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

Geoffrey W. Cox, Ph.D., ?

Section 9.5 Summary & Exercises

In this section, we introduced the concept of the forward Laplace transform and derived some common Laplace transforms that we will use throughtout this chapter. The following points summarize the essential concepts from the forward Laplace transform section:

Summary of the Key Ideas.

  • Differential \(\to\) Algebraic Equations. The Laplace transform converts a differential equation into an algebraic equation, simplifying the solution process by eliminating derivatives.
  • Laplace Transform Concept. Applying the Laplace transform to a differential equation involves transforming each term by multiplying by \(e^{-st}\) and integrating with respect to \(t\) from \(0\) to \(\infty\text{,}\) but is often simplified by directly applying the Laplace operator, \(\laplacesym\text{.}\)
  • Linearity Property. The Laplace transform is linear, meaning it distributes across addition and subtraction, and allows for constants to be factored out. This property is essential for transforming complex equations.
  • Transforming Initial Conditions. Initial conditions are incorporated directly into the Laplace-transformed equation, modifying the transformed terms to include initial values, making it easier to solve the resulting algebraic equation.
  • Common Function Transforms. The Laplace transforms of common functions, such as exponentials, sines, cosines, and polynomials, are essential tools in transforming differential equations and are summarized in the provided table.
  • Transforming Derivatives. The Laplace transform of a derivative, \(y'(t)\) or higher, transfers the derivative onto the Laplace variable \(s\text{,}\) reducing the order of the equation while introducing initial condition terms.
  • Multiplication by \(e^{at}\) and \(t^n\). When multiplying a function by an exponential \(e^{at}\text{,}\) the Laplace transform shifts by \(a\) in the \(s\)-domain, and multiplying by \(t^n\) corresponds to differentiating the transform \(n\) times with respect to \(s\text{,}\) introducing a sign change.
  • Transforming the Entire Equation. The process of applying the Laplace transform to an entire differential equation with initial conditions involves systematically transforming each term and leads to a simplified algebraic equation in the \(s\)-domain, ready for solving.
The table below summarizes the common Laplace transforms that we derived in this section.

Common Laplace Transforms \(L_1-L_8\).

Let \(a\) and \(b\) be constant.
\({\LARGE \vphantom{\int}}L_1\)
\(\ds \lap{ 1 } = \frac{1}{s}, \quad s >0 \)
\({\LARGE \vphantom{\int}}L_2\)
\(\ds \lap{ e^{at} } = \frac{1}{s - a}, \quad s >a \)
\({\LARGE \vphantom{\int}}L_3\)
\(\ds \lap{ t^n } = \frac{n!}{s^{n+1}}, \quad s >0, \quad n = 1, 2, 3, \ldots \)
\({\LARGE \vphantom{\int}}L_4\)
\(\ds \lap{ \sin(bt) } = \frac{b}{s^2+b^2}, \quad s >0\)
\({\LARGE \vphantom{\int}}L_5\)
\(\ds \lap{ \cos(bt) } = \frac{s}{s^2+b^2}, \quad s >0 \)
\({\LARGE \vphantom{\int}}L_6\)
\(\lap{ t^n e^{at} } = \ds\frac{n!}{(s-a)^{n+1}}, \quad s >a \)
\({\LARGE \vphantom{\int}}L_7.\)
\(\ds \lap{ e^{at}\sin(bt) } = \frac{b}{(s-a)^2+b^2}, \quad s >a \)
\({\LARGE \vphantom{\int}}L_8.\)
\(\ds \lap{ e^{at}\cos(bt) } = \frac{s-a}{(s-a)^2+b^2}, \quad s >a\)

Laplace Transform Property \(P_1\).

Let \(F(s) = \lap{f(t)}\) and \(G(s) = \lap{g(t)}\text{.}\)
\(P_1\)
\(\ds \lap{ a f(t) \pm b g(t) } = a F(s) \pm b G(s), \quad a, b \) are constants
\(P_2.\)
\(\ds \lap{ e^{at} f(t) } = F(s-a), \quad a \) is a constant
\(P_3.\)
\(\ds \lap{ f'(t) } = sF(s) - f(0) \)
\(P_4.\)
\(\ds \lap{ f''(t) } = s^2F(s) - sf(0) - f'(0) \)
\(P_5.\)
\(\ds \lap{ f'''(t) } = s^3F(s) - s^2f(0) - sf'(0) - f''(0) \)
\({\LARGE \vphantom{\int}}P_6.\)
\(\ds \lap{ t^n f(t) } = (-1)^n \frac{d^{(n)}}{ds^{(n)}}F(s), \quad n = 1, 2, 3, \ldots \)

Exercises Exercises

Definition of the Laplace Transform.

Use the definition to compute each of the following Laplace Transforms.

1.

\(\ds \lap{7t}\)
Answer. Answer
\(\ds \frac{7}{s^2} \)

2.

\(\ds \lap{7t + e^{5t}}\)
Answer. Answer
\(\ds \frac{7}{s^2} + \frac{1}{s-5} \)

3.

\(\ds \lap{e^{at}}\)
Answer. Answer
\(\ds W(s) = \frac{1}{s-a} \)

4.

\(\ds \lap{-40e^{3t}}\)
Answer. Answer
\(\ds \frac{40}{s-3}\)

5.

\(\ds \lap{15}\)
Answer. Answer
\(\ds \frac{15}{s}\)

6.

\(\ds \lap{e^{7t}}\)
Answer. Answer
\(\ds \frac{1}{s - 7}\)

7.

\(\ds \lap{11t}\)
Answer. Answer
\(\ds \frac{11}{s^2}\)

Forward Transform.

Compute the forward Laplace Transforms using the common Laplace transform table.

8.

\(\ds \lap{e^{2t}} \)
Answer. Answer
\(\ds L_2\ (a=2)\)
\begin{equation*} \lap{e^{2t}} = \frac{1}{s-2} \end{equation*}

9.

\(\ds \lap{e^{-9t}}\)
Answer. Answer
\(\ds L_2\ (a=-9)\)
\begin{equation*} \lap{e^{-9t}} = \frac{1}{s+9} \end{equation*}

10.

\(\ds \lap{t^2} \)
Answer. Answer
\(\ds L_3\ (n=2)\)
\begin{equation*} \lap{t^2} = \frac{2!}{s^{2+1}} = \frac{2}{s^3} \end{equation*}

11.

\(\ds \lap{t^9}\)
Answer. Answer
\(\ds L_3\ (n=9)\)
\begin{equation*} \lap{t^9} = \frac{9!}{s^{9+1}} = \frac{9!}{s^{10}} \end{equation*}

12.

\(\ds \lap{\sin (5t)} \)
Answer. Answer
\(\ds L_4\ (b=5)\)
\begin{align*} \lap{\sin (5t)} =\amp \frac{5}{s^2 + 5^2} \\ =\amp \frac{5}{s^2 + 25} \end{align*}

13.

\(\ds \lap{\cos (-\pi t)} \)
Answer. Answer
\(\ds L_4\ (b=-\pi)\)
\begin{align*} \lap{\cos (-\pi t)} =\amp \frac{-\pi}{s^2 + (-\pi)^2} \\ =\amp \frac{-\pi}{s^2 + \pi^2} \end{align*}

14.

\(\ds \lap{t^{599}} \)
Answer. Answer
\(\ds L_3\ (n=599)\)
\begin{equation*} \lap{t^{599}} = \frac{599!}{s^{599+1}} = \frac{599!}{s^{600}} \end{equation*}

15.

\(\ds \lap{e^{0.0001 t}} \)
Answer. Answer
\(\ds L_2\ (a=0.0001)\)
\begin{equation*} \lap{e^{0.0001 t}} = \frac{1}{s-0.0001} \end{equation*}

Exploring the Forward Laplace Transform.

Use the concepts discussed in the section to solve the following exercises.

16.

Answer the following questions related to the Laplace transform of \(y'\text{.}\)
  1. Show that
    \begin{equation*} \lap{y'(t)} = \lim_{b\to \infty}\ub{\int_0^{b} y'(t) \cdot e^{-st} \ dt}_{I}\text{.} \end{equation*}
    Solution. Solution
    This follows directly from the definition of the Laplace transform.
    \begin{equation*} \lap{y'(t)} = \int_0^{\infty} y'(t) \cdot e^{-st} \ dt = \lim_{b\to \infty}\ub{\int_0^{b} y'(t) \cdot e^{-st} \ dt}_{I}\text{.} \end{equation*}
  2. Use integration by parts to show that
    \begin{equation*} I = e^{-sb} \cdot y(b) - y(0) + s \int_0^{b} y(t) \cdot e^{-st} \ dt\text{.} \end{equation*}
    Solution. Solution
    Select \(u = e^{-st}\) and \(dv = y'(t) \ dt\text{.}\) Then
    \begin{align*} \int_0^{b} e^{-st} \cdot y' \ dt \amp = e^{-st} \cdot y(t)\bigg|_0^{b} - \int_0^{b} y(t) \cdot (-s e^{-st}) \ dt \\ \amp = e^{-sb} \cdot y(b) - e^{0} \cdot y(0) - (-s) \int_0^{b} y(t) \cdot e^{-st} \ dt \\ \amp = e^{-sb} \cdot y(b) - \cdot y(0) + s \int_0^{b} y(t) \cdot e^{-st} \ dt \end{align*}
  3. Plug \(I\) from (b) into the limit from (a) to show that
    \begin{equation*} \lap{y'} = \ub{\lim_{b \to \infty} \left(\frac{y(b)}{e^{sb}}\right)}_L - y(0) + s \lap{y}\text{.} \end{equation*}
    Solution. Solution
    Substitute \(I\) from (b) into the limit from (a).
    \begin{align*} \lap{ \amp y'}\\ =\amp \lim_{b\to \infty} \left( e^{-sb} \cdot y(b) - y(0) + s \int_0^{b} y(t) \cdot e^{-st} \ dt \right) \\ =\amp \lim_{b\to \infty} \left( e^{-sb} \cdot y(b) \right) - \lim_{b\to \infty} y(0) + \lim_{b\to \infty} \left(s \int_0^{b} y(t) \cdot e^{-st} \ dt \right) \\ =\amp \lim_{b\to \infty} \left( e^{-sb} \cdot y(b) \right) - y(0) + s\lim_{b\to \infty} \left(\int_0^{b} y(t) \cdot e^{-st} \ dt \right) \\ =\amp \ub{\lim_{b \to \infty} \left(\frac{y(b)}{e^{sb}}\right)}_L - y(0) + s \lap{y} \end{align*}
  4. In order for \(\lap{y'}\) to exist, what must be true?
    Answer. Answer
    The limit, \(L\text{,}\) must converge. That is, as \(t\) gets large, the ratio, \(\ds\frac{y(t)}{e^{st}}\text{,}\) flattens out to some number. In order to maintain this ratio, growth rate of \(y(t)\) must be less than or equal to \(e^{st}\text{.}\)

Forward Transform.

Compute the forward Laplace Transforms using the common transform table and properties.

17.

\(\ds \lap{t^3 e^{-9t}} \)
Answer. Answer
\begin{align*} \lap{ t^3 e^{-9t} } =\amp \frac{3!}{\big(s-(-9)\big)^{3+1}} \quad \knowl{./knowl/xref/L6.html}{\text{\({\LARGE \vphantom{\int}}L_6\)}}\ (a=2, n=4)\\ =\amp \frac{6}{(s+9)^4} \end{align*}

18.

\(\ds \lap{t^3 e^{-2t}}\)
Answer. Answer
\(\ds \frac{6}{(s + 2)^4}\)

19.

\(\ds \lap{t\sin(3t)}\)
Answer. Answer
\(\ds -\frac{2s}{(s^2 + 9)^2}\)

20.

\(\ds \lap{e^{5t}\cos(4t)}\)
Answer. Answer
\(\ds \frac{s - 5}{(s - 5)^2 + 16}\)

21.

\(\ds \lap{e^{2t} - t^3 - \sin (5t)}\)

22.

\(\ds \lap{e^{-2t}\sin(2t) + t^2 e^{3t}} \)
Answer. Answer
\(\ds L_7\ (a=2), L_9\ (n=2, a=3)\)
\begin{align*} \lap{e^{-2t}\sin(2t) + t^2 e^{3t}} =\amp \frac{2}{(s+2)^2 + 4} + \frac{2}{(s-3)^3} \\ =\amp \frac{2}{(s+2)^2 + 4} + \frac{2}{(s-3)^3} \end{align*}

23.

\(\ds \lap{8t\cos(6t) + e^{3t}\sin(4t)}\)
Answer. Answer
\(\ds L_3\ (n=9)\)
\begin{equation*} \lap{t^9} = \frac{9!}{s^{9+1}} = \frac{9!}{s^{10}} \end{equation*}

24.

\(\ds \lap{t^2 \sin(3t)} \)
Answer. Answer
\(\ds P(s) = \frac{18\left[s^2 - 3\right]}{(s^2 + 9)^3} \)

Forward Transforming Equations.

Use the forward Laplace transform to solve the following differential equations.

25.

Solve the differential equation \(y'' + 3y' + 2y = 0\) given the initial conditions \(y(0) = 2\) and \(y'(0) = -1\text{.}\)
Answer. Answer
\(y(t) = e^{-t} - 3e^{-2t}\)

26.

Solve the differential equation \(y' + 4y = 10e^{-2t}\) with the initial condition \(y(0) = 3\text{.}\)
Answer. Answer
\(y(t) = \frac{7}{2}e^{-4t} - \frac{5}{2}e^{-2t}\)
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