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Contents Index Calc Dark Mode Prev Next Profile \(\newcommand\DLGray{\color{Gray}}
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Section 12.1 Step 1 — Into the Laplace Domain
The first step of the Laplace transform method is to convert a differential equation—where the unknown appears through derivatives—into an algebraic equation in the Laplace domain. This transformation makes it easier to isolate the unknown and move toward a solution.
Subsection Crossing Into the Laplace Domain
In terms of the roadmap, we can visualize this first step as follows:
Step 1️⃣: Into the Laplace Domain (Forward Transform).
\begin{equation*}
\ul{\qquad\textbf{Original Domain}\qquad}
\end{equation*}
\begin{equation*}
\DLBa\textbf{Laplace Domain}
\end{equation*}
\begin{gather*}
\os{\vphantom{m}}{y'' - 9y = 10e^{2t}}\\
y(0) = 1,\quad y'(0) = -7\\
\os{\large }{\text{Differential Equation}}
\end{gather*}
\begin{gather*}
\underrightarrow{\text{ Forward}}\\
\text{Apply}\ \laplacesym\\
\end{gather*}
\begin{gather*}
\DLBa s^2Y - 9Y - s + 7 = \frac{10}{s - 2}\\
\DLBa\dotsm
\end{gather*}
Simply put, forward transforming a differential equation into the Laplace domain means applying the Laplace transform to both sides of the equation. The result is an algebraic equation where the derivatives have been converted into powers of
\(s\) and the new unknown is
\(Y\text{.}\)
Subsection Basic Mechanics of the Forward Transform
Symbolically, applying the forward Laplace transform to the differential equation
\begin{equation*}
y'' - 9y = 10e^{2t}, \quad y(0) = 1,\quad y'(0) = -7
\end{equation*}
amounts to applying it to both sides of the equation as follows:
\begin{equation*}
\lap{y'' - 9y} = \lap{10e^{2t}}\text{.}
\end{equation*}
\begin{equation*}
\lap{y''} - 9\lap{y} = 10\lap{e^{2t}}.
\end{equation*}
Letting
\(Y(s) = \lap{y(t)}\) and applying rule
R\(_2\) for
\(\lap{y''}\) and common transform
L\(_2\) , we get:
\begin{equation*}
s^2 Y(s) - s y(0) - y'(0) - 9 Y(s) = 10 \left(\frac{1}{s-2}\right).
\end{equation*}
This is where the initial conditions come into play. Substituting the initial conditions and simplifying gives the desired Laplace domain equation:
\begin{equation*}
s^2Y - 9Y - s + 7 = \frac{10}{s - 2}.
\end{equation*}
Our new unknown is
\(Y\text{,}\) and since the derivatives have been eliminated, we can use algebra to solve for it. In fact, the solution to our differential equation is actually hiding inside
\(Y\text{.}\) So, by solving for
\(Y\) you are also solving for
\(y(t)\) at the same time!
Let’s practice this with one more example.
🌌 Example 198 . Forward Transforming Equations.
Forward transform the following equations into the Laplace domain:
\begin{equation*}
\begin{array}{l}
y'' - 4y' + 6y = t^2,\\
y(0) = 2, \quad y'(0) = 0
\end{array}
\end{equation*}
Solution .
First applying the Laplace transform to both sides:
\begin{equation*}
\lap{y'' - 4y' + 6y} = \lap{t^2}\text{.}
\end{equation*}
\begin{equation*}
\lap{y''} - 4\lap{y'} + 6\lap{y} = \lap{t^2}\text{.}
\end{equation*}
\begin{equation*}
\left[s^2 Y(s) - s y(0) - y'(0)\right] -4\left[sY(s) - y(0)\right] + 6 Y(s) = \frac{1}{s^3}\text{.}
\end{equation*}
Finally, plug in the initial conditions, \(y(0) = 2,\ y'(0) = 0\) and clean up:
\begin{equation*}
s^2 Y(s) - 4sY(s) + 6Y(s) - 2s + 8 = \frac{2}{s^3}\text{.}
\end{equation*}
Checkpoint 199 . 📖❓ Select the Forward Transform.
Which of the following represents the correct Laplace transform of the equation
\begin{equation*}
y'' - 4y' + 6y = t^2, \quad y(0) = 2, \quad y'(0) = -1\text{?}
\end{equation*}
\(\ds s^2Y - 4sY + 6Y - 2s + 9 = \frac{2}{s^3}\) Correct: apply the transform rules for derivatives, substitute \(y(0)=2\text{,}\) \(y'(0)=-1\text{,}\) and use \(\lap{t^2}=\dfrac{2}{s^3}\text{.}\)
\(\ds s^2Y - 4sY + 6Y - 2s = \frac{2}{s^3}\) This answer is missing the constant term that comes from applying the initial conditions.
\(\ds s^2Y - 4sY + 6Y - 2s + 7 = \frac{2}{s^3}\) Check the sign of the \(y'(0)\) term in \(\lap{y''}\text{;}\) with \(y'(0)=-1\) it contributes \(+1\text{,}\) not \(-1\text{.}\)
\(\ds s^2Y - 4sY + 6Y + 9 = \frac{2}{s^3}\) This answer is missing the \(-2s\) term that comes from the \(-s y(0)\) part of \(\lap{y''}\text{.}\)
Now that you’ve seen the mechanical steps involved with applying the Laplace transform to a differential equation, it is important to take a moment to see what’s happening behind the scenes during this process.
Subsection What is Going on Behind the Scenes?
Checkpoint 200 . 📖❓ Applying the Laplace Transform to an Equation.
The process of applying a Laplace transform to a differential equation involves multiplying the equation by
and
with respect to
\(t\text{.}\)
\(e^{-t},\) integrating
No, this misses the \(s\) parameter in the exponential. It should be \(e^{-st}\text{.}\)
\(e^{-st},\) integrating
Correct! The Laplace transform involves multiplying the original function by \(e^{-st}\) and then integrating with respect to \(t\text{.}\)
\(e^{st},\) integrating
No, the exponent should have a negative sign: \(e^{-st}\text{.}\)
\(e^{-st},\) differentiating
No, the Laplace transform requires integration, not differentiation.
\(s,\) differentiating
No, the transform uses \(e^{-st}\) and integration, not the variable \(s\) directly or differentiation.
Since the Laplace transform only applies to functions, applying a “forward transform to a differential equation” really means we are taking the Laplace transform of both sides of that equation. So applying a forward transformation to the initial-value problem
\begin{equation*}
y' - 9y = 10 e^{7t}, \quad y(0) = 1
\end{equation*}
implies that we multiply both sides of the equation by \(e^{-st}\) and integrate from \(0\) to \(\infty\text{:}\)
\begin{equation*}
\ul{\qquad\text{Symbolically}\qquad}
\end{equation*}
\begin{equation*}
\ul{\qquad\text{Behind the Scenes Math}\qquad}
\end{equation*}
\begin{equation*}
\small\lap{y' - 9y} = \lap{10 e^{7t}}
\end{equation*}
\begin{align*}
\small\left(y' - 9y\right) e^{-st} \amp = (10 e^{7t}) e^{-st}\\
\small\int_0^{\infty} \left(y' - 9y\right) e^{-st} dt \amp = \int_0^{\infty} (10 e^{7t}) e^{-st} dt
\end{align*}
Next, we use the linearity of the integral:
\begin{equation*}
\small\lap{y'} - 9\lap{y} = 10\lap{e^{7t}}
\end{equation*}
\begin{equation*}
\small\us{\large I_1}{\ul{\int_0^{\infty} (y') e^{-st} dt}} - 9 \int_0^{\infty} (y) e^{-st} dt = 10 \us{\large I_2}{\ul{\int_0^{\infty} (e^{7t}) e^{-st} dt}}
\end{equation*}
After applying integration by parts to
\(I_1\text{,}\) the derivative is transferred from
\(y(t)\) to
\(e^{-st}\text{,}\) resulting in a factor of
\(s\) and an evaluation of
\(y(t)\) at
\(t=0\) and
\(t=b\text{:}\)
\begin{equation*}
\small
-y(0) + sY(s) - 9 Y(s) = 10 \left(\frac{1}{s-7}\right)
\end{equation*}
\begin{equation*}
\leftarrow\ \text{Symbolically}
\end{equation*}
\begin{equation*}
\text{Mathematically}\ \rightarrow
\end{equation*}
\begin{gather*}
\small
\us{\large\text{as}\ b\ \to\ \infty}{\ul{\cancelto{\large 0}{e^{-sb}y(b)}}} - y(0)
+ s\int_0^{\infty} e^{-st}y\, dt - 9\int_0^{\infty} e^{-st}y\, dt = \frac{10}{s-7}
\end{gather*}
The details for computing
\(I_2 = \lap{e^{7t}}\) are given
here .
Since the two integrals are \(\lap{y} = Y(s)\text{,}\) the symbolic and mathematical paths come back together as
\begin{equation*}
sY(s) - 9 Y(s) - y(0) = \frac{10}{s-7}\text{.}
\end{equation*}
By transforming our equation into this new form, we have momentarily stored
\(y(t)\) inside the function
\(Y\text{.}\) Therefore, solving for
\(Y\) indirectly also solves for
\(y(t)\text{.}\)
Step 1 — Into the Laplace Domain Summary. Apply
\(\laplacesym\) to each term of the differential equation; derivatives become powers of
\(s\) and initial conditions drop in as constants, leaving an algebraic equation in
\(Y(s)\text{.}\)
You have attempted
of
activities on this page.
