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Section 12.5 Exercises
Exercises 💡 Conceptual Quiz
📝: Abbreviations.
1.
(a) Technique not used to prepare for the inverse transform.
Which of the following is NOT a technique mentioned for preparing
\(Y(s)\) for the backward transform?
- Completing the square
Incorrect. Completing the square is a technique used to rewrite \(Y(s)\) as a sum of known Laplace transforms.
- Partial fraction decomposition
Incorrect. Partial fraction decomposition is another technique used to prepare \(Y(s)\) for the inverse transform.
- Integration by Parts
Correct! Integration by parts is not a technique used to prepare \(Y(s)\) for the backward transform.
- Rewriting as a sum of \(s \)-functions
Incorrect. Rewriting \(Y(s)\) is a technique used in Step 2b.
(b) Applying initial conditions.
Similar to other methods, the Laplace transform method applies the initial conditions to the general solution to find a particular solution.
(c) Where would you complete the square?
Which part of the Laplace transform method might require you to complete the square?
Applying the Forward Transform
Completing the square is not part of applying the forward transform.
Solving for \(Y(s)\)
Completing the square is not required to solve for \(Y(s)\text{.}\)
Preparing \(Y(s)\) for an inverse transform
Completing the square is a technique used to prepare \(Y(s)\text{.}\)
Applying the inverse transform
Completing the square should be done before applying the backward transform.
(d) Splitting Fractions.
Select the next step to prepare
\begin{equation*}
Y(s) = \dfrac{3s + 6}{s^2 + 9}
\end{equation*}
for the inverse transform?
\(\quad\dfrac{3s}{s^2 + 9} + \dfrac{6}{s^2 + 9} \)
- Yes! Now, each term corresponds to a separate entry in the Laplace table.
\(\quad\dfrac{3(s + 2)}{s^2 + 9} \)
- This doesn’t help you match any table entries more clearly.
\(\quad\dfrac{3s + 3}{s^2 + 9} + \dfrac{3}{s^2 + 9} \)
- This breaks the numerator improperly.
\(\quad\dfrac{3}{s + 3} + \dfrac{6}{s^2 + 9} \)
- This has an incorrect decomposition.
(e) What’s the Best Next Step?
What is the next “best” step needed to compute
\begin{equation*}
\ilap{\dfrac{s+3}{(s - 1)^2 + 9}}\text{?}
\end{equation*}
- Look up the inverse Laplace transform in the table.
Incorrect. This function is not directly in the table.
- Factor the denominator.
Incorrect. Factoring the denominator is not necessary at this stage.
-
Rewrite the numerator, then split the fraction like so:
\begin{equation*}
\dfrac{s-1+4}{(s - 1)^2 + 9} = \dfrac{s-1}{(s - 1)^2 + 9} + \dfrac{4}{(s - 1)^2 + 9}\text{.}
\end{equation*}
Correct! The numerator needs an
\(s-1\) to match with
L\(_8\).
-
Split the fraction directly, like so:
\begin{equation*}
\dfrac{s+3}{(s - 1)^2 + 9} = \dfrac{s}{(s - 1)^2 + 9} + \dfrac{3}{(s - 1)^2 + 9}
\end{equation*}
Incorrect. The next step is to decompose the function into simpler forms.
(f) Completing the Square.
(g) Match the Next Step in the Method.
Match the statements on the left that complete the statements on the right.- you get the particular solution, \(y(t)\text{.}\)
- After applying the inverse Laplace transform to \(Y(s)\text{,}\)
- you isolate \(Y(s)\text{.}\)
- After solving the Laplace-domain equation,
- you get an algebraic equation involving \(Y(s)\)
- After forward transforming the differential equation involving \(y(t)\text{,}\)
- you need to prepare \(Y(s)\) for an inverse Laplace transform.
- Once you have isolated \(Y(s)\text{,}\)
- apply the Laplace transform to its terms.
- Given an initial-valued problem,
Exercises 🏋️♂️ Practice Drills
1.
(a) Select the Forward Transform.
What is the Laplace transform of the initial-value problem
\begin{equation*}
y'' + 3y' + 2y = -40e^{3t}, \quad y(0) = 1, \quad y'(0) = 0\text{?}
\end{equation*}
\(s^2Y + 3sY + 2Y - s = \dfrac{-40}{s-3}\)
- This is the correct transformation of the given differential equation.
\(s^2Y + 3sY + 2Y = \dfrac{-40}{s-3}\)
- Where are the initial conditions?
\(s^2Y + 3sY + 2Y + s = \dfrac{-40}{s - 3}\)
- Close, but this answer is off by a sign.
\(s^2Y + 3sY + 2Y - 1 = \dfrac{-40}{s - 3}\)
- Look closely at the initial conditions.
(b) Select the Forward Transform.
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*}
\(s^2Y - 4sY + 6Y - 2s + 8 = \dfrac{1}{s^2 + 1}\)
- This is the correct transformation of the initial-value problem.
\(s^2Y + 4sY + 6Y = \dfrac{1}{s^2 + 1}\)
- Double-check the signs and the initial condition terms.
\(s^2Y - 4sY + 6Y = \dfrac{1}{s^2 + 1}\)
- This answer is missing the initial condition terms on the left-hand side.
\(Y = \dfrac{1}{s^2 + 1}\)
- This only represents the transform of \(t^2\) and does not account for the left-hand side of the equation.
(c) Select the Inverse.
\(\quad \ilap{\dfrac{10}{(s - 2)^2 + 25}} = \)
\(\quad 2e^{2t}\sin(5t)\)
Correct! Factoring out \(10\) and placing the missing constant gives the correct form: \(2e^{2t}\sin(5t)\text{.}\)
\(\quad 5e^{2t}\sin(5t)\)
Incorrect. The correct answer requires factoring and rebalancing, giving \(2e^{2t}\sin(5t)\text{.}\)
\(\quad e^{2t}\cos(5t)\)
Incorrect. The sine form, not cosine, matches this function.
(d) Select the Inverse.
\begin{equation*}
Y(s) = \dfrac{7}{s^2} \quad\Rightarrow\quad y(t) = \fillinmath{XXXXXXXXXXXXXXX}
\end{equation*}
\(\dfrac{t}{7}\)
Incorrect.
\(7t^2\)
Incorrect.
\(7t\)
Correct!
\(\dfrac{7}{t}\)
Incorrect.
(e) Select the Inverse.
\(\ilap{\dfrac{1}{s - 1} - \dfrac{1}{s + 1} + \dfrac{4s + 3}{s^2 + 1}} = \fillinmath{XXXXXXXXXX}\)
\(\quad e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\)
Correct! The inverse Laplace transform matches the forms in the table.
\(\quad e^{t} + e^{t} + \cos(t) + \sin(t)\)
No, this does not match the correct inverse Laplace transform.
\(\quad e^{-t} - e^{t} + \cos(t)\)
Incorrect. This does not account for all terms in the inverse Laplace transform.
(f) Select the Inverse.
What is the Laplace transform of
\(e^{3t}\text{?}\)
\(\quad\dfrac{1}{s-3}\)
-
\(\quad\dfrac{1}{s+3}\)
-
\(\quad\dfrac{1}{s-2}\)
-
\(\quad\dfrac{1}{s}\)
-
(g) Select the Inverse.
\(\quad \ilap{\dfrac{3}{s^2 + 9}} = \fillinmath{XXXXXX}\)
\(\quad \sin(3t)\)
Correct! The inverse Laplace transform of \(\dfrac{3}{s^2 + 9}\) is \(\sin(3t)\text{.}\)
\(\quad \cos(3t)\)
No, the correct transform for \(\dfrac{3}{s^2 + 9}\) is \(\sin(3t)\text{,}\) not \(\cos(3t)\text{.}\)
\(\quad e^{3t}\)
No, this is not the correct inverse transform for the given expression.
\(\quad \dfrac{3}{s-3}\)
No, this is not an inverse transform expression.
(h) Select the Inverse.
\(\quad \ilap{\dfrac{1}{s+3}} = \fillinmath{XXXXXX}\)
\(\quad e^{-3t}\)
Correct! The inverse Laplace transform of \(\dfrac{1}{s+3}\) is indeed \(e^{-3t}\text{.}\)
\(\quad e^{3t}\)
No, the correct answer is \(e^{-3t}\text{,}\) not \(e^{3t}\text{.}\)
\(\quad e^{-t}\)
No, the exponent should be \(-3t\text{,}\) not \(-t\text{.}\)
\(\quad\dfrac{1}{s-3}\)
No, this is not the correct inverse Laplace transform.
(i) Select the Inverse.
\(\quad \ilap{\dfrac{24}{s^5}} = \fillinmath{XXXXXX}\)
\(\quad t^3\)
No, this is incorrect. The correct answer is \(t^4\text{.}\)
\(\quad t^4\)
Correct! The inverse Laplace transform of \(\dfrac{24}{s^5}\) is \(t^4\text{.}\)
\(\quad t^2\)
No, the correct answer is \(t^4\text{,}\) not \(t^2\text{.}\)
\(\quad \dfrac{1}{s^5}\)
No, this is the original function in the \(s\)-domain, not its inverse transform.
(j) Select the Inverse.
\(\quad \ilap{\dfrac{2}{s^2 + 4}} = \fillinmath{XXXXXX}\)
\(\quad \sin(2t)\)
Correct! The inverse Laplace transform of \(\dfrac{2}{s^2 + 4}\) is \(\sin(2t)\text{.}\)
\(\quad \cos(2t)\)
No, the correct inverse Laplace transform is \(\sin(2t)\text{,}\) not \(\cos(2t)\text{.}\)
\(\quad e^{2t}\)
No, the correct inverse transform is \(\sin(2t)\text{,}\) not \(e^{2t}\text{.}\)
\(\quad t^2\)
No, the correct answer is \(\sin(2t)\text{,}\) not \(t^2\text{.}\)
(k) Select the Inverse.
\(\quad \ilap{\dfrac{1}{s - 5}} = \fillinmath{XXXXXX}\)
\(\quad e^{5t}\)
Correct! The inverse Laplace transform of \(\dfrac{1}{s - 5}\) is \(e^{5t}\text{.}\)
\(\quad e^{-5t}\)
No, the correct answer is \(e^{5t}\text{,}\) not \(e^{-5t}\text{.}\)
\(\quad \cos(5t)\)
No, this is not the correct inverse transform for the given expression.
\(\quad \sin(5t)\)
No, the correct inverse transform for \(\dfrac{1}{s - 5}\) is \(e^{5t}\text{.}\)
(l) Select the Inverse.
\(\quad \ilap{\dfrac{1}{(s+4)^2}} = \fillinmath{XXXXXX}\)
\(\quad te^{-4t}\)
Correct! The inverse Laplace transform of \(\dfrac{1}{(s+4)^2}\) is \(te^{-4t}\text{.}\)
\(\quad e^{4t}\)
No, the correct answer is \(te^{-4t}\text{,}\) not \(e^{4t}\text{.}\)
\(\quad t^2e^{-4t}\)
No, the correct inverse transform is \(te^{-4t}\text{,}\) not \(t^2e^{-4t}\text{.}\)
\(\quad e^{-4t}\)
No, the correct answer is \(te^{-4t}\text{,}\) not \(e^{-4t}\text{.}\)
(m) Select the Inverse.
\(\ilap{\dfrac{s-1}{(s-1)^2 + 4}} = \fillinmath{XXXXXX}\)
\(\quad e^{t}\cos(2t)\)
Correct! The inverse Laplace transform of this expression is \(e^{t}\cos(2t)\text{.}\)
\(\quad e^{-t}\sin(2t)\)
Incorrect. The correct inverse Laplace transform is \(e^{t}\cos(2t)\text{.}\)
\(\quad \cos(2t)\)
Incorrect. The correct inverse Laplace transform is \(e^{t}\cos(2t)\text{.}\)
2. Forward Transform the Equation.
Forward transform the following initial-value problem into the Laplace Domain. Don’t Solve for \(Y(s)\text{.}\)
\begin{equation*}
y'' - 4y' + 6y = e^{2t}, \quad y(0) = 1,\quad y'(0) = 0
\end{equation*}
Match the Sine Form.
Fill in the boxes with the values needed to match the Laplace sine form:
\begin{equation*}
A \cdot \dfrac{b}{s^2 + b^2}\text{.}
\end{equation*}
3.
\(\dfrac{1}{s^2 + 49} = \boxed{\phantom{ \left|\dfrac{1}{7}\right|}} \cdot \dfrac{\boxed{\phantom{ |7| }}}{s^2 + \boxed{\phantom{ |7| }}^2}\)
4.
\(\dfrac{12}{s^2 + 16} = \boxed{\phantom{ \left|\dfrac{1}{4}\right|}} \cdot \dfrac{\boxed{\phantom{ |4| }}}{s^2 + \boxed{\phantom{ |4| }}^2}\)
5.
\(\dfrac{3}{s^2 + 5} = \boxed{\phantom{ \dfrac{3}{\sqrt{5}}}} \cdot \dfrac{\boxed{\phantom{ \sqrt{5} }}}{s^2 + \boxed{\phantom{ \sqrt{5} }}^2}\)
Match the Cosine Form.
Fill in the boxes with the values needed to match the Laplace cosine form:
\begin{equation*}
A \cdot \dfrac{s}{s^2 + b^2}\text{.}
\end{equation*}
6.
\(\dfrac{4s}{s^2 + 9} = \boxed{\phantom{ \left|\dfrac{1}{3}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |3| }}^2}\)
7.
\(\dfrac{s}{2s^2 + 8} = \boxed{\phantom{ \left|\dfrac{1}{2}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |2| }}^2}\)
Match the Power Form.
Fill in the boxes with the values needed to match the Laplace power form:
\begin{equation*}
A \cdot \dfrac{n!}{s^{n+1}}\text{.}
\end{equation*}
8.
\(\dfrac{1}{s^5} = \boxed{\phantom{ \dfrac{1}{24} }} \cdot \dfrac{\boxed{\phantom{ |4| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 4\ }} + 1}}\)
9.
\(\dfrac{7}{s^3} = \boxed{\phantom{ \left|\dfrac{7}{2}\right| }} \cdot \dfrac{\boxed{\phantom{ |2| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 2\ }} + 1}}\)
10.
\(\dfrac{2}{5s^4} = \boxed{\phantom{ \dfrac{1}{15} }} \cdot \dfrac{\boxed{\phantom{ |3| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 3\ }} + 1}}\)
Inverse Transforms.
Apply the inverse transform to revert each Laplace solution back into a function of
\(t\text{.}\)
11.
12.
13.
14.
15.
\(\dfrac{10}{(s - 3)^2 + 11}\)
16.
17.
\(Y(s) = \dfrac{12}{s} + \dfrac{7s}{s^2 + 25}\)
18.
\(Y(s) = \dfrac{4}{(s - 2)^2 + 16} + \dfrac{2}{(s+7)^5}\)
19.
\(Y(s) = \dfrac{s + 5}{(s + 3)^2 + 16}\)
20.
Drag the function on the left to the form it best matches on the right.
\begin{equation*}
\end{equation*}
\(\color{RoyalBlue}\small\text{Function}\ ⤵︎ \)
\(\color{BurntOrange}\small\text{Form}\ ⤵︎\)
- \(\dfrac{7}{1-s}\)
- \(\dfrac{1}{s-a}\)
- \(\dfrac{7}{2s^2}\)
- \(\dfrac{n!}{s^{n+1}}\)
- \(\dfrac{12}{s^2 + 15}\)
- \(\dfrac{b}{s^2 + b^2}\)
- \(\dfrac{1}{(s-3)^5}\)
- \(\dfrac{n!}{(s-a)^{n+1}}\)
- \(\dfrac{-1}{s^2 + 4s + 1}\)
- \(\dfrac{b}{(s-a)^2 + b^2}\)
Backward Transforms- Level 1.
Prepare each Laplace-domain solution for an inverse transformation, then transform it back into a function of
\(t\text{.}\)
21.
\(Y(s) = \dfrac{6}{(s-1)^4} \)22.
\(X(s) = \dfrac{1}{s^5} \)23.
\(G(s) = \dfrac{4}{s^2 + 9} \)24.
\(Y(s) = \dfrac{5s}{s^2 + 7} \)25.
\(X(s) = \dfrac{21}{(s+2)^2 + 16} \)26.
\(Y(s) = \dfrac{10}{s-10} \)27.
\(Q(s) = \dfrac{7s+7}{(s+1)^2 + 12} \)28.
\(R(s) = \dfrac{3s + 2}{s^2 + 4s + 4}\)
29.
\(S(s) = \dfrac{4s + 1}{s^2 + 6s + 9} \)
30.
Backward Transforms- Level 2.
31.
\(G(s) = \dfrac{1}{s^2 + 4s + 8} \)32.
\(\dfrac{2s + 7}{(s + 3)^2 + 4}\)
33.
\(F(s) = \dfrac{2s+16}{s^2 + 4s + 13} \)34.
\(Y(s) = \dfrac{3s-15}{2s^2 - 4s + 8} \)35.
\(Q(s) = \dfrac{1}{s^2 + 6s + 9} \)36.
\(Y(s) = \dfrac{11}{s^2 - 6s + 14}\)
37.
\(Y(s) = \dfrac{s+3}{s^2 + 2s + 10}\)
38.
\(P(s) = \dfrac{s}{s^2 - s - 6}\)
39.
\(H(s) = \dfrac{5}{s-6} - \dfrac{6s}{s^2 + 9} + \dfrac{3}{2s^2 + 8s + 10} \)40.
\(Y(s) = \dfrac{2s + 7}{(s + 1)(s + 3)}\)
41.
\(Y(s) = \dfrac{5}{s(s + 4)}\)
42.
\(Y(s) = \dfrac{s+9}{s^2 - 2s - 3}\)
43.
\(Y(s) = \dfrac{s-6}{s^2 - 4s + 29}\)
44.
\(X(s) = \dfrac{7s-4}{s^2 + 36} \)45.
\(G(s) = \dfrac{2s-19}{s^2 - 4s+13} + \dfrac{5}{s-1} \)46.
\(F(s) = \dfrac{s}{s^2 + 6s + 11} \)
Backward Transforms- Level 3.
47.
\(X(s) = \dfrac{90s^2 - 195s + 30}{s^3 - 7s^2 + 6s} \)48.
\(\dfrac{3s + 4}{s^2 + 2s + 1} \)
49.
\(I(s) = \dfrac{5s^2 + 34s + 53}{(s+3)^2(s+1)} \)50.
\(F(s) = \dfrac{7s^3 - 2s^2 - 3s + 6}{s^3(s-2)} \)51.
\(R(s) = \dfrac{4s^3 - 13s^2 + 74s + 27}{s^4 - 4s^3 + 14s^2 + 44s + 25} = \dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}\)
In the Laplace Domain: Isolate, Prepare & Invert.
Isolate
\(Y(s)\text{,}\) prepare it for inversion, and invert it.
52.
\(s^2 Y(s) - 4Y(s) = \dfrac{60}{s+1} \)53.
\(s^2 Y(s) + sY(s) - 6Y(s) = \dfrac{30s^2 + 120}{s^2 + s} \)
Match the Laplace Form.
Manipulate each of the expressions so that it matches one of the forms:
\begin{equation*}
\dfrac{b}{(s-a)^2 + b^2}\quad \text{or}\quad \dfrac{s-a}{(s-a)^2 + b^2}.
\end{equation*}
54.
\(\dfrac{s+3}{s^2 - 6s + 10}\)
55.
\(\dfrac{s}{s^2 - 8s + 25}\)
56.
\(\dfrac{s}{s^2 + 8s + 25}\)
57.
\(\dfrac{s-5}{s^2 +12s + 72}\)
Exercises ✍🏻 Problems
Solve each of the following initial-value problems using Laplace Transforms.
1.
\(x'' + 6x' + 9x = 64e^{5t},\quad x(0) = 3,\quad x'(0) = 0\)
2.
\(y''-2y'+5y = -8e^{-t},\quad y(0) = 2,\quad y'(0) = 12\)
3.
\(y'' + 3y' + 2y = 0,\quad y(0) = 2,\quad y'(0) = -1\)
4.
\(y' + 4y = 10e^{-2t},\quad y(0) = 3\)
5.
\(y' + 3y = 6e^{2t}, \quad y(0) = 1\)
6.
\(y'' + 4y = \cos(2t), \quad y(0) = 0, \quad y'(0) = 1\)
7.
\(y' - 2y = 3e^{4t},\quad y(0) = 1\)
8.
\(y'' + 5y' + 6y = \sin(2t),\quad y(0) = 0,\quad y'(0) = 1\)
9.
\(y'' + 3y' + 2y = 4t,\quad y(0) = 1,\quad y'(0) = 0\)
10.
\(y'' + 3y' + 2y = 0, \quad y(0) = 1,\quad y'(0) = 0\)
11.
\(y'' - y = e^{2t}, \quad y(0) = 0,\quad y'(0) = 1\)
12.
\(y' + y = 4, \quad y(0) = 2\)
13.
\(y'' - 3y' + 2y = e^{2t}, \quad y(0) = 1, \quad y'(0) = 0\)
14.
\(x'' - 4x' + 13x = 54e^{-t}, \quad x(0) = 0, \quad x'(0) = 0\)
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