Problem 14 - Differential Equations and Linear Algebra

Question

Use properties of the Laplace transform and the table of Laplace transforms to determine \(L[f]\). $$f(t)=6 t^{4} e^{-2 t}-2 t e^{t+1}+\sqrt{10 t}$$

Step-by-Step Solution

Answer

The Laplace transform of the given function, \(f(t)\), is: $$L[f(t)] = 6\frac{d^4}{ds^4}\left(\frac{1}{s+2}\right)-2(-1)^1\frac{d}{ds}\left(\frac{1}{(s-1)^2}\right) + \sqrt{10}\frac{\Gamma(3/2)}{s^{3/2}}$$

Step 1: Break the function into separate terms

We have the function $$f(t)=6 t^{4} e^{-2 t}-2 t e^{t+1}+\sqrt{10 t}$$ Break it down into 3 separate terms: $$f(t)=g(t) - h(t) + i(t)$$ where $$g(t)=6 t^{4} e^{-2 t},\ h(t)=2 t e^{t+1},\ i(t)=\sqrt{10 t}$$

Step 2: Find the Laplace transform of term g(t)

To find the Laplace transform of \(g(t)\), we will use the property of Laplace transforms: \[L[tf(t)]=(-1)^nL^{(n)}[F(s)]\] for \(t^n f(t)\), where \(n\) is an integer. In this case, the function g(t) is \(t^4 e^{-2t}\), and we have: $$L[g(t)] = L[6 t^4 e^{-2t}] = 6L[t^4 e^{-2t}]$$ Using the property and by referring to the Laplace transforms table, we have: $$L[g(t)] = 6 (-1)^4 \frac{d^4}{ds^4}\left(\frac{1}{s+2}\right) = 6\frac{d^4}{ds^4}\left(\frac{1}{s+2}\right)$$

Step 3: Find the Laplace transform of term h(t)

To find the Laplace transform of \(h(t)\), we will use the properties of Laplace transforms, as well as the first shifting theorem. The first shifting theorem states that \[L[e^{at}f(t)] = F(s-a)\] In this case, the function h(t) is \(2 t e^{t+1}\), and we have: $$L[h(t)] = L[2 t e^{(t+1)}] = 2L[t e^{(t+1)}]$$ Using the first shifting theorem, we have: $$L[h(t)] = 2 (-1)^1 \frac{d}{ds}\left(\frac{1}{(s-1)^2}\right)$$

Step 4: Find the Laplace transform of term i(t)

To find the Laplace transform of \(i(t)\), we will use the property of Laplace transforms for \(t^a\), where \(a\) is not an integer. In this case, the function i(t) is \(\sqrt{10t}\), and we have: $$L[i(t)] = L[\sqrt{10t}] = L[10^{1/2} \cdot t^{1/2}]$$ Using the table of Laplace transforms and applying the property, we get: $$L[i(t)] = 10^{1/2} \frac{\Gamma(1/2+1)}{s^{1/2+1}} = \sqrt{10}\frac{\Gamma(3/2)}{s^{3/2}}$$

Step 5: Combine the Laplace transforms of all the terms

Now, we will combine the Laplace transforms of all the terms using the following relationship: $$L[f(t)] = L[g(t)] - L[h(t)] + L[i(t)]$$ By calculating the derivatives for the L[g(t)] and L[h(t)], we get: $$L[f(t)] = 6\frac{d^4}{ds^4}\left(\frac{1}{s+2}\right)-2(-1)^1\frac{d}{ds}\left(\frac{1}{(s-1)^2}\right) + \sqrt{10}\frac{\Gamma(3/2)}{s^{3/2}}$$ This is the final Laplace transform of the given function, f(t).

Key Concepts

Properties of Laplace Transform

The Laplace Transform is a potent tool in solving differential equations, especially for engineers and physicists. Understanding its properties can simplify complex transformations. Here are some key properties:

• Linearity: This property means that the Laplace Transform of a sum is the sum of the Laplace Transforms. Mathematically, if you have functions \(f(t)\) and \(g(t)\), then \(L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]\).
• Frequency Shifting: It states that multiplying a function by an exponential term affects the Laplace Transform parameter. If \(f(t)\) has a Laplace Transform \(F(s)\), then \(e^{at}f(t)\) has a Laplace Transform \(F(s-a)\).
• Time Shifting: Allows us to handle Laplace Transforms of delayed functions. If \(f(t-a)\) for \(a > 0\) is considered, it modifies the transform.

These properties are crucial when breaking down complex expressions into manageable parts, as seen in the given exercise.

First Shifting Theorem

The First Shifting Theorem is particularly useful when dealing with terms in the form of \(e^{at}f(t)\). The theorem establishes that multiplying by an exponential shifts the transform in the frequency domain.Consider the formula:

• \[L[e^{at}f(t)] = F(s-a)\]

This rule is invaluable in simplifying the computation of Laplace Transforms for shifted inputs, streamlining the solution. In the exercise, \(h(t) = 2t e^{t+1}\) was handled by recognizing a shift due to the exponentials, allowing us to simplify its complex form into something more manageable.Without this theorem, transforming such functions would be significantly more complicated.

Gamma Function

The Gamma Function, denoted \(\Gamma(n)\), extends the factorial concept to non-integer values. It is defined by the integral:

• \[\Gamma(n) = \int_0^{\infty} x^{n-1}e^{-x} \, dx\]

For positive integers, \(\Gamma(n)\) simplifies to \((n-1)!\). It plays a crucial role in Laplace Transforms of power functions like \(t^a\) where \(a\) is not an integer.In the exercise, calculating the Laplace Transform of \(i(t) = \sqrt{10t}\) required the Gamma Function due to the presence of \(t^{1/2}\). Understanding how to apply it in these contexts aids in solving complex expressions in calculus.

Differentiation in Laplace Transform

Differentiation plays a key role in the Laplace Transform, particularly through the differentiation property, which involves taking derivatives within the transform context. When working with a term like \(t^n f(t)\), we often use:

• \[L[t^n f(t)] = (-1)^n\frac{d^n}{ds^n}[F(s)]\]

This expression highlights that differentiation with respect to \(s\) corresponds to time multiplication with \(t^n\). In the exercise, for \(g(t) = 6t^4 e^{-2t}\), this property enabled us to perform a high-order derivative.Understanding this relationship is crucial, as it allows us to handle polynomials multiplied by other functions efficiently, breaking complex tasks down to straightforward operations in tackling Laplace Transforms.