r(t), we get: $$ R(s) = 5\frac{1}{s-(-7)} + \frac{1}{s^2} + 2\frac{1}{s-2} $$

Simplify the Laplace transform

Simplify the above expression for R(s) to get: $$ R(s) = \frac{5}{s+7} + \frac{1}{s^2} + \frac{2}{s-2} $$

Determine the values of s for which Laplace transform exists

According to the given Laplace transforms, the existence of the Laplace transform depends on the value of s: $$ \frac{1}{s-a}, \quad s>a $$ In our case, we have two exponential terms with different a values: 1. For the term with a = -7: s > -7 2. For the term with a = 2: s > 2 As for the linear term, the Laplace transform exists when s > 0. Now, we take the intersection of these conditions to find the values of s for which R(s) exists: Intersection of s > -7, s > 2, and s > 0 is s > 2. So, the Laplace transform R(s) exists for values of s > 2. Final Answer: $$ R(s) = \frac{5}{s+7} + \frac{1}{s^2} + \frac{2}{s-2}, \quad s > 2 $$

Key concepts

Linearity Property

One of the most essential concepts in understanding the Laplace transform is its linearity property. This is a rule that significantly simplifies the calculation of the Laplace transform for a combination of functions.

According to this principle, if you have a function that can be expressed as a sum of simpler functions, each scaled by a constant, the Laplace transform of the entire function is simply the sum of the Laplace transforms of each individual, simpler function, scaled by the same constants. In mathematical terms, this is represented as:

• \[\mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}\]

This property is widely applied because it breaks down complex functions into more manageable pieces, allowing for straightforward calculation of their Laplace transforms. The step-by-step solution illustrates this by decomposing the function into exponential and polynomial parts, treating each separately with the linearity property.

Laplace Transform Calculation

To compute the Laplace transform of a given function, certain basic transforms, like those of exponentials and polynomials, can be utilized as building blocks. The process involves recognizing components of the function that match these basic forms and then applying known transform equations to those parts.

For example, the exercise guided us to use:

• \[\mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \quad s>a\]
• \[\mathcal{L}\{t\} = \frac{1}{s^{2}}, \quad s>0\]

After applying the linearity property, one can plug in the known transforms for each term. Building on the properties of exponentials and polynomials, we can find a simplified expression for the Laplace transform of the entire function. Ensuring that the pieces of the function satisfy the existence conditions for their respective transforms is a critical part of this process.

Transform Existence Conditions

Not all functions have a Laplace transform. For a transform to exist, certain criteria which are dependent on the complex frequency variable 's' must be met. These existence conditions are critical to determine not just if a transform is possible, but also the domain in 's' where the transform is valid.

The key conditions for basic functions like \(e^{at}\) and \(t\) involve inequalities that define where the real part of 's' must lie. When dealing with multiple terms in a function like \(r(t)\), we must consider the most restrictive condition among all the terms for the transform to exist. In the provided solution, we see the intersection of the conditions for each term - this intersection then dictates the domain of 's' where the Laplace transform for the entire function is defined. In this exercise, the transform exists for \(s > 2\), as this covers the most restrictive condition from among the function's parts.