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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section 10.3 Summary & Exercises

This section covers the final step in solving differential equations using the Laplace Transform Method: finding the inverse Laplace transform. The focus is on converting the algebraic equation with unknown function \(Y(s)\) back to the original function \(y(t)\text{.}\) Several methods and techniques are introduced to handle different cases where direct computation isn’t possible.

Summary of the Key Ideas.

Common Forms: A table of common Laplace transforms is provided, which doubles as a reference for inverse transforms. The focus is on recognizing forms that match the table entries for functions like \(\sin(bt), \cos(bt)\text{,}\) and others.
Direct Computation: When the function of \(s\) directly matches a form in the common Laplace transform table, the inverse Laplace transform can be easily computed.
Modifying Functions: When a function doesn’t match a known form, minor modifications, such as multiplying by missing constants or splitting fractions, can help.
Completing the Square: When dealing with quadratic expressions in the denominator, especially when the discriminant is negative, completing the square can transform the expression into a form that matches known inverse Laplace transforms. Several examples demonstrate this technique.
Partial Fraction Decomposition: For more complex rational functions, partial fraction decomposition breaks down the function into simpler fractions that match the common transform forms.

Exercises Exercises

1. \(\ds \ilap{\frac{1}{s - 1} - \frac{1}{s + 1} + \frac{4s + 3}{s^2 + 1}} = \fillinmath{XXXXXXXXXX}\).

  • \(\ds e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\)
  • 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.

2. What is the Laplace transform of \(e^{3t}\text{?}\)

  • \(\ds\frac{1}{s-3}\)
  • \(\ds\frac{1}{s+3}\)
  • \(\ds\frac{1}{s-2}\)
  • \(\ds\frac{1}{s}\)

3. In the equation \(y'' + 4y = \cos(2t)\text{,}\) what is the Laplace transform of the right-hand side?

  • \(\ds\frac{s}{s^2 + 4}\)
  • \(\ds\frac{4}{s^2 + 4}\)
  • \(\ds\frac{s}{s^2 + 1}\)
  • \(\ds\frac{1}{s^2 + 4}\)

4. The expression \(\ds\frac{s-1}{(s-1)^2 + 4}\) matches the form required to apply the inverse Laplace transform found in the common transform table.

  • True.

  • True. This expression matches the form in the table and can be directly transformed.

  • False.

  • True. This expression matches the form in the table and can be directly transformed.

Match the Sine Form.

Rewrite each of the following into the form
\begin{equation*} A \cdot \frac{b}{s^2 + b^2}\text{,} \end{equation*}
by filling in the appropriate values in the boxes.

5.

\(\ds \frac{1}{s^2 + 49} = \boxed{\phantom{ \left|\frac{1}{7}\right|}} \cdot \frac{\boxed{\phantom{ |7| }}}{s^2 + \boxed{\phantom{ |7| }}^2}\)
Answer.
\begin{equation*} \frac{1}{s^2 + 49} = \frac{1}{s^2 + 7^2} = \boxed{\frac{1}{7}} \cdot \frac{\boxed{7}}{s^2 + \boxed{7}^2} \end{equation*}

6.

\(\ds \frac{12}{s^2 + 16} = \boxed{\phantom{ \left|\frac{1}{4}\right|}} \cdot \frac{\boxed{\phantom{ |4| }}}{s^2 + \boxed{\phantom{ |4| }}^2}\)
Answer. Answer
\begin{equation*} \frac{12}{s^2 + 16} = 3 \cdot \frac{4}{s^2 + 4^2} = \boxed{3} \cdot \frac{\boxed{4}}{s^2 + \boxed{4}^2} \end{equation*}

7.

\(\ds \frac{3}{s^2 + 5} = \boxed{\phantom{ \frac{3}{\sqrt{5}}}} \cdot \frac{\boxed{\phantom{ \sqrt{5} }}}{s^2 + \boxed{\phantom{ \sqrt{5} }}^2}\)
Answer. Answer
\begin{equation*} \frac{3}{s^2 + 5} = \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{s^2 + (\sqrt{5})^2} = \boxed{\frac{3}{\sqrt{5}}} \cdot \frac{\boxed{\sqrt{5}}}{s^2 + \boxed{\sqrt{5}}^2} \end{equation*}

Match the Cosine Form.

Rewrite each of the following into the form
\begin{equation*} A \cdot \frac{s}{s^2 + b^2}\text{,} \end{equation*}
by filling in the appropriate values in the boxes.

8.

\(\ds \frac{4s}{s^2 + 9} = \boxed{\phantom{ \left|\frac{1}{3}\right|}} \cdot \frac{s}{s^2 + \boxed{\phantom{ |3| }}^2}\)
Answer. Answer
\begin{equation*} \frac{4s}{s^2 + 9} = 4 \cdot \frac{s}{s^2 + 3^2} = \boxed{4} \cdot \frac{s}{s^2 + \boxed{3}^2} \end{equation*}

9.

\(\ds \frac{s}{2s^2 + 8} = \boxed{\phantom{ \left|\frac{1}{2}\right|}} \cdot \frac{s}{s^2 + \boxed{\phantom{ |2| }}^2}\)
Answer. Answer
\begin{equation*} \frac{s}{2s^2 + 8} = \frac{1}{2} \cdot \frac{s}{s^2 + 2^2} = \boxed{\frac{1}{2}} \cdot \frac{s}{s^2 + \boxed{2}^2} \end{equation*}

Match the Power Form.

Rewrite each of the following into the form
\begin{equation*} A \cdot \frac{n!}{s^{n+1}}\text{,} \end{equation*}
by filling in the appropriate values in the boxes.

10.

\(\ds \frac{1}{s^5} = \boxed{\phantom{ \frac{1}{24} }} \cdot \frac{\boxed{\phantom{ |4| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 4\ }} + 1}}\)
Answer. Answer
\begin{equation*} \frac{1}{s^5} = \frac{1}{24} \cdot \frac{4!}{s^{4+1}} = \boxed{\frac{1}{24}} \cdot \frac{\boxed{4}!}{s^{\boxed{4}+1}} \end{equation*}

11.

\(\ds \frac{7}{s^3} = \boxed{\phantom{ \left|\frac{7}{2}\right| }} \cdot \frac{\boxed{\phantom{ |2| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 2\ }} + 1}}\)
Answer. Answer
\begin{equation*} \frac{7}{s^3} = \frac{7}{2} \cdot \frac{2!}{s^{2+1}} = \boxed{\frac{7}{2}} \cdot \frac{\boxed{2}!}{s^{\boxed{2}+1}} \end{equation*}

12.

\(\ds \frac{2}{5s^4} = \boxed{\phantom{ \frac{1}{15} }} \cdot \frac{\boxed{\phantom{ |3| }}\ {\large !}}{s\vphantom{S|}^{\boxed{\phantom{ 3\ }} + 1}}\)
Answer. Answer
\begin{equation*} \frac{2}{5s^4} = \frac{1}{15} \cdot \frac{3!}{s^{3+1}} = \boxed{\frac{1}{15}} \cdot \frac{\boxed{3}!}{s^{\boxed{3}+1}} \end{equation*}

Backward Transforms- Level 1.

13.

\(\ds F(s) = \frac{6}{(s-1)^4} \)
Answer. Answer

14.

\(\ds X(s) = \frac{1}{s^5} \)
Answer. Answer

15.

\(\ds G(s) = \frac{4}{s^2 + 9} \)
Answer. Answer

16.

\(\ds Y(s) = \frac{5s}{s^2 + 7} \)
Answer. Answer

17.

\(\ds X(s) = \frac{21}{(s+2)^2 + 16} \)
Answer. Answer

18.

\(\ds F(s) = \frac{10}{s-10} \)
Answer. Answer

19.

\(\ds Q(s) = \frac{7s+7}{(s+1)^2 + 12} \)
Answer. Answer

Backward Transforms- Level 2.

20.

\(\ds G(s) = \frac{1}{s^2 + 4s + 8} \)
Answer. Answer

21.

\(\ds F(s) = \frac{2s+16}{s^2 + 4s + 13} \)
Answer. Answer

22.

\(\ds Y(s) = \frac{3s-15}{2s^2 - 4s + 8} \)
Answer. Answer

23.

\(\ds Q(s) = \frac{1}{s^2 + 6s + 9} \)
Answer. Answer

24.

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

25.

\(\ds X(s) = \frac{7s-4}{s^2 + 36} \)
Answer. Answer

26.

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

27.

\(\ds F(s) = \frac{s}{s^2 + 6s + 11} \)
Answer. Answer

Backward Transforms- Level 3.

28.

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

29.

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

30.

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

Steps 3 & 4.

31.

\(\ds s^2 F(s) - 4F(s) = \frac{60}{s+1} \)
Answer. Answer

32.

\(\ds \ds s^2 F(s) + sF(s) - 6F(s) = \frac{30s^2 + 120}{s^2 + s} \)
Answer. Answer

Inverse Laplace Transforms.

Determine the original time-domain functions from the given Laplace-transformed functions.

33.

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

34.

\(\ds F(s) = \frac{5}{s(s + 4)}\)
Answer. Answer
\(\ds f(t) = \frac{5}{4}(1 - e^{-4t})\)

Exercise Group.

Find the inverse Laplace transforms of each function.

35.

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

36.

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

Conceptual Questions.

37.

True or False: The inverse Laplace transform is always a unique function.
Answer. Answer
True

38.

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.
Answer. Answer
It converts a function from the frequency domain to the time domain.

39.

Explain the significance of completing the square in finding inverse Laplace transforms.
Answer. Answer
Completing the square is a technique used to transform quadratic expressions in the denominator into a form that matches known inverse Laplace transforms, particularly those involving sine and cosine functions.

40.

Why is partial fraction decomposition helpful in finding inverse Laplace transforms?
Answer. Answer
Partial fraction decomposition breaks down complex rational functions into simpler fractions that correspond to known inverse Laplace transforms, making the transformation process easier.

Computational Problems.

41.

Find the inverse Laplace transform of \(\ds \frac{5}{s^2 + 16}\text{.}\)

42.

Compute the inverse Laplace transform of \(\ds \frac{2s + 7}{(s + 3)^2 + 4} \text{.}\)

43.

Use partial fraction decomposition to find the inverse Laplace transform of \(\ds \frac{3s + 4}{s^2 + 2s + 1} \text{.}\)

Application-Based Problems.

44.

Given the function \(\ds Y(s) = \frac{7}{(s - 2)^2 + 9} \text{,}\) describe the behavior of the corresponding time-domain function.

45.

Find the inverse Laplace transform of the function \(\ds F(s) = \frac{6s + 10}{(s + 4)(s^2 + 9)} \) using partial fraction decomposition.

46. \(\ds \ilap{\frac{4s^3 + 3s^2 - 2s + 1}{(s - 1)(s + 1)(s^2 + 1)}} = \fillinmath{XXXXXX}\).

  • \(\ds e^{t} - e^{-t} + 4\cos(t) + 3\sin(t)\)
  • 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.

47. \(\ds \ilap{\frac{s-1}{(s-1)^2 + 4}} = \fillinmath{XXXXXX}\).

  • \(\ds e^{t}\cos(2t)\)
  • 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 \(e^{t}\cos(2t)\text{.}\)

48. \(\ds \ilap{\frac{11}{(s - 3)^2 + 5}} = \frac{11}{\sqrt{5}}e^{3t}\sin(\sqrt{5}t)\text{,}\) 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.
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