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Section Chapter 10 Exercises
Reading Questions α―β
β Quick-Answer Questions
1. True-False.
(a) Which of the following is NOT a technique mentioned for preparing \(Y(s) \) for the backward 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) Which of the statements are true?
Which of the statements are true?
The forward Laplace transform converts a differential equation into an algebraic equation This statement is true. The forward Laplace transform simplifies the differential equation by converting it into an algebraic equation in terms of \(Y(s)\text{.}\)
In Step 2, you isolate \(y(t)\text{.}\) This statement is false. In Step 2, you solve for \(Y(s)\text{,}\) not \(y(t)\text{.}\)
Step 2 involves applying the backward Laplace transform
This statement is false. Step 2 involves solving for \(Y(s)\) as a function of \(s\text{.}\)
Step 2 involves rewriting \(Y(s)\) to match forms in the common Laplace transform table
This statement is true. Step 2 prepares \(Y(s)\) for the backward transform by breaking it into known forms found in the table of common Laplace transforms.
The final step involves applying the inverse Laplace transform to recover the solution \(y(t) \)
This statement is true. The final step of the Laplace Transform Method involves applying the inverse Laplace transform to recover the solution \(y(t) \text{.}\)
(c) Similar to other methods, this method applies the initial conditions to the general solution to find a particular solution .
Similar to other methods, this method applies the initial conditions to the general solution to find a particular solution
(d) The expression \(\ds\dfrac{s-1}{(s-1)^2 + 4}\) matches the form required to apply the inverse Laplace transform found in the common transform table .
(e) True or False: The inverse Laplace transform is always a unique function.
2. Multiple-Choice.
(a) What is the goal of Step 2 in the Laplace Transform Method?
What is the goal of Step 2 in the Laplace Transform Method?
To apply the forward Laplace transform
Incorrect. Applying the forward transform is done in Step 1.
To solve for \(y(t)\)
Incorrect. Solving for \(y(t)\) is not the goal of Step 2.
To rewrite \(Y(s)\) as a sum of \(s\) -functions found in the table of common transforms
Correct! Step 2 involves breaking down \(Y(s)\) into simpler components that match known Laplace transforms.
To find the particular solution to the differential equation
Incorrect. Finding the particular solution is the ultimate goal of the Laplace Transform Method, not just Step 2.
(b) Which of the following is the main purpose of Step 2 in the Laplace Transform Method?
Which of the following is the main purpose of Step 2 in the Laplace Transform Method?
To solve the algebraic equation for \(Y(s) \)
Incorrect. Solving for \(Y(s) \) happens in Step 2.
To prepare \(Y(s) \) for the inverse Laplace transform by rewriting it as a sum of known forms
Correct! Step 2 involves breaking down \(Y(s) \) into simpler components that match known Laplace transforms.
To apply the forward Laplace transform
Incorrect. Applying the forward transform happens in Step 1.
To account for initial conditions
Incorrect. Initial conditions are accounted for in Step 1.
(c) Which step is likely to involve completing the square?
Which step is likely to involve completing the square?
Step 1: Apply the Forward Transform
Incorrect. Completing the square is not part of applying the forward transform.
Step 2: Solve for \(Y(s) \)
Incorrect. Completing the square typically happens in Step 2.
Step 2: Prepare for the Backward Transform
Correct! Completing the square is a technique used in Step 2 to simplify \(Y(s) \text{.}\)
Step 3: Apply the Backward Transform
Incorrect. Completing the square should be done before applying the backward transform.
(d) πβ Select the Forward Transform.
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*}
\(\ds s^2Y + 3sY + 2Y - s = \dfrac{-40}{s-3}\) This is the correct transformation of the given differential equation.
\(\ds s^2Y + 3sY + 2Y = \dfrac{-40}{s-3}\) Where are the initial conditions?
\(\ds s^2Y + 3sY + 2Y + s = \dfrac{-40}{s - 3}\) Close, but this answer is off by a sign.
\(\ds s^2Y + 3sY + 2Y - 1 = \dfrac{-40}{s - 3}\) Look closely at the initial conditions.
(e) πβ 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.
(f) πβ Splitting Fractions.
Which expression helps reveal the table match for
\(Y(s) = \dfrac{3s + 6}{s^2 + 9}\text{?}\)
\(\dfrac{3s}{s^2 + 9} + \dfrac{6}{s^2 + 9} \) Yes! Now each term matches a separate entry in the Laplace table.
\(\dfrac{3(s + 2)}{s^2 + 9} \) This doesnβt help you match any table entries more clearly.
\(\dfrac{3s + 3}{s^2 + 9} + \dfrac{3}{s^2 + 9} \) This breaks the numerator improperly.
\(\dfrac{3}{s + 3} + \dfrac{6}{s^2 + 9} \) This has an incorrect decomposition.
(g) πβ.
In the Laplace Transform Method, the inverse transform
\(\fillinmath{XXXXX}\)
converts a differential equation into an algebraic equation.
No, this describes the forward Laplace transform.
solves the algebraic equation for \(Y(s)\text{.}\)
No, solving for \(Y(s)\) happens after applying the forward transform.
recovers the original function \(y(t)\) from \(Y(s)\text{.}\)
Correct! The inverse Laplace transform brings us back to the original function \(y(t)\text{.}\)
eliminates initial conditions from the equation.
No, initial conditions are incorporated into the transformed equation, not eliminated.
(h) πβ.
What is the next step needed to compute \(\ds\ilap{\dfrac{s+3}{(s - 1)^2 + 9}}\text{?}\)
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: \(\ds\dfrac{s-1+4}{(s - 1)^2 + 9} = \dfrac{s-1}{(s - 1)^2 + 9} + \dfrac{4}{(s - 1)^2 + 9}\text{.}\)
Correct! The numerator needs an
\(s-1\) to match with
L\(_8\) .
Split the fraction directly, like so: \(\ds\dfrac{s+3}{(s - 1)^2 + 9} = \dfrac{s}{(s - 1)^2 + 9} + \dfrac{3}{(s - 1)^2 + 9}\)
Incorrect. The next step is to decompose the function into simpler forms.
(i) πβ.
Which of the following
\(s\) -functions require adjustment to match one of the common inverse Laplace transforms?
\(\ds\dfrac{s}{s^2 + 16}\) Incorrect. This function already matches a known cosine form and does not require any missing constants.
\(\ds\dfrac{8}{s^2 + 16}\) Correct! This function would require a missing constant adjustment to match the sine form, with the numerator needing to be \(4\text{.}\)
\(\ds\dfrac{s - 3}{(s - 3)^2 + 25}\) Incorrect. This function already matches a known form and does not require any missing constants.
(j) πβ.
\(\ds\quad \ilap{\dfrac{10}{(s - 2)^2 + 25}} = \)
\(2e^{2t}\sin(5t)\) Correct! Factoring out \(10\) and placing the missing constant gives the correct form: \(2e^{2t}\sin(5t)\text{.}\)
\(5e^{2t}\sin(5t)\) Incorrect. The correct answer requires factoring and rebalancing, giving \(2e^{2t}\sin(5t)\text{.}\)
\(e^{2t}\cos(5t)\) Incorrect. The sine form, not cosine, matches this function.
(k) πβ.
\begin{equation*}
Y(s) = \dfrac{7}{s^2} \quad\Rightarrow\quad y(t) = \fillinmath{XXXXXXXXXXXXXXX}
\end{equation*}
\(\ds\dfrac{t}{7}\) Incorrect.
\(7t^2\) Incorrect.
\(7t\) Correct!
\(\ds\dfrac{7}{t}\) Incorrect.
(l) \(\ds \ilap{\dfrac{1}{s - 1} - \dfrac{1}{s + 1} + \dfrac{4s + 3}{s^2 + 1}} = \fillinmath{XXXXXXXXXX}\)
\(\ds e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\) Correct! The inverse Laplace transform matches the forms in the table.
\(\ds e^{t} + e^{t} + \cos(t) + \sin(t)\) No, this does not match the correct inverse Laplace transform.
\(\ds e^{-t} - e^{t} + \cos(t)\) Incorrect. This does not account for all terms in the inverse Laplace transform.
(m) What is the Laplace transform of \(e^{3t}\text{?}\)
\(\ds\dfrac{1}{s-3}\)
\(\ds\dfrac{1}{s+3}\)
\(\ds\dfrac{1}{s-2}\)
\(\ds\dfrac{1}{s}\)
(n) In the equation \(y'' + 4y = \cos(2t)\text{,}\) what is the Laplace transform of the right-hand side?
\(\ds\dfrac{s}{s^2 + 4}\)
\(\ds\dfrac{4}{s^2 + 4}\)
\(\ds\dfrac{s}{s^2 + 1}\)
\(\ds\dfrac{1}{s^2 + 4}\)
(o) πβ Direct Inverse.
\(\ds\quad \ilap{\dfrac{3}{s^2 + 9}} = \fillinmath{XXXXXX}\)
\(\sin(3t)\) Correct! The inverse Laplace transform of \(\ds\dfrac{3}{s^2 + 9}\) is \(\sin(3t)\text{.}\)
\(\cos(3t)\) No, the correct transform for \(\ds\dfrac{3}{s^2 + 9}\) is \(\sin(3t)\text{,}\) not \(\cos(3t)\text{.}\)
\(e^{3t}\) No, this is not the correct inverse transform for the given expression.
\(\ds\dfrac{3}{s-3}\) No, this is not an inverse transform expression.
(p) πβ.
\(\ds\quad \ilap{\dfrac{1}{s+3}} = \fillinmath{XXXXXX}\)
\(e^{-3t}\) Correct! The inverse Laplace transform of \(\ds\dfrac{1}{s+3}\) is indeed \(e^{-3t}\text{.}\)
\(e^{3t}\) No, the correct answer is \(e^{-3t}\text{,}\) not \(e^{3t}\text{.}\)
\(e^{-t}\) No, the exponent should be \(-3t\text{,}\) not \(-t\text{.}\)
\(\ds\dfrac{1}{s-3}\) No, this is not the correct inverse Laplace transform.
(q) πβ.
\(\ds\quad \ilap{\dfrac{24}{s^5}} = \fillinmath{XXXXXX}\)
\(t^3\) No, this is incorrect. The correct answer is \(t^4\text{.}\)
\(t^4\) Correct! The inverse Laplace transform of \(\ds\dfrac{24}{s^5}\) is \(t^4\text{.}\)
\(t^2\) No, the correct answer is \(t^4\text{,}\) not \(t^2\text{.}\)
\(\ds\dfrac{1}{s^5}\) No, this is the original function in the \(s\) -domain, not its inverse transform.
(r) πβ.
\(\ds\quad \ilap{\dfrac{2}{s^2 + 4}} = \fillinmath{XXXXXX}\)
\(\sin(2t)\) Correct! The inverse Laplace transform of \(\ds\dfrac{2}{s^2 + 4}\) is \(\sin(2t)\text{.}\)
\(\cos(2t)\) No, the correct inverse Laplace transform is \(\sin(2t)\text{,}\) not \(\cos(2t)\text{.}\)
\(e^{2t}\) No, the correct inverse transform is \(\sin(2t)\text{,}\) not \(e^{2t}\text{.}\)
\(t^2\) No, the correct answer is \(\sin(2t)\text{,}\) not \(t^2\text{.}\)
(s) πβ.
\(\ds\quad \ilap{\dfrac{1}{s - 5}} = \fillinmath{XXXXXX}\)
\(e^{5t}\) Correct! The inverse Laplace transform of \(\ds\dfrac{1}{s - 5}\) is \(e^{5t}\text{.}\)
\(e^{-5t}\) No, the correct answer is \(e^{5t}\text{,}\) not \(e^{-5t}\text{.}\)
\(\cos(5t)\) No, this is not the correct inverse transform for the given expression.
\(\sin(5t)\) No, the correct inverse transform for \(\ds\dfrac{1}{s - 5}\) is \(e^{5t}\text{.}\)
(t) πβ.
\(\ds\quad \ilap{\dfrac{1}{(s+4)^2}} = \fillinmath{XXXXXX}\)
\(te^{-4t}\) Correct! The inverse Laplace transform of \(\ds\dfrac{1}{(s+4)^2}\) is \(te^{-4t}\text{.}\)
\(e^{4t}\) No, the correct answer is \(te^{-4t}\text{,}\) not \(e^{4t}\text{.}\)
\(t^2e^{-4t}\) No, the correct inverse transform is \(te^{-4t}\text{,}\) not \(t^2e^{-4t}\text{.}\)
\(e^{-4t}\) No, the correct answer is \(te^{-4t}\text{,}\) not \(e^{-4t}\text{.}\)
(u) \(\ds \ilap{\dfrac{4s^3 + 3s^2 - 2s + 1}{(s - 1)(s + 1)(s^2 + 1)}} = \fillinmath{XXXXXX}\)
\(\ds e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\) Correct! The inverse Laplace transform is \(e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\text{.}\)
\(\ds 1 + e^{-t} + \cos(t)\) Incorrect. This does not match the correct inverse Laplace transform.
\(\ds 2\cos(t) + 3\sin(t)\) Incorrect. This does not include the full inverse Laplace transform.
(v) \(\ds \ilap{\dfrac{s-1}{(s-1)^2 + 4}} = \fillinmath{XXXXXX}\)
\(\ds e^{t}\cos(2t)\) Correct! The inverse Laplace transform of this expression is \(e^{t}\cos(2t)\text{.}\)
\(\ds e^{-t}\sin(2t)\) Incorrect. The correct inverse Laplace transform is \(e^{t}\cos(2t)\text{.}\)
\(\ds \cos(2t)\) Incorrect. The correct inverse Laplace transform is \(e^{t}\cos(2t)\text{.}\)
(w) \(\ds \ilap{\dfrac{11}{(s - 3)^2 + 5}} = \dfrac{11}{\sqrt{5}}e^{3t}\sin(\sqrt{5}t)\text{,}\) which matches the form in the table when \(b = \fillinmath{X}\)
\(\sqrt{5}\) Correct! The correct value for \(b\) is \(\sqrt{5}\text{.}\)
\(5\) Incorrect. The correct value is \(\sqrt{5}\text{,}\) not 5.
\(3\) Incorrect. The correct value is \(\sqrt{5}\text{,}\) not 3.
(x)
Which of the following is true about the inverse Laplace transform?
It converts a function from the frequency domain to the time domain.
It is a method to solve algebraic equations.
It is used to differentiate a function with respect to time.
It only works for polynomials.
3. Other.
(a) .
After applying the backward Laplace transform in Step 3, you obtain
\(y(t)\text{,}\) the
\(\ul{\hspace{3em}}\) to the differential equation
(b) ?
After completing the square of the denominator:
\(\ds\dfrac{11}{s^2 + 18s + 400} = \dfrac{11}{(s + 9)^2 + \fillinmath{X}} \) missing number
\(=\)
(c) πβ Match the Next Step in the Method.
Drag the task, on the left, that directly follows the task on the right. ... The result is your solution.
After applying the inverse laplace transform to \(Y(s)\) ...
... Use algebra to isolate it.
A forward transform you get an algebraic equation involving \(Y(s)\) ...
... Rewrite it as a sum of expressions resembling common Laplace transforms.
Once you have isolated \(Y(s)\text{...}\)
... Apply the Laplace Transform to it
Given an initial-valued problem ...
(d) Explain the significance of completing the square in finding inverse Laplace transforms.
(e) Why is partial fraction decomposition helpful in finding inverse Laplace transforms?
(f) Why is the Laplace Transform Method considered a powerful tool for solving differential equations?
(g) What role does the table of common Laplace transforms play in the Laplace Transform Method?
(h) Explain how the Laplace transform simplifies solving linear differential equations. Include a discussion of how initial conditions are incorporated into the transformed equation.
(i) Multiple Choice: Which of the following is NOT a property of the Laplace transform?
The Laplace transform is linear.
The Laplace transform converts a derivative into a polynomial in \(s\text{.}\)
The Laplace transform can only be applied to functions defined for all time.
The Laplace transform of \(e^{at}f(t)\) shifts by \(a\) in the \(s\) -domain.
(j) Fill in the blank: The Laplace transform of the second derivative \(y''(t)\) is given by
Exercises ποΈ Warm-ups & Drills
1. Forward Transform the Equation.
Forward transform the following initial-value problem into the Laplace Domain.
Donβt 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.
Rewrite each of the following into the form
\begin{equation*}
A \cdot \dfrac{b}{s^2 + b^2}\text{,}
\end{equation*}
by filling in the appropriate values in the boxes.
2. \(\ds \dfrac{1}{s^2 + 49} = \boxed{\phantom{ \left|\dfrac{1}{7}\right|}} \cdot \dfrac{\boxed{\phantom{ |7| }}}{s^2 + \boxed{\phantom{ |7| }}^2}\) 3. \(\ds \dfrac{12}{s^2 + 16} = \boxed{\phantom{ \left|\dfrac{1}{4}\right|}} \cdot \dfrac{\boxed{\phantom{ |4| }}}{s^2 + \boxed{\phantom{ |4| }}^2}\) 4. \(\ds \dfrac{3}{s^2 + 5} = \boxed{\phantom{ \dfrac{3}{\sqrt{5}}}} \cdot \dfrac{\boxed{\phantom{ \sqrt{5} }}}{s^2 + \boxed{\phantom{ \sqrt{5} }}^2}\)
Match the Cosine Form.
Rewrite each of the following into the form
\begin{equation*}
A \cdot \dfrac{s}{s^2 + b^2}\text{,}
\end{equation*}
by filling in the appropriate values in the boxes.
5. \(\ds \dfrac{4s}{s^2 + 9} = \boxed{\phantom{ \left|\dfrac{1}{3}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |3| }}^2}\) 6. \(\ds \dfrac{s}{2s^2 + 8} = \boxed{\phantom{ \left|\dfrac{1}{2}\right|}} \cdot \dfrac{s}{s^2 + \boxed{\phantom{ |2| }}^2}\)
Match the Power Form.
Rewrite each of the following into the form
\begin{equation*}
A \cdot \dfrac{n!}{s^{n+1}}\text{,}
\end{equation*}
by filling in the appropriate values in the boxes.
7. \(\ds \dfrac{1}{s^5} = \boxed{\phantom{ \dfrac{1}{24} }} \cdot \dfrac{\boxed{\phantom{ |4| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 4\ }} + 1}}\) 8. \(\ds \dfrac{7}{s^3} = \boxed{\phantom{ \left|\dfrac{7}{2}\right| }} \cdot \dfrac{\boxed{\phantom{ |2| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 2\ }} + 1}}\) 9. \(\ds \dfrac{2}{5s^4} = \boxed{\phantom{ \dfrac{1}{15} }} \cdot \dfrac{\boxed{\phantom{ |3| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 3\ }} + 1}}\)
Inverse Transforms. Apply the inverse transform to revert each Laplace solution back into a function of
\(t\text{.}\)
10. \(\ds \dfrac{12}{s}\) 11. \(\ds \dfrac{5}{s^2 + 4}\) 12. \(\ds \dfrac{17}{s^4}\) 13. \(\ds \dfrac{7s}{s^2 + 25}\) 14. \(\ds \dfrac{10}{(s - 3)^2 + 11}\) 15. \(\ds \dfrac{2}{(s+7)^5}\) 16. \(Y(s) = \dfrac{12}{s} + \dfrac{7s}{s^2 + 25}\) 17. \(Y(s) = \dfrac{4}{(s - 2)^2 + 16} + \dfrac{2}{(s+7)^5}\) 18. \(Y(s) = \dfrac{s + 5}{(s + 3)^2 + 16}\)
Exercises ποΈ Drill: Preparing & Invert
Prepare each Laplace-domain solution for an inverse transformation, then transform it back into a function of
\(t\text{.}\)
Backward Transforms- Level 1.
1. \(\ds Y(s) = \dfrac{6}{(s-1)^4} \) 2. \(\ds X(s) = \dfrac{1}{s^5} \) 3. \(\ds G(s) = \dfrac{4}{s^2 + 9} \) 4. \(\ds Y(s) = \dfrac{5s}{s^2 + 7} \) 5. \(\ds X(s) = \dfrac{21}{(s+2)^2 + 16} \) 6. \(\ds Y(s) = \dfrac{10}{s-10} \) 7. \(\ds Q(s) = \dfrac{7s+7}{(s+1)^2 + 12} \) 8. \(\ds R(s) = \dfrac{3s + 2}{s^2 + 4s + 4} \) 9. \(\ds S(s) = \dfrac{4s + 1}{s^2 + 6s + 9} \) 10. \(\dfrac{5}{s^2 + 16}\)
Backward Transforms- Level 2.
11. \(\ds G(s) = \dfrac{1}{s^2 + 4s + 8} \) 12. \(\dfrac{2s + 7}{(s + 3)^2 + 4}\) 13. \(\ds F(s) = \dfrac{2s+16}{s^2 + 4s + 13} \) 14. \(\ds Y(s) = \dfrac{3s-15}{2s^2 - 4s + 8} \) 15. \(\ds Q(s) = \dfrac{1}{s^2 + 6s + 9} \) 16. \(\ds Y(s) = \dfrac{11}{s^2 - 6s + 14}\) 17. \(\ds Y(s) = \dfrac{s+3}{s^2 + 2s + 10}\) 18. \(\ds P(s) = \dfrac{s}{s^2 - s + 6}\) 19. \(\ds H(s) = \dfrac{5}{s-6} - \dfrac{6s}{s^2 + 9} + \dfrac{3}{2s^2 + 8s + 10} \) 20. \(\ds Y(s) = \dfrac{2s + 7}{(s + 1)(s + 3)}\) 21. \(Y(s) = \dfrac{5}{s(s + 4)}\) 22. \(Y(s) = \dfrac{s+9}{s^2 - 2s - 3}\) 23. \(Y(s) = \dfrac{s-6}{s^2 - 4s + 29}\) 24. \(\ds X(s) = \dfrac{7s-4}{s^2 + 36} \) 25. \(\ds G(s) = \dfrac{2s-19}{s^2 - 4s+13} + \dfrac{5}{s-1} \) 26. \(\ds F(s) = \dfrac{s}{s^2 + 6s + 11} \)
Backward Transforms- Level 3.
27. \(\ds X(s) = \dfrac{90s^2 - 195s + 30}{s^3 - 7s^2 + 6s} \) 28. \(\ds \dfrac{3s + 4}{s^2 + 2s + 1} \) 29. \(\ds I(s) = \dfrac{5s^2 + 34s + 53}{(s+3)^2(s+1)} \) 30. \(\ds F(s) = \dfrac{7s^3 - 2s^2 - 3s + 6}{s^3(s-2)} \) 31. \(R(s) = \dfrac{4s^3 - 13s^2 + 74s + 27}{s^4 - 4s^3 + 14s^2 + 44s + 25} = \dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}\)
In the Laplace Domain: Isolate, Prepare & Invert. Isolate
\(Y(s)\text{,}\) prepare it for inversion, and invert it.
32. \(\ds s^2 Y(s) - 4Y(s) = \dfrac{60}{s+1} \) 33. \(\ds \ds s^2 Y(s) + sY(s) - 6Y(s) = \dfrac{30s^2 + 120}{s^2 + s} \)
Exercises Solving Differential Equations
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. \(x'' - 4x' + 13x = 54e^{-t}, \quad x(0) = 0, \quad x'(0) = 0\) You have attempted
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