Section 11.1 Derivative Transfer via Integration by Parts
Before we apply the Laplace transform to solve differential equations, we need to understand its fundamental mechanism: transferring derivatives within an integral. This section begins with a quick refresher on integration by parts (IBP), emphasizing how derivatives applied to one function can be shifted to another. This transfer process is the key that unlocks the Laplace transform method.
Subsection Review of Integration by Parts (IBP)
Checkpoint 172. Differentiation and Integration Essentials.
Integration by parts is a core calculus technique that allows us to transfer derivatives from one function to another inside an integral. Its most common form is:
\begin{equation*}
\int_{a}^{b} u\ dv = uv\big|_{a}^{b} - \int_{a}^{b} v\ du\text{.}
\end{equation*}
When \(u\) and \(v\) are functions of \(t\text{,}\) we get a version of IBP which makes the transfer of derivatives more clear:
\begin{equation*}
\int_{a}^{b} \os{{\DLBa\large\text{derivative on }v}}{
\us{\large u}{ \us{\uparrow}{\ul{
u(t)
}}} \
\us{\large dv}{\us{\uparrow}{\ul{
{\DLBa\os{\downarrow}{v'\vphantom{\Large\Gamma}}(t)} dt
}}}
}
= \ub{u(b)v(b) - u(a)v(a)}_{\large\text{evaluates to a number}}
- \int_{a}^{b} \os{{\DLBa\large\text{derivative on }u}}{
\us{\large v}{ \us{\uparrow}{\ul{ v(t) }}} \
\us{\large du}{\us{\uparrow}{\ul{ {\DLBa\os{\downarrow}{u'\vphantom{\Large\Gamma}}(t)} dt }}}
}
\end{equation*}
This simple idea of shifting derivatives underlies the Laplace transform method. We’ll see it in action when one of the functions is \(e^{-st}\text{.}\)
Checkpoint 173. 📖❓ Derivative Transfer Technique.
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 function compositions, not transferring derivatives within an integral.
- L’Hôpital’s Rule
- Incorrect. L’Hôpital’s Rule helps evaluate indeterminate limits, not manipulate derivatives inside integrals.
- Integration by Parts
- Correct! Integration by parts is the key tool for shifting derivatives between functions in integrals.
- Implicit Differentiation
- Incorrect. Implicit differentiation is useful for equations where the dependent variable isn’t isolated.
Subsection Using Exponentials to Transfer Derivatives
Checkpoint 174. 👀 Review: Derivative of \(e^{-st}\).
Let \(s\) be a constant. Which function is equal to the derivative
\begin{equation*}
\frac{d}{dt}\left[e^{-st}\right] =\text{?}
\end{equation*}
- \(\ds\quad -se^{-st}\)
- Correct! The derivative of \(e^{-st}\) with respect to \(t\) is \(-se^{-st}\) by the chain rule.
- \(\ds\quad e^{-st}\)
- Incorrect. You’re missing the \(-s\) from the chain rule.
- \(\ds\quad se^{-st}\)
- Incorrect. The sign should be negative because the exponent is \(-st\text{.}\)
- \(\ds\quad -\frac{1}{s}e^{-st}\)
- Incorrect. This form appears when integrating, not differentiating. Try applying the chain rule.
Let’s now see why exponentials make this derivative transfer especially useful. Suppose we choose \(u(t) = e^{-st}\) and \(v(t) = y(t)\text{.}\) Differentiating gives \(u'(t) = -s e^{-st}\) and \(v'(t) = y'(t)\text{.}\) Applying integration by parts yields:
\begin{align*}
\int_{0}^{b} \os{{\DLBa\large\text{derivative on }y}}{
e^{-st} \ {\DLBa\os{\downarrow}{y'\vphantom{\Large\Gamma}}(t)}\ dt
}
\amp = e^{-st} \ y(t) \Big|_{0}^{b} - \int_{0}^{b} \os{{\DLBa\large\text{derivative on }e^{-st}}}{
{\DLBa\os{\downarrow}{\left(-s e^{-st}\right)\vphantom{\Large\Gamma}}}\ y(t)\ dt
}\\
\amp = e^{-sb} y(b) - y(0) + {\DLBa s} \int_{0}^{b} {\DLBa e^{-st}} y(t)\, dt\text{.}
\end{align*}
Notice the key point: the derivative on \(y\) has been completely removed. Instead, we get two terms: a boundary term \(e^{-st} y(t)\Big|_0^b\) and a new integral without \(y'(t)\text{.}\)
This trade is usually worth it since in the context of Laplace transforms, the term \(e^{-sb} y(b)\) will often vanish and \(y(0)\) is the given initial condition.
Checkpoint 175. 🤔💭 Derivative Transfer Reading Questions.
(a) 📖❓ Key Concepts Behind Laplace Transforms.
Which two core mathematical ideas does the Laplace transform method rely on to solve differential equations?
- Chain Rule
- Incorrect. The chain rule is not central to the Laplace transform method. Think about a process that lets you transfer derivatives.
- Product Rule
- No, the product rule is not a key idea in the Laplace transform method. Consider an integration-based technique.
- Integration by Parts
- Correct! Integration by parts allows derivatives to be transferred between functions inside an integral, a foundational technique in the Laplace method.
- Properties of Exponential Functions
- Correct! The method uses properties of exponential functions, especially their predictable behavior under differentiation and integration.
(b) 📖❓ IBP on \(t^2 y'\).
Select the result of applying integration by parts to
\begin{equation*}
\int t^2 \ y'\ dt\text{.}
\end{equation*}
Don’t think of this as a Calculus exercise. Instead, try to think about transferring derivatives.
- \(\ds \int t^2 \ y'\ dt = t^2 y - \int y\ dt\)
- Incorrect. A correct application of integration by parts requires the proper choice of \(u\) and \(dv\text{.}\) Try again.
- \(\ds \int t^2 \ y'\ dt = t^2 y - \int t^2 y\ dt\)
- Incorrect. Be sure you’re transferring the derivative and integrating the correct part.
- \(\ds \int t^2 \ y'\ dt = 2\int t y\ dt\)
- Incorrect. This result skips important parts of the integration by parts formula.
- \(\ds \int t^2 \ y'\ dt = t^2 y - 2\int t y\ dt\)
- Correct! Integration by parts with \(u = t^2\) and \(dv = y'\ dt\) gives this result.
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