Question

This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of $F$ is the set of all real numbers s such that the improper integral converges. Find the Laplace transform $F$ of each of the following functions and give the domain of $F$.$f(x)=x$

Short answer

The Laplace transform is $F(s)=\frac{1}{s^2}$ with domain $s>0$.

Step-by-step solution

Definition of Laplace Transform

The Laplace transform of a function $f(t)$ is defined as $F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt$. In this exercise, we need to find $F(s)$ for $f(t)=t$.

Set Up the Integral

Write the integral for the Laplace transform of $f(t)=t$: $F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}tdt$.

Apply Integration by Parts

To solve the integral $\int e^{-st}tdt$, use integration by parts. Let $u=t$ and $dv=e^{-st}dt$. Then $du=dt$ and $v=\frac{-1}{s}{e}^{-st}$.

Calculate Integral Parts

Substitute the parts into integration by parts formula: $\int udv=uv-\int vdu$. So, $F(s)={[-\frac{t}{s}{e}^{-st}]}_0^{\mathrm{\infty }}+\frac{1}{s}{\int }_0^{\mathrm{\infty }}{e}^{-st}dt$.

Evaluate First Term

Evaluate the boundary term ${[-\frac{t}{s}{e}^{-st}]}_0^{\mathrm{\infty }}$. As $t\to \mathrm{\infty }$, $e^{-st}\to 0$, making the term zero if $s>0$. At $t=0$, the term is 0.

Integrate Remaining Part

Integrate the remaining part: ${\int }_0^{\mathrm{\infty }}{e}^{-st}dt={\left[\frac{-1}{s}{e}^{-st}\right]}_0^{\mathrm{\infty }}=\frac{1}{s}$ when $s>0$.

Combine and Simplify

Combine results: $F(s)=0+\frac{1}{s^2}=\frac{1}{s^2}$.

Determine Domain of F

The domain of the Laplace transform $F(s)=\frac{1}{s^2}$ is $s>0$ because the integral converges only for positive $s$.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They play a crucial role in modeling real-world phenomena where change is involved. Here's a simple breakdown:

• These equations are tools to describe how a quantity changes over time.
• They can be used in various fields such as physics, engineering, biology, and economics.
• A solution to a differential equation is a function, not just a single value.

When solving initial-value problems in differential equations, the Laplace transform is often used for simplification. It helps to turn complex differential equations into simpler algebraic equations, which are easier to tackle and solve. This transformation is especially handy for interpreting systems that react to unknown or sudden changes. Essentially, recognizing and working with differential equations allows you to predict future behavior based on current trends.

Improper Integral

An improper integral is a type of definite integral where either the interval of integration is infinite or the integrand becomes infinite within the interval. When we're dealing with improper integrals, understanding convergence is critical.

• Convergence means that as you add up an infinite number of infinitely small parts, the sum approaches a finite value.
• If the integral converges, it equals a number. If it diverges, it doesn’t sum to a finite value.

In the context of the Laplace transform, you often encounter improper integrals since the upper limit extends to infinity. To affirm that the Laplace transform makes sense (converges), certain conditions on the parameter must be satisfied. For example, in the provided example, the convergence is only guaranteed when the variable $s$ is greater than zero.

Integration by Parts

Integration by parts is a useful technique derived from the product rule for differentiation. It's specifically applied to simplify integrals involving products of functions. Here is a brief guide to its application:

• Choose parts of the integral as $u$ and $dv$.
• Differentiate $u$ to get $du$ and integrate $dv$ to get $v$.
• Apply the formula: $\int udv=uv-\int vdu$.

For example, when solving the integral for the Laplace transform of $f(t)=t$, knowing what to choose for $u$ and $dv$ is crucial. A common choice is letting $u=t$ (making differentiation straightforward) and choosing $dv=e^{-st}dt$ (making integration similarly straightforward). As you work through the parts, always check back to ensure your choices simplify rather than complicate the integral. This process often reveals otherwise hidden solutions and illuminates integral relationships.