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Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Section 12.5 Exercises

Exercises 💡 Conceptual Quiz

📝: Abbreviations.

1.

(a) Technique not used to prepare from the inverse.
Which of the following is NOT a technique mentioned for preparing \(Y(s)\) for the backward transform?
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  • 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.
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(b) Applying intial conditions.
Similar to other methods, the Laplace transform method applies the initial conditions to the general solution to find a particular solution.
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  • True
  • Incorrect.
  • False
  • Correct! The Laplace Transform Method accounts for initial conditions in Step 1.
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(c) Where would you complete the square?
Which part of the Laplace transform method might require you to complete the square?
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  • 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.
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(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?
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  • \(\quad\dfrac{3s}{s^2 + 9} + \dfrac{6}{s^2 + 9} \)
  • Yes! Now each term matches 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.
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(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*}
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  • 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.
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(f) ?
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(g) Match the Next Step in the Method.
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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*}
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  • \(s^2Y + 3sY + 2Y - s = \dfrac{-40}{s-3}\)
  • 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.
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(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*}
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  • \(s^2Y - 4sY + 6Y - 2s + 8 = \dfrac{1}{s^2 + 1}\)
  • 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.
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(c) Select the Inverse.
\(\quad \ilap{\dfrac{10}{(s - 2)^2 + 25}} = \)
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  • \(\quad 2e^{2t}\sin(5t)\)
  • 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.
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(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.
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(e) Select the Inverse.
\(\ilap{\dfrac{1}{s - 1} - \dfrac{1}{s + 1} + \dfrac{4s + 3}{s^2 + 1}} = \fillinmath{XXXXXXXXXX}\)
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  • \(\quad e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\)
  • 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.
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(f) Select the Inverse.
What is the Laplace transform of \(e^{3t}\text{?}\)
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  • \(\quad\dfrac{1}{s-3}\)
  • \(\quad\dfrac{1}{s+3}\)
  • \(\quad\dfrac{1}{s-2}\)
  • \(\quad\dfrac{1}{s}\)
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(g) Select the Inverse.
\(\quad \ilap{\dfrac{3}{s^2 + 9}} = \fillinmath{XXXXXX}\)
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  • \(\quad \sin(3t)\)
  • 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.
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(h) Select the Inverse.
\(\quad \ilap{\dfrac{1}{s+3}} = \fillinmath{XXXXXX}\)
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  • \(\quad e^{-3t}\)
  • 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.
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(i) Select the Inverse.
\(\quad \ilap{\dfrac{24}{s^5}} = \fillinmath{XXXXXX}\)
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  • \(\quad t^3\)
  • 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.
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(j) Select the Inverse.
\(\quad \ilap{\dfrac{2}{s^2 + 4}} = \fillinmath{XXXXXX}\)
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  • \(\quad \sin(2t)\)
  • 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 \(t^2\text{.}\)
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(k) Select the Inverse.
\(\quad \ilap{\dfrac{1}{s - 5}} = \fillinmath{XXXXXX}\)
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  • \(\quad e^{5t}\)
  • 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 \(e^{5t}\text{.}\)
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(l) Select the Inverse.
\(\quad \ilap{\dfrac{1}{(s+4)^2}} = \fillinmath{XXXXXX}\)
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  • \(\quad te^{-4t}\)
  • 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 \(e^{-4t}\text{.}\)
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(m) Select the Inverse.
\(\ilap{\dfrac{s-1}{(s-1)^2 + 4}} = \fillinmath{XXXXXX}\)
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  • \(\quad e^{t}\cos(2t)\)
  • 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 \(e^{t}\cos(2t)\text{.}\)
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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*}
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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*}
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3.
\(\dfrac{1}{s^2 + 49} = \boxed{\phantom{ \left|\dfrac{1}{7}\right|}} \cdot \dfrac{\boxed{\phantom{ |7| }}}{s^2 + \boxed{\phantom{ |7| }}^2}\)
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Answer.
\begin{equation*} \dfrac{1}{s^2 + 49} = \dfrac{1}{s^2 + 7^2} = \boxed{\dfrac{1}{7}} \cdot \dfrac{\boxed{7}}{s^2 + \boxed{7}^2} \end{equation*}
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4.
\(\dfrac{12}{s^2 + 16} = \boxed{\phantom{ \left|\dfrac{1}{4}\right|}} \cdot \dfrac{\boxed{\phantom{ |4| }}}{s^2 + \boxed{\phantom{ |4| }}^2}\)
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Answer.
\begin{equation*} \dfrac{12}{s^2 + 16} = 3 \cdot \dfrac{4}{s^2 + 4^2} = \boxed{3} \cdot \dfrac{\boxed{4}}{s^2 + \boxed{4}^2} \end{equation*}
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5.
\(\dfrac{3}{s^2 + 5} = \boxed{\phantom{ \dfrac{3}{\sqrt{5}}}} \cdot \dfrac{\boxed{\phantom{ \sqrt{5} }}}{s^2 + \boxed{\phantom{ \sqrt{5} }}^2}\)
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Answer.
\begin{equation*} \dfrac{3}{s^2 + 5} = \dfrac{3}{\sqrt{5}} \cdot \dfrac{\sqrt{5}}{s^2 + (\sqrt{5})^2} = \boxed{\dfrac{3}{\sqrt{5}}} \cdot \dfrac{\boxed{\sqrt{5}}}{s^2 + \boxed{\sqrt{5}}^2} \end{equation*}
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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*}
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6.
\(\dfrac{4s}{s^2 + 9} = \boxed{\phantom{ \left|\dfrac{1}{3}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |3| }}^2}\)
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Answer.
\begin{equation*} \dfrac{4s}{s^2 + 9} = 4 \cdot \dfrac{s}{s^2 + 3^2} = \boxed{4} \cdot \dfrac{s}{s^2 + \boxed{3}^2} \end{equation*}
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7.
\(\dfrac{s}{2s^2 + 8} = \boxed{\phantom{ \left|\dfrac{1}{2}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |2| }}^2}\)
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Answer.
\begin{equation*} \dfrac{s}{2s^2 + 8} = \dfrac{1}{2} \cdot \dfrac{s}{s^2 + 2^2} = \boxed{\dfrac{1}{2}} \cdot \dfrac{s}{s^2 + \boxed{2}^2} \end{equation*}
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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*}
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8.
\(\dfrac{1}{s^5} = \boxed{\phantom{ \dfrac{1}{24} }} \cdot \dfrac{\boxed{\phantom{ |4| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 4\ }} + 1}}\)
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Answer.
\begin{equation*} \dfrac{1}{s^5} = \dfrac{1}{24} \cdot \dfrac{4!}{s^{4+1}} = \boxed{\dfrac{1}{24}} \cdot \dfrac{\boxed{4}!}{s^{\boxed{4}+1}} \end{equation*}
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9.
\(\dfrac{7}{s^3} = \boxed{\phantom{ \left|\dfrac{7}{2}\right| }} \cdot \dfrac{\boxed{\phantom{ |2| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 2\ }} + 1}}\)
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Answer.
\begin{equation*} \dfrac{7}{s^3} = \dfrac{7}{2} \cdot \dfrac{2!}{s^{2+1}} = \boxed{\dfrac{7}{2}} \cdot \dfrac{\boxed{2}!}{s^{\boxed{2}+1}} \end{equation*}
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10.
\(\dfrac{2}{5s^4} = \boxed{\phantom{ \dfrac{1}{15} }} \cdot \dfrac{\boxed{\phantom{ |3| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 3\ }} + 1}}\)
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Answer.
\begin{equation*} \dfrac{2}{5s^4} = \dfrac{1}{15} \cdot \dfrac{3!}{s^{3+1}} = \boxed{\dfrac{1}{15}} \cdot \dfrac{\boxed{3}!}{s^{\boxed{3}+1}} \end{equation*}
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Inverse Transforms.

Apply the inverse transform to revert each Laplace solution back into a function of \(t\text{.}\)
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11.
\(\dfrac{12}{s}\)
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Answer.
The \(s\) in the denominator tells us that we need \(L_1\text{.}\) Before we do, let’s factor out \(12\text{:}\)
\begin{equation*} \ilap{\dfrac{12}{s}} = 12 \cdot \ilap{\dfrac{1}{s}} = 12 \cdot 1 = 12\text{.} \end{equation*}
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12.
\(\dfrac{5}{s^2 + 4}\)
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Answer.
Since the denominator has the form, \(s^2 + \text{number}\text{,}\) and there is no \(s\) in the numerator, we should apply \(L_4\text{.}\) As before, it is helpful to first factor out the constant \(5\text{,}\)
\begin{equation*} \ilap{\dfrac{5}{s^2 + 4}} = 5\ \ilap{\dfrac{1}{s^2 + 4}} \quad \leftarrow \knowl{./knowl/xref/lt-L4-table.html}{\text{L_4}} (b=2)\text{.} \end{equation*}
According to \(L_4\text{,}\) we are missing \(2\) in the numerator. Let’s put it there by multiplying by \(2/2\text{,}\) like so
\begin{align*} = 5\ \ilap{\dfrac{2}{2}\dfrac{1}{s^2 + 4}} \amp = 5\ \ilap{\dfrac{1}{2}\dfrac{2}{s^2 + 4}}\\ \amp = \dfrac{5}{2} \ilap{\dfrac{2}{s^2 + 4}}\\ \amp = \dfrac{5}{2} \sin(2t) \end{align*}
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13.
\(\dfrac{17}{s^4}\)
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Answer.
This denominator has the form \(s^{\text{power}}\text{,}\) which matches \(L_3\) with \(n = 3\text{.}\)
\begin{equation*} \ilap{\dfrac{17}{s^4}} = 17\ \ilap{\dfrac{1}{s^4}} \quad \leftarrow \knowl{./knowl/xref/lt-L3-table.html}{\text{L_3}} (n=3)\text{.} \end{equation*}
In this case, the numerator is missing a \(3!\text{.}\) We can introduce it by multiplying by \(3!/3!\text{,}\) like so
\begin{gather*} = 17\ \ilap{\dfrac{3!}{3!}\dfrac{1}{s^4}} = \dfrac{17}{3!} \ilap{\dfrac{3!}{s^4}} = \dfrac{17}{6} t^3 \end{gather*}
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14.
\(\dfrac{7s}{s^2 + 25}\)
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Answer.
\begin{equation*} \knowl{./knowl/xref/lt-L5-table.html}{\text{L_5}}, b=5 \end{equation*}
\begin{equation*} \ilap{\dfrac{7s}{s^2 + 25}} = 7\ \ilap{\dfrac{s}{s^2 + 25}} = 7\ \cos(5t) \end{equation*}
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15.
\(\dfrac{10}{(s - 3)^2 + 11}\)
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Answer.
The form of this denominator is \(\ul{(s \pm \text{shift})^{2} + \text{number}}\) and has no \(s\) in the numerator. Therefore,
\begin{align*} \ilap{\dfrac{10}{(s - 3)^2 + 11}} \amp\ \quad \knowl{./knowl/xref/lt-L7-table.html}{\text{\(L_7\)}} (a=3,\ b=\sqrt{11})\\ \amp = 10\ \ilap{\dfrac{\sqrt{11}}{\sqrt{11}}\dfrac{1}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} \ilap{\dfrac{\sqrt{11}}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} e^{3t} \sin(\sqrt{11}t) \end{align*}
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16.
\(\dfrac{2}{(s+7)^5}\)
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Answer.
The denominator, \(\ul{(s \pm \text{shift})^{\text{power}}}\text{,}\) indicates an \(e^{at}\) and a \(t^n\) term. Therefore,
\begin{align*} \ilap{\dfrac{2}{(s+7)^5}} \amp\ \quad \knowl{./knowl/xref/lt-L6-table.html}{\text{\(L_6\)}} (a=-7, n=4)\\ \amp = 2\ \ilap{\dfrac{4!}{4!}\dfrac{1}{(s+7)^5}}\\ \amp = \dfrac{2}{4!} \ilap{\dfrac{4!}{(s+7)^5}}\\ \amp = \dfrac{1}{12} t^4 e^{-7t} \end{align*}
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17.
\(Y(s) = \dfrac{12}{s} + \dfrac{7s}{s^2 + 25}\)
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Answer.
The \(s\) in the denominator tells us that we need \(L_1\text{.}\) Before we do, let’s factor out \(12\text{:}\)
\begin{equation*} \ilap{\dfrac{12}{s}} = 12 \cdot \ilap{\dfrac{1}{s}} = 12 \cdot 1 = 12\text{.} \end{equation*}
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\begin{equation*} \knowl{./knowl/xref/lt-L5-table.html}{\text{L_5}}, b=5 \end{equation*}
\begin{equation*} \ilap{\dfrac{7s}{s^2 + 25}} = 7\ \ilap{\dfrac{s}{s^2 + 25}} = 7\ \cos(5t) \end{equation*}
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18.
\(Y(s) = \dfrac{4}{(s - 2)^2 + 16} + \dfrac{2}{(s+7)^5}\)
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Answer.
The form of this denominator is \(\ul{(s \pm \text{shift})^{2} + \text{number}}\) and has no \(s\) in the numerator. Therefore,
\begin{align*} \ilap{\dfrac{10}{(s - 3)^2 + 11}} \amp\ \quad \knowl{./knowl/xref/lt-L7-table.html}{\text{\(L_7\)}} (a=3,\ b=\sqrt{11})\\ \amp = 10\ \ilap{\dfrac{\sqrt{11}}{\sqrt{11}}\dfrac{1}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} \ilap{\dfrac{\sqrt{11}}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} e^{3t} \sin(\sqrt{11}t) \end{align*}
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The denominator, \(\ul{(s \pm \text{shift})^{\text{power}}}\text{,}\) indicates an \(e^{at}\) and a \(t^n\) term. Therefore,
\begin{align*} \ilap{\dfrac{2}{(s+7)^5}} \amp\ \quad \knowl{./knowl/xref/lt-L6-table.html}{\text{\(L_6\)}} (a=-7, n=4)\\ \amp = 2\ \ilap{\dfrac{4!}{4!}\dfrac{1}{(s+7)^5}}\\ \amp = \dfrac{2}{4!} \ilap{\dfrac{4!}{(s+7)^5}}\\ \amp = \dfrac{1}{12} t^4 e^{-7t} \end{align*}
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19.
\(Y(s) = \dfrac{s + 5}{(s + 3)^2 + 16}\)
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Answer.
Rewrite \(s + 5\) as \((s + 3) + 2\) to get:
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\begin{equation*} Y(s) = \dfrac{s + 3}{(s + 3)^2 + 16} + \dfrac{2}{(s + 3)^2 + 16} \end{equation*}
The first term matches L\(_8\) with \(a = -3\text{,}\) \(b = 4\text{,}\) and the second term matches L\(_7\) after multiplying by \(\sfrac{4}{4}\text{:}\)
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\begin{equation*} \ilap{Y(s)} = e^{-3t}\cos(4t) + \dfrac{2}{4}e^{-3t}\sin(4t) = e^{-3t}\cos(4t) + \dfrac{1}{2}e^{-3t}\sin(4t) \end{equation*}
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20.

    Drag the function, on the left, to the form it best matches, on the right.
    \begin{equation*} \end{equation*}
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    \(\color{RoyalBlue}\small\text{Function}\ ⤵︎ \)
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    \(\color{BurntOrange}\small\text{Form}\ ⤵︎\)
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  • \(\dfrac{7}{1-s}\)
  • \(\dfrac{1}{s-a}\)
  • \(\dfrac{7}{2s^2}\)
  • \(\dfrac{n!}{s^{n+1}}\)
  • \(\dfrac{12}{s^2 + 15}\)
  • \(\dfrac{b}{s^2 + b^2}\)
  • \(\dfrac{1}{(s-3)^5}\)
  • \(\dfrac{n!}{(s-a)^{n+1}}\)
  • \(\dfrac{-1}{s^2 + 4s + 1}\)
  • \(\dfrac{b}{(s-a)^2 + b^2}\)
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Backward Transforms- Level 1.

Prepare each Laplace-domain solution for an inverse transformation, then transform it back into a function of \(t\text{.}\)
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25.
\(X(s) = \dfrac{21}{(s+2)^2 + 16} \)
Answer.
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27.
\(Q(s) = \dfrac{7s+7}{(s+1)^2 + 12} \)
Answer.
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28.
\(R(s) = \dfrac{3s + 2}{s^2 + 4s + 4}\)
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Answer.
\(r(t) = 3e^{-2t} + 2te^{-2t}\)
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29.
\(S(s) = \dfrac{4s + 1}{s^2 + 6s + 9} \)
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Answer.
\(s(t) = 4e^{-3t} + (t + 1)e^{-3t} \)
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30.
\(\dfrac{5}{s^2 + 16}\)
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Answer.
This expression matches the form \(\dfrac{b}{s^2 + b^2} \) with \(b = 4 \text{,}\) so the inverse Laplace transform is \(5 \cdot \sin(4t) \text{.}\)
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Backward Transforms- Level 2.

31.
\(G(s) = \dfrac{1}{s^2 + 4s + 8} \)
Answer.
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32.
\(\dfrac{2s + 7}{(s + 3)^2 + 4}\)
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33.
\(F(s) = \dfrac{2s+16}{s^2 + 4s + 13} \)
Answer.
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34.
\(Y(s) = \dfrac{3s-15}{2s^2 - 4s + 8} \)
Answer.
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35.
\(Q(s) = \dfrac{1}{s^2 + 6s + 9} \)
Answer.
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36.
\(Y(s) = \dfrac{11}{s^2 - 6s + 14}\)
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37.
\(Y(s) = \dfrac{s+3}{s^2 + 2s + 10}\)
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38.
\(P(s) = \dfrac{s}{s^2 - s + 6}\)
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39.
\(H(s) = \dfrac{5}{s-6} - \dfrac{6s}{s^2 + 9} + \dfrac{3}{2s^2 + 8s + 10} \)
Answer.
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40.
\(Y(s) = \dfrac{2s + 7}{(s + 1)(s + 3)}\)
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Answer.
\(y(t) = e^{-t} + \dfrac{1}{2}e^{-3t}\)
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41.
\(Y(s) = \dfrac{5}{s(s + 4)}\)
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Answer.
\(f(t) = \dfrac{5}{4}(1 - e^{-4t})\)
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42.
\(Y(s) = \dfrac{s+9}{s^2 - 2s - 3}\)
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43.
\(Y(s) = \dfrac{s-6}{s^2 - 4s + 29}\)
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44.
\(X(s) = \dfrac{7s-4}{s^2 + 36} \)
Answer.
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45.
\(G(s) = \dfrac{2s-19}{s^2 - 4s+13} + \dfrac{5}{s-1} \)
Answer.
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46.
\(F(s) = \dfrac{s}{s^2 + 6s + 11} \)
Answer.
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Backward Transforms- Level 3.

47.
\(X(s) = \dfrac{90s^2 - 195s + 30}{s^3 - 7s^2 + 6s} \)
Answer.
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48.
\(\dfrac{3s + 4}{s^2 + 2s + 1} \)
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49.
\(I(s) = \dfrac{5s^2 + 34s + 53}{(s+3)^2(s+1)} \)
Answer.
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50.
\(F(s) = \dfrac{7s^3 - 2s^2 - 3s + 6}{s^3(s-2)} \)
Answer.
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51.
\(R(s) = \dfrac{4s^3 - 13s^2 + 74s + 27}{s^4 - 4s^3 + 14s^2 + 44s + 25} = \dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}\)
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In the Laplace Domain: Isolate, Prepare & Invert.

Isolate \(Y(s)\text{,}\) prepare it for inversion, and invert it.
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52.
\(s^2 Y(s) - 4Y(s) = \dfrac{60}{s+1} \)
Answer.
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53.
\(s^2 Y(s) + sY(s) - 6Y(s) = \dfrac{30s^2 + 120}{s^2 + s} \)
Answer.
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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*}
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54.
\(\dfrac{s+3}{s^2 - 6s + 10}\)
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Answer.
\(\dfrac{s-3}{(s - 3)^2 + 1^2} + 6\cdot \dfrac{1}{(s - 3)^2 + 1^2}\)
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55.
\(\dfrac{s}{s^2 - 8s + 25}\)
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Answer.
\(\dfrac{s - 4}{(s - 4)^2 +3^2} + \dfrac{4}{3}\cdot \dfrac{3}{(s - 4)^2 +3^2}\)
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56.
\(\dfrac{s}{s^2 + 8s + 25}\)
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Answer.
\(\dfrac{s + 4}{(s + 4)^2 +3^2} - \dfrac{4}{3}\cdot \dfrac{3}{(s + 4)^2 +3^2}\)
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57.
\(\dfrac{s-5}{s^2 +12s + 72}\)
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Answer.
\begin{equation*} \dfrac{s+6}{(s + 6)^2 + 6^2} - \dfrac{11}{6}\cdot \dfrac{6}{(s + 6)^2 + 6^2} \end{equation*}
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Exercises ✍🏻 Problems

Solve each of the following initial-value problems using Laplace Transforms.
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1.

\(x'' + 6x' + 9x = 64e^{5t},\quad x(0) = 3,\quad x'(0) = 0\)
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Answer.
\(x = e^{5t} + 2e^{-3t} + te^{-3t} \)
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2.

\(y''-2y'+5y = -8e^{-t},\quad y(0) = 2,\quad y'(0) = 12\)
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Answer.
\(y(t) = -e^{-t} + 3e^t \cos(2t) + 4e^t \sin(2t) \)
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3.

\(y'' + 3y' + 2y = 0,\quad y(0) = 2,\quad y'(0) = -1\)
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Answer.
\(y(t) = e^{-t} - 3e^{-2t}\)
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4.

\(y' + 4y = 10e^{-2t},\quad y(0) = 3\)
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Answer.
\(y(t) = \dfrac{7}{2}e^{-4t} - \dfrac{5}{2}e^{-2t}\)
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5.

\(y' + 3y = 6e^{2t}, \quad y(0) = 1\)
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6.

\(y'' + 4y = \cos(2t), \quad y(0) = 0, \quad y'(0) = 1\)
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7.

\(y' - 2y = 3e^{4t},\quad y(0) = 1\)
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8.

\(y'' + 5y' + 6y = \sin(2t),\quad y(0) = 0,\quad y'(0) = 1\)
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9.

\(y'' + 3y' + 2y = 4t,\quad y(0) = 1,\quad y'(0) = 0\)
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10.

\(y'' + 3y' + 2y = 0, \quad y(0) = 1,\quad y'(0) = 0\)
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11.

\(y'' - y = e^{2t}, \quad y(0) = 0,\quad y'(0) = 1\)
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12.

\(y' + y = 4, \quad y(0) = 2\)
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13.

\(y'' - 3y' + 2y = e^{2t}, \quad y(0) = 1, \quad y'(0) = 0\)
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14.

\(x'' - 4x' + 13x = 54e^{-t}, \quad x(0) = 0, \quad x'(0) = 0\)
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