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Section 11.6 Exercises
Exercises 💡 Conceptual Quiz
📝: Abbreviations.
Differential Equation
Initial Value Problem
1.
(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) Limit of \(b^3/e^{6b}\) .
What is the value of
\(\lim_{b \to \infty} \dfrac{b^3}{e^{6b}}\text{?}\)
\(\infty\) No, exponential growth (\(e^{6b}\) ) grows faster than polynomial growth (\(b^3\) ) as \(b \to \infty\text{.}\) This pulls the ratio, \(\dfrac{b^3}{e^{6b}}\text{,}\) to zero.
\(0\) Correct, exponential growth (\(e^{6b}\) ) grows faster than polynomial growth (\(b^3\) ) as \(b \to \infty\text{.}\) This pulls the ratio, \(\dfrac{b^3}{e^{6b}}\text{,}\) to zero. This can also be verified with L’Hôpital’s rule.
\(b^3\) No, \(b\) goes to \(\infty\text{.}\) So the limit cannot have \(b\) in its value.
\(6\) No, exponential growth (\(e^{6b}\) ) grows faster than polynomial growth (\(b^3\) ) as \(b \to \infty\text{.}\) This pulls the ratio, \(\dfrac{b^3}{e^{6b}}\text{,}\) to zero.
(c) Fill in the Missing Value.
\(\ds\quad \lap{e^{3t}} = \dfrac{1}{s - \fillinmath{X}}\)
3
Correct! The Laplace transform of \(e^{3t}\) is \(\ds\dfrac{1}{s - 3}\text{.}\)
1
No, the exponent of \(e^{3t}\) is 3, not 1.
0
No, \(s - 0\) would imply a transform of 1, not \(e^{3t}\text{.}\)
-3
No, that would correspond to \(e^{-3t}\text{.}\) Watch the sign.
(d) Laplace Transform of \(t\) .
\(\ds\dfrac{1}{s^2}\) Correct! This matches the rule: \(\lap{t^n} = \dfrac{n!}{s^{n+1}}\) with \(n = 1\text{.}\)
\(\ds\dfrac{1}{s}\) No, that’s the transform of \(1\text{,}\) not \(t\text{.}\)
\(1\) No, the transform of \(t\) is not a constant.
\(\ds\dfrac{2}{s^2}\) No, the numerator should be \(1\text{,}\) not \(2\text{.}\)
(e) Laplace Transform of \(t^2\) .
\(\ds\dfrac{2}{s^3}\) Correct! The Laplace transform of \(t^2\) is \(\ds\dfrac{2!}{s^{2+1}} = \dfrac{2}{s^3}\text{.}\)
\(\ds\dfrac{1}{s^2}\) No, that’s the transform of \(t\text{,}\) not \(t^2\text{.}\)
\(1\) No, \(\lap{t^2}\) is not a constant.
\(\ds\dfrac{2}{s^2}\) No, double-check the exponent in the denominator, it should be \(3\text{,}\) not \(2\text{.}\)
Exercises 🏋️♂️ Practice Drills
Transforming Single Functions.
1. \(\lap{e^{2t}}\) Answer .
\(L_2\ (a=2)\) \(\lap{e^{2t}} = \dfrac{1}{s-2}\)
2. \(\lap{e^{-9t}}\) Answer .
\(L_2\ (a=-9)\) \(\lap{e^{-9t}} = \dfrac{1}{s+9}\)
3. \(\lap{t^2}\) Answer .
\(L_3\ (n=2)\) \(\lap{t^2} = \dfrac{2!}{s^{2+1}} = \dfrac{2}{s^3} \)
4. \(\lap{t^9}\) Answer .
\(L_3\ (n=9)\) \(\lap{t^9} = \dfrac{9!}{s^{9+1}} = \dfrac{9!}{s^{10}} \)
5. \(\lap{\sin (5t)}\) Answer .
\(L_4\ (b=5)\)
\begin{align*}
\lap{\sin (5t)} =\amp \dfrac{5}{s^2 + 5^2} \\
=\amp \dfrac{5}{s^2 + 25}
\end{align*}
6. \(\lap{\cos (-\pi t)}\) Answer .
\(L_4\ (b=-\pi)\)
\begin{align*}
\lap{\cos (-\pi t)} =\amp \dfrac{-\pi}{s^2 + (-\pi)^2} \\
=\amp \dfrac{-\pi}{s^2 + \pi^2}
\end{align*}
7. \(\lap{t^{599}}\) Answer .
\(L_3\ (n=599)\) \(\lap{t^{599}} = \dfrac{599!}{s^{599+1}} = \dfrac{599!}{s^{600}} \)
8. \(\lap{e^{0.0001 t}}\) Answer .
\(L_2\ (a=0.0001)\) \(\lap{e^{0.0001 t}} = \dfrac{1}{s-0.0001} \)
Transforming Combined Functions.
9. \(\lap{15 - 4e^{9t} + 11t^3}\) 10. \(\lap{e^{3t}\sin(6t) - t^3e^{-5t}}\) 11. \(\lap{t^2 \cos(8t)}\) 12. \(\lap{t^3 e^{-9t}}\) Answer .
\begin{align*}
\lap{ t^3 e^{-9t} }
=\amp \dfrac{3!}{\big(s-(-9)\big)^{3+1}} \quad \knowl{./knowl/xref/lt-L6-table.html}{\text{L\(_6\)}}\ (a=2, n=4)\\
=\amp \dfrac{6}{(s+9)^4}
\end{align*}
13. \(\lap{t^3 e^{-2t}}\)
14. \(\lap{t\sin(3t)}\) Answer .
\(-\dfrac{2s}{(s^2 + 9)^2}\)
15. \(\lap{e^{5t}\cos(4t)}\) Answer .
\(\dfrac{s - 5}{(s - 5)^2 + 16}\)
16. \(\lap{e^{2t} - t^3 - \sin (5t)}\) 17. \(\lap{e^{-2t}\sin(2t) + t^2 e^{3t}}\) Answer .
\(L_7\ (a=2), L_9\ (n=2, a=3)\)
\begin{align*}
\lap{e^{-2t}\sin(2t) + t^2 e^{3t}} =\amp \dfrac{2}{(s+2)^2 + 4} + \dfrac{2}{(s-3)^3} \\
=\amp \dfrac{2}{(s+2)^2 + 4} + \dfrac{2}{(s-3)^3}
\end{align*}
18. \(\lap{8t\cos(6t) + e^{3t}\sin(4t)}\) Answer .
\(L_3\ (n=9)\) \(\lap{t^9} = \dfrac{9!}{s^{9+1}} = \dfrac{9!}{s^{10}} \)
19. \(\lap{t^2 \sin(3t)}\) Answer .
\(\ds P(s) = \dfrac{18\left[s^2 - 3\right]}{(s^2 + 9)^3}\)
Exercises ✍🏻 Problems
Using the Definition.
1. \(\lap{11t}\)
2. \(\lap{7t + e^{5t}}\) Answer .
\(\ds \dfrac{7}{s^2} + \dfrac{1}{s-5}\)
3. \(\lap{2.3 t^2}\)
4. \(\lap{-40e^{3t}}\)
5. \(\lap{15}\)
6. \(\lap{e^{7t}}\)
7.
\begin{equation*}
\cos(3t) = \dfrac12\left(e^{3it} + e^{-3it}\right)
\end{equation*}
to show that
\begin{equation*}
\lap{\cos(3t)} = \dfrac{s}{s^2 + 9}\text{.}
\end{equation*}
8.
Compute the Laplace transform of \(\sin(-4t)\text{.}\)
Use (a) as a guide to show
\begin{equation*}
\lap{\sin(bt)} = \dfrac{b}{s^2 + b^2}, \quad s \gt 0\text{,}
\end{equation*}
for any constant
\(b\text{.}\)
9.
\begin{equation*}
\lap{y'(t)} = \int_0^{\infty} y'(t) \cdot e^{-st} \ dt = \lim_{b\to \infty}\ub{\int_0^{b} y'(t) \cdot e^{-st} \ dt}_{I}\text{.}
\end{equation*}
Answer the following questions related to the Laplace transform of \(y'\text{.}\)
Use integration by parts to show that
\begin{equation*}
I = e^{-sb} \cdot y(b) - y(0) + s \int_0^{b} y(t) \cdot e^{-st} \ dt\text{.}
\end{equation*}
Solution .
Select \(u = e^{-st}\) and \(dv = y'(t) \ dt\text{.}\) Then
\begin{align*}
\int_0^{b} e^{-st} \cdot y' \ dt \amp = e^{-st} \cdot y(t)\bigg|_0^{b} - \int_0^{b} y(t) \cdot (-s e^{-st}) \ dt \\
\amp = e^{-sb} \cdot y(b) - e^{0} \cdot y(0) - (-s) \int_0^{b} y(t) \cdot e^{-st} \ dt \\
\amp = e^{-sb} \cdot y(b) - \cdot y(0) + s \int_0^{b} y(t) \cdot e^{-st} \ dt
\end{align*}
Use this \(I\) to show that
\begin{equation*}
\lap{y'} = \ub{\lim_{b \to \infty} \left(\dfrac{y(b)}{e^{sb}}\right)}_L - y(0) + s \lap{y}\text{.}
\end{equation*}
Solution .
Substitute \(I\) from (b) into the limit from (a).
\begin{align*}
\lap{y'} =\amp \lim_{b\to \infty} \left( e^{-sb} \cdot y(b) - y(0) + s \int_0^{b} y(t) \cdot e^{-st} \ dt \right)\\
=\amp \lim_{b\to \infty} \left( e^{-sb} \cdot y(b) \right) - \lim_{b\to \infty} y(0) + \lim_{b\to \infty} \left(s \int_0^{b} y(t) \cdot e^{-st} \ dt \right)\\
=\amp \lim_{b\to \infty} \left( e^{-sb} \cdot y(b) \right) - y(0) + s\lim_{b\to \infty} \left(\int_0^{b} y(t) \cdot e^{-st} \ dt \right)\\
=\amp \ub{\lim_{b \to \infty} \left(\dfrac{y(b)}{e^{sb}}\right)}_L - y(0) + s \lap{y}
\end{align*}
In order for
\(\lap{y'}\) to exist, what must be true?
Answer .
The limit,
\(L\text{,}\) must converge. That is, as
\(t\) gets large, the ratio,
\(\ds\dfrac{y(t)}{e^{st}}\text{,}\) flattens out to some number. In order to maintain this ratio, growth rate of
\(y(t)\) must be less than or equal to
\(e^{st}\text{.}\)
