- Chain Rule
- Incorrect. The chain rule is not a key feature of the Laplace transform method. Think about a process that allows you to transfer derivatives.
- Product Rule
- No, the product rule is not a key concept in the Laplace transform method. Consider a method related to integration.
- Integration by Parts
- Correct! The Laplace transform method leverages integration by parts to move derivatives from one function to another inside integrals.
- Properties of Exponential Functions
- Correct! The Laplace transform method uses the properties of exponential functions, especially their behavior under differentiation and integration, to solve differential equations.
Subsection 9.1.1 Motivation
The Laplace transform is a powerful tool for solving differential equations by exploiting integration by parts and the properties of exponential functions. Before we dive into the details, let’s recall the basic formula for integration by parts:
\begin{equation*}
\int u\ dv = uv - \int v\ du\text{.}
\end{equation*}
In this equation, \(u\) and \(v\) are functions of \(t\text{.}\) So \(dv = v' \, dt\text{,}\) and \(du = u' \, dt\text{.}\) Now, we can rewrite this formula as:
\begin{equation*}
\ub{\int u v'\, dt}_{\text{derivative on }v} = u v - \ub{\int v u'\, dt}_{\text{derivative on }u}\text{.}
\end{equation*}
Notice how the derivative initially applied to \(v\) on the left-hand side shifts over to \(u\) on the right-hand side. This ability to transfer derivatives between functions within an integral forms the core idea behind the Laplace transform.
Let’s see this concept in action. Consider an integral of the form:
\begin{equation*}
\int e^{-st} y' \, dt\text{,}
\end{equation*}
where \(s\) is a constant. By setting \(u = e^{-st}\) and \(dv = y' \, dt\) in our integration by parts formula, we get:
\begin{equation*}
\os{\quad (1)}{\int \us{u}{\us{\uparrow}{\ul{e^{-st}\vphantom{y'}}}} \ \us{dv}{\us{\uparrow}{\ul{y' dt}}}} =
\us{u}{\us{\uparrow}{\ul{e^{-st}\vphantom{y}}}} \ \us{v}{\us{\uparrow}{\ul{y}}} -
\os{(2)}{\int \us{v}{\us{\uparrow}{\ul{y}}} \ \us{du}{\us{\uparrow}{\ul{(-s e^{-st} \ dt \vphantom{y})}}}} =
e^{-st} \, y + s \os{\quad (3)}{\int e^{-st} \, y \, dt}\text{.}
\end{equation*}
What’s happening here? The derivative on \(y'\) in integral \((1)\) is effectively transferred to the exponential function \(e^{-st}\) in integral \((2)\text{,}\) leaving an extra factor of \(s\) next to integral \((3)\text{.}\)
This process of transferring derivatives between functions is a reoccuring theme in the Laplace transform method and in the coming sections, we will uncover how it uses this idea to solve a wide range of differential equations.
Reading Questions Check-Point Questions
1. The Laplace transform method is leverages which concepts to solve differential equations?
2. Which concept from calculus allows you to transfer the derivative of one function to another inside an integral?
- Chain Rule
- Incorrect. The chain rule is used for compositions of functions, not for transferring derivatives within integrals. Try again.
- L’Hopital’s Rule
- Incorrect. L’Hopital’s Rule is used to evaluate limits of indeterminate forms, not to transfer derivatives within integrals.
- Integration by Parts
- Correct! Integration by Parts allows the transfer of a derivative from one function to another inside an integral, a fundamental tool in the Laplace transform method.
- Implicit Differentiation
- Incorrect. Implicit differentiation is used to differentiate equations where the variable isn’t isolated, not to transfer derivatives within integrals.
3. \(\ds\frac{d}{dt}\left[e^{-st}\right] =\) .
- \(-se^{-st}\)
- Correct! The derivative of \(e^{-st}\) with respect to \(t\) is \(-se^{-st}\text{.}\)
- \(e^{-st}\)
- Incorrect. The derivative should include a factor of \(-s\) due to the chain rule.
- \(se^{-st}\)
- Incorrect. The derivative should be negative because the exponent of \(e^{-st}\) has a negative sign.
- \(-\frac{1}{s}e^{-st}\)
- Incorrect. This expression incorrectly represents the derivative of \(e^{-st}\text{.}\) Think about applying the chain rule to the exponent.
4. Select the result of applying integration by parts to the integral, \(\ \ \ds \int t^2 \ y'\ dt\).
- \(\ds \int t^2 \ y'\ dt = t^2\, y - \int y \ dt\)
- Incorrect. Pay close attention to how integration by parts is applied. Try again.
- \(\ds \int t^2 \ y'\ dt = t^2 y - \int t^2\, y \ dt\)
- Incorrect. You need to transfer the derivative correctly by choosing the appropriate \(u\) and \(dv\text{.}\)
- \(\ds \int t^2 \ y'\ dt = 2\int t\, y \ dt\)
- Incorrect. This result is incorrect due to a missing term. Try applying integration by parts again.
- \(\ds \int t^2 \ y'\ dt = t^2\, y - 2\int t\, y \ dt\)
- Correct! The result of applying integration by parts to \(\ds \int t^2 \ y'\ dt\) is \(t^2 y - 2\int t \ y \ dt\text{.}\)
