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Exercises 12.6 Exercises

💡 Conceptual Quiz.

Answer the following questions to test your understanding of the Laplace transform method.

📝: 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.
Answer.
Integration by Parts.

(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.
  • True
  • Incorrect.
  • False
  • Correct! The Laplace Transform Method accounts for initial conditions in Step 1.
Answer.
False.

(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.
Answer.
Preparing \(Y(s)\) for inverse 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.
Answer.
\(\dfrac{3s}{s^2 + 9} + \dfrac{6}{s^2 + 9}\text{.}\)

(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.
Answer.
Split into \(\dfrac{s-1}{(s - 1)^2 + 9} + \dfrac{4}{(s - 1)^2 + 9}\text{.}\)

(f) Completing the Square.

(g) Match the Next Step in the Method.

Answer.
See solution for correct sequence.

🏋️‍♂️ Practice Drills.

More practice to strengthen your understanding.

2.

(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) Inverse of a Shifted Sine.

\(\quad \ilap{\dfrac{10}{(s - 2)^2 + 25}} = \)
  • \(\quad 2e^{2t}\sin(5t)\)
  • 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) Inverse of a Power of s.

\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) Inverse of a Sum of Terms.

\(\ilap{\dfrac{1}{s - 1} - \dfrac{1}{s + 1} + \dfrac{4s + 3}{s^2 + 1}} = \fillinmath{XXXXXXXXXX}\)
  • \(\quad e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\)
  • 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 Forward Transform.

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) Inverse of a Sine.

\(\quad \ilap{\dfrac{3}{s^2 + 9}} = \fillinmath{XXXXXX}\)
  • \(\quad \sin(3t)\)
  • 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) Inverse of a Decaying Exponential.

\(\quad \ilap{\dfrac{1}{s+3}} = \fillinmath{XXXXXX}\)
  • \(\quad e^{-3t}\)
  • 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) Inverse of a Higher Power.

\(\quad \ilap{\dfrac{24}{s^5}} = \fillinmath{XXXXXX}\)
  • \(\quad t^3\)
  • 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) Inverse of a Sine Wave.

\(\quad \ilap{\dfrac{2}{s^2 + 4}} = \fillinmath{XXXXXX}\)
  • \(\quad \sin(2t)\)
  • 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) Inverse of a Growing Exponential.

\(\quad \ilap{\dfrac{1}{s - 5}} = \fillinmath{XXXXXX}\)
  • \(\quad e^{5t}\)
  • 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) Inverse of a Repeated Factor.

\(\quad \ilap{\dfrac{1}{(s+4)^2}} = \fillinmath{XXXXXX}\)
  • \(\quad te^{-4t}\)
  • 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) Inverse of a Shifted Cosine.

\(\ilap{\dfrac{s-1}{(s-1)^2 + 4}} = \fillinmath{XXXXXX}\)
  • \(\quad e^{t}\cos(2t)\)
  • 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 \(e^{t}\cos(2t)\text{.}\)

3. 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*}
Answer.
\begin{equation*} (s^2 - 4s + 6)Y(s) = \dfrac{1}{s-2} + s + 4 \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*}

4.

\(\dfrac{1}{s^2 + 49} = \boxed{\phantom{ \left|\dfrac{1}{7}\right|}} \cdot \dfrac{\boxed{\phantom{ |7| }}}{s^2 + \boxed{\phantom{ |7| }}^2}\)
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*}

5.

\(\dfrac{12}{s^2 + 16} = \boxed{\phantom{ \left|\dfrac{1}{4}\right|}} \cdot \dfrac{\boxed{\phantom{ |4| }}}{s^2 + \boxed{\phantom{ |4| }}^2}\)
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*}

6.

\(\dfrac{3}{s^2 + 5} = \boxed{\phantom{ \dfrac{3}{\sqrt{5}}}} \cdot \dfrac{\boxed{\phantom{ \sqrt{5} }}}{s^2 + \boxed{\phantom{ \sqrt{5} }}^2}\)
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*}

🏋️‍♂️ 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*}

7.

\(\dfrac{4s}{s^2 + 9} = \boxed{\phantom{ \left|\dfrac{1}{3}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |3| }}^2}\)
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*}

8.

\(\dfrac{s}{2s^2 + 8} = \boxed{\phantom{ \left|\dfrac{1}{2}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |2| }}^2}\)
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*}

🏋️‍♂️ 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*}

9.

\(\dfrac{1}{s^5} = \boxed{\phantom{ \dfrac{1}{24} }} \cdot \dfrac{\boxed{\phantom{ |4| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 4\ }} + 1}}\)
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*}

10.

\(\dfrac{7}{s^3} = \boxed{\phantom{ \left|\dfrac{7}{2}\right| }} \cdot \dfrac{\boxed{\phantom{ |2| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 2\ }} + 1}}\)
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*}

11.

\(\dfrac{2}{5s^4} = \boxed{\phantom{ \dfrac{1}{15} }} \cdot \dfrac{\boxed{\phantom{ |3| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 3\ }} + 1}}\)
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*}

🏋️‍♂️ Inverse Transforms.

Apply the inverse transform to revert each Laplace solution back into a function of \(t\text{.}\)

12.

\(\dfrac{12}{s}\)
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*}

13.

\(\dfrac{5}{s^2 + 4}\)
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*}

14.

\(\dfrac{17}{s^4}\)
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*}

15.

\(\dfrac{7s}{s^2 + 25}\)
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*}

16.

\(\dfrac{10}{(s - 3)^2 + 11}\)
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*}

17.

\(\dfrac{2}{(s+7)^5}\)
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*}
Adding the two terms gives the full inverse transform:
\begin{equation*} y(t) = e^{2t}\sin(4t) + \dfrac{1}{12}t^4 e^{-7t}\text{.} \end{equation*}

18.

\(Y(s) = \dfrac{12}{s} + \dfrac{7s}{s^2 + 25}\)
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*}
\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*}

19.

\(Y(s) = \dfrac{4}{(s - 2)^2 + 16} + \dfrac{2}{(s+7)^5}\)
Answer.
The form of the first denominator is \(\ul{(s \pm \text{shift})^{2} + \text{number}}\) and has no \(s\) in the numerator. Here the numerator already matches \(b\text{,}\) so
\begin{align*} \ilap{\dfrac{4}{(s - 2)^2 + 16}} \amp\ \quad \knowl{./knowl/xref/lt-L7-table.html}{\text{\(L_7\)}} (a=2,\ b=4)\\ \amp = e^{2t} \sin(4t) \end{align*}
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*}

20.

\(Y(s) = \dfrac{s + 5}{(s + 3)^2 + 16}\)
Answer.
Rewrite \(s + 5\) as \((s + 3) + 2\) to get:
\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 \(\sfrac{4}{4}\text{:}\)
\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*}

21.

🏋️‍♂️ Backward Transforms- Level 1.

Prepare each Laplace-domain solution for an inverse transformation, then transform it back into a function of \(t\text{.}\)

29.

\(R(s) = \dfrac{3s + 2}{s^2 + 4s + 4}\)
Answer.
\(r(t) = 3e^{-2t} + 2te^{-2t}\)

30.

\(S(s) = \dfrac{4s + 1}{s^2 + 6s + 9} \)
Answer.
\(s(t) = 4e^{-3t} + (t + 1)e^{-3t} \)

31.

\(\dfrac{5}{s^2 + 16}\)
Answer.
This expression matches the form \(\dfrac{b}{s^2 + b^2} \) with \(b = 4 \text{,}\) so the inverse Laplace transform is \(5 \cdot \sin(4t) \text{.}\)

🏋️‍♂️ Backward Transforms- Level 2.

Prepare each Laplace-domain solution for an inverse transformation, then transform it back into a function of \(t\text{.}\)

33.

\(\dfrac{2s + 7}{(s + 3)^2 + 4}\)

37.

\(Y(s) = \dfrac{11}{s^2 - 6s + 14}\)

38.

\(Y(s) = \dfrac{s+3}{s^2 + 2s + 10}\)

39.

\(P(s) = \dfrac{s}{s^2 - s - 6}\)
Answer.
\begin{equation*} \ilap{P(s)} = \dfrac25 e^{-2t} + \dfrac35 e^{3t}. \end{equation*}

40.

\(H(s) = \dfrac{5}{s-6} - \dfrac{6s}{s^2 + 9} + \dfrac{3}{2s^2 + 8s + 10} \)
Answer.

41.

\(Y(s) = \dfrac{2s + 7}{(s + 1)(s + 3)}\)
Answer.
\(y(t) = e^{-t} + \dfrac{1}{2}e^{-3t}\)

42.

\(Y(s) = \dfrac{5}{s(s + 4)}\)
Answer.
\(f(t) = \dfrac{5}{4}(1 - e^{-4t})\)

43.

\(Y(s) = \dfrac{s+9}{s^2 - 2s - 3}\)

44.

\(Y(s) = \dfrac{s-6}{s^2 - 4s + 29}\)

46.

\(G(s) = \dfrac{2s-19}{s^2 - 4s+13} + \dfrac{5}{s-1} \)
Answer.

🏋️‍♂️ Backward Transforms- Level 3.

Prepare each Laplace-domain solution for an inverse transformation, then transform it back into a function of \(t\text{.}\)

48.

\(X(s) = \dfrac{90s^2 - 195s + 30}{s^3 - 7s^2 + 6s} \)
Answer.

49.

\(\dfrac{3s + 4}{s^2 + 2s + 1} \)

50.

\(I(s) = \dfrac{5s^2 + 34s + 53}{(s+3)^2(s+1)} \)
Answer.

51.

\(F(s) = \dfrac{7s^3 - 2s^2 - 3s + 6}{s^3(s-2)} \)
Answer.

52.

\(R(s) = \dfrac{4s^3 - 13s^2 + 74s + 27}{s^4 - 4s^3 + 14s^2 + 44s + 25} = \dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}\)

🏋️‍♂️ In the Laplace Domain: Isolate, Prepare & Invert.

Isolate \(Y(s)\text{,}\) prepare it for inversion, and invert it.

54.

\(s^2 Y(s) + sY(s) - 6Y(s) = \dfrac{30s^2 + 120}{s^2 + s} \)
Answer.

🏋️‍♂️ 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*}

55.

\(\dfrac{s+3}{s^2 - 6s + 10}\)
Answer.
\(\dfrac{s-3}{(s - 3)^2 + 1^2} + 6\cdot \dfrac{1}{(s - 3)^2 + 1^2}\)

56.

\(\dfrac{s}{s^2 - 8s + 25}\)
Answer.
\(\dfrac{s - 4}{(s - 4)^2 +3^2} + \dfrac{4}{3}\cdot \dfrac{3}{(s - 4)^2 +3^2}\)

57.

\(\dfrac{s}{s^2 + 8s + 25}\)
Answer.
\(\dfrac{s + 4}{(s + 4)^2 +3^2} - \dfrac{4}{3}\cdot \dfrac{3}{(s + 4)^2 +3^2}\)

58.

\(\dfrac{s-5}{s^2 +12s + 72}\)
Answer.
\begin{equation*} \dfrac{s+6}{(s + 6)^2 + 6^2} - \dfrac{11}{6}\cdot \dfrac{6}{(s + 6)^2 + 6^2} \end{equation*}

✍🏻 Problems.

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

59.

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

60.

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

61.

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

62.

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

63.

\(y' + 3y = 6e^{2t}, \quad y(0) = 1\)
Answer.
\begin{equation*} y(t) = 5e^{-3t} + 2e^{2t} \end{equation*}

64.

\(y'' + 4y = \cos(2t), \quad y(0) = 0, \quad y'(0) = 1\)
Answer.
\begin{equation*} y(t) = \dfrac{t+2}{4}\sin(2t) \end{equation*}

65.

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

66.

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

67.

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

68.

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

69.

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

70.

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

71.

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

72.

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