Section 9.4 Forward Transforming Equations
In the previous sections, we’ve focused on applying Laplace transforms to individual functions. Now, we shift our focus to solving differential equations by transforming the entire equation, along with its initial conditions. This process brings us back to the first step of the Laplace Transform Method roadmap.
To solve a differential equation using Laplace transforms, we apply the Laplace transform to both sides of the equation, leveraging the common transforms and properties we’ve discussed so far. Applying the transform to a function can be thought of as multiplying the function by \(e^{-st}\) and integrating with respect to \(t\text{.}\) However, instead of rewriting this process for each term, we simply apply the Laplace operator, \(\laplacesym\text{,}\) to both sides of the equation. Let’s illustrate this with an example.
Example 13.
Apply the Laplace Transform to the initial-value problem:
\begin{equation*}
y'' + 3y' + 2y = -40e^{3t}, \quad y(0) = 0, \quad y'(0) = 1.
\end{equation*}
Solution. Solution
First, we apply \(\laplacesym\) to both sides:
\begin{equation*}
\lap{y'' + 3y' + 2y} = \lap{-40e^{3t}}.
\end{equation*}
Using the linearity property
P1, we break this into four separate transforms:
\begin{equation*}
\ub{\lap{y''}}_{(1)} + 3\ub{\lap{y'}}_{(2)} + 2\ub{\lap{y}}_{(3)} = -40\ub{\lap{e^{3t}}}_{(4)}.
\end{equation*}
Now, let’s compute each transform:
\begin{align*}
(1):\quad \amp\ \lap{y''} = s^2Y - s\os{\large 0}{\os{\Large \shortparallel}{y(0)}} - \os{\large 1}{\os{\Large \shortparallel}{y'(0)}} = s^2Y - 1 \quad \knowl{./knowl/xref/P4.html}{\text{\(P_4\)}}.\\
(2):\quad \amp\ \lap{y'} = sY - y(0) = sY \quad \knowl{./knowl/xref/P3.html}{\text{\(P_3\)}}.\\
(3):\quad \amp\ \lap{y} = Y.\\
(4):\quad \amp\ \lap{e^{3t}} = \frac{1}{s - 3} \quad \knowl{./knowl/xref/L2.html}{\text{\(L_2\)}}.
\end{align*}
Substituting these results back into the equation, we get:
\begin{align*}
s^2Y - 1 + 3sY + 2Y =\amp\ \frac{-40}{s - 3}, \quad \text{or}\\
s^2Y + 3sY + 2Y - 1 =\amp\ \frac{-40}{s - 3}.
\end{align*}
Let’s look at one more example for additional clarity.
Example 14.
Apply the Laplace Transform to the initial-value problem:
\begin{equation*}
y'' - 4y' + 6y = t^2, \quad y(0) = 2, \quad y'(0) = 0.
\end{equation*}
Solution. Solution
Applying the Laplace transform to both sides:
\begin{align*}
\lap{y'' - 4y' + 6y} =\amp\ \lap{t^2}, \\
\lap{y''} - 4\lap{y'} + 6\lap{y} =\amp\ \lap{t^2}, \\
s^2Y - s\os{\large 2}{\os{\Large \shortparallel}{y(0)}} - \os{\large 0}{\os{\Large \shortparallel}{y'(0)}} - 4(sY - \os{\large 2}{\os{\Large \shortparallel}{y(0)}}) + 6Y =\amp\ \frac{2}{s^3}, \\
s^2Y - 2s - 4sY + 8 + 6Y =\amp\ \frac{2}{s^3}, \\
s^2Y - 4sY + 6Y - 2s + 8 =\amp\ \frac{2}{s^3}.
\end{align*}
In both examples, the differential equations have been transformed into algebraic equations. Solving for \(Y(s)\) is the next step in the Laplace Transform Roadmap.
Reading Questions Check-Point Questions
1. Applying the Laplace transform to a differential equation transforms the equation into an algebraic equation involving functions of \(s\).
Applying the Laplace transform to a differential equation transforms the equation into an algebraic equation involving functions of \(s\)
True.
True. The Laplace transform converts a differential equation into an algebraic equation in \(s\text{.}\)
False.
True. The Laplace transform converts a differential equation into an algebraic equation in \(s\text{.}\)
2. Which of the following represents the correct Laplace transform of the equation
\begin{equation*}
y'' + 3y' + 2y = -40e^{3t}, \quad y(0) = 1, \quad y'(0) = 0\text{?}
\end{equation*}
- \(\ds s^2Y + 3sY + 2Y = \frac{-40}{s-3}\)
No, this is incorrect. Where are the initial conditions?
- \(\ds s^2Y + 3sY + 2Y - s = \frac{-40}{s-3}\)
Correct! This is the correct transformation of the given differential equation.
- \(\ds s^2Y + 3sY + 2Y + s = \frac{-40}{s - 3}\)
No, this is close to the answer, but it is off by a sign.
- \(\ds s^2Y + 3sY + 2Y - 1 = \frac{-40}{s - 3}\)
Incorrect, look closely at the initial conditions.
3. 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*}
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*}
- \(\ds s^2Y - 4sY + 6Y - 2s = \frac{1}{s^3}\)
No, this is incorrect. You are missing a term.
- \(\ds s^2Y - 4sY + 6Y + 2s - 1 = \frac{2}{s^3}\)
No, this is close to the answer. Double check the signs of the initial conditions.
- \(\ds s^2Y - 4sY + 6Y + 1 = \frac{1}{s^3}\)
No, this is incorrect. You are missing a term.
- \(\ds s^2Y - 4sY + 6Y - 2s + 1 = \frac{2}{s^3}\)
Correct! This is the correct transformation using the Laplace transform and applying initial conditions.
4. What is the Laplace transform of the initial-value problem
\begin{equation*}
y'' - 4y' + 6y = \sin t, \quad y(0) = 2, \quad y'(0) = 0\text{?}
\end{equation*}
What is the Laplace transform of the initial-value problem
\begin{equation*}
y'' - 4y' + 6y = \sin t, \quad y(0) = 2, \quad y'(0) = 0\text{?}
\end{equation*}
- \(s^2Y + 4sY + 6Y = \frac{1}{s^2 + 1}\)
No, this is incorrect. Double-check the signs and the initial condition terms.
- \(s^2Y - 4sY + 6Y - 2s + 8 = \frac{1}{s^2 + 1}\)
Correct! This is the correct transformation of the initial-value problem.
- \(s^2Y - 4sY + 6Y = \frac{1}{s^2 + 1}\)
No, this is incorrect. You are missing the initial condition terms on the left-hand side.
- \(Y = \frac{1}{s^2 + 1}\)
No, this is incorrect. This only represents the transform of \(t^2\) and does not account for the left-hand side of the equation.
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