Ordinary Differential Equations

To solve initial value problems, we need an idea of an inverse Laplace transform.

Definition 3.2.1. Inverse Laplace transform.

Suppose \(f\) and \(f'\) are piecewise continuous on \(t \geq 0\) and let \(f\) be of exponential order \(c\) for \(t \geq 0\text{.}\) Since \(F\) is the Laplace transform of \(f\text{,}\) the corresponding inverse Laplace transform of \(F\) is

\begin{equation} f(t) = \mathscr{L}^{-1}\{F\}(t) = \dfrac{1}{2\pi i} \displaystyle\lim_{R\rightarrow\infty} \displaystyle\int_{\gamma-iR}^{\gamma+iR} e^{st}F(s)\mathrm{d}s,\tag{3.2.1} \end{equation}

where \(\gamma>c\text{.}\)

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The integral (3.2.1) is called a Bromwich contour integral and understanding it requires an understanding of residues in complex analysis. However, we are still able to use and apply it to solve differential equations without using that deeper theory. For our purposes, we will have a function \(F\) and ask the question "what function is this \(F\) the Laplace transform of?". To answer the question, we will use a table instead of trying to compute the integral (3.2.1).

Theorem 3.2.2. Uniqueness of inverse Laplace transforms.

Suppose that the function \(f\) and \(g\) satisfy the assumptiuons of Theorem 3.1.10 so we know their respective Laplace transforms \(F\) and \(G\) exist. If \(F(s)=G(s)\) for all \(s> c\text{,}\) then \(f(t)=g(t)\) whenever both \(f\) and \(g\) are continuous.

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Theorem 3.2.3. Linear properties of the inverse Laplace transform.

Suppose that the inverse Laplace transforms of \(F\) and \(G\) are, respectively, \(f\) and \(g\) and let \(\alpha \in \mathbb{R}\text{.}\) Then,

1. \(\displaystyle \mathcal{L}^{-1}\{\alpha F\}(t) = \alpha \mathcal{L}^{-1}\{F\}(t),\)
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2. \(\displaystyle \mathcal{L}^{-1}\{F \pm \}(t) = \mathcal{L}^{-1}\{F\}(t) \pm \mathcal{L}^{-1}\{G\}(t).\)
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Proof.

later!

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Some inverse Laplace transforms you need to take to solve differential equations don’t always appear exactly as they are in the table. It sometimes requires careful algebraic manipulation, see Section 3.7 to turn a given expression into something we can use the table in Section 3.8 to compute.

Example 3.2.4.

Find the inverse Laplace transform of the following functions.

1. \(\displaystyle F(s)=\dfrac{1}{s^4} - \dfrac{1}{2s-1} + \dfrac{2}{s^2+9}\)
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2. \(\displaystyle G(s)=\dfrac{s}{s^2+2s-3}\)
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3. \(\displaystyle H(s)=\dfrac{2s+1}{s(s^2+4)}\)
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Theorem 3.2.5. Laplace transforms of derivatives.

Suppose \(f\) is piecewise continuous and smooth on \(t\geq 0\) with exponential order \(c\text{.}\) Let \(F\) be the Laplace transform of \(f\text{.}\) Then,

\begin{equation*} \mathcal{L}\{f'\}(s)=sF(s)-f(0). \end{equation*}

Furthermore, if \(f\) and \(f'\) are piecewise continuous and smooth on \(t\geq 0\) and are of exponential order \(c\text{,}\) then

\begin{equation*} \mathcal{L}\{f''\}(s)=s^2F(s)-sf(0)-f'(0). \end{equation*}

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Proof.

later!

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Corollary 3.2.6. Laplace transform of higher derivatives.

Suppose that \(f, f', \ldots, f^{(n-1)}\) are piecewise continuous and smooth for \(t \geq 0\) with exponential order \(c\text{.}\) Then,

\begin{equation*} \mathcal{L}\{f^{(n)}\}(s)=s^nF(s)-s^{n-1}f(0) - s^{n-2}f'(0)-\ldots - sf^{(n-2)}(0)-f^{(n-1)}(0). \end{equation*}

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Example 3.2.7.

The transfer function \(H\) of a linear system is defined as the ratio of the Laplace transform of the output function \(y\) to the Laplace transform of the input function \(g\) under the assumption that all initial conditions are zero. That is, \(H(s)=\dfrac{Y(s)}{G(s)}\text{.}\) Find the transfer function associated with the differential equaiton \(ay''+by'+cy=g(t)\) for \(t\geq 0\text{,}\) where \(y\) and \(g\) are both of exponential order.

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Algorithm 3.2.8. Solving IVPs with the Laplace transform.

There are four basic steps to solving any initial value problem using the Laplace transform.

1. Take the Laplace transform of both sides to get an algebraic expression in the frequency variable \(s\text{.}\)
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2. Solve the algebraic expression for \(Y(s)\text{.}\)
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3. Use our algebraic techniques to simplify \(Y(s)\) into more familiar terms.
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4. Take the inverse Laplace transform of both sides to find the solution \(y\text{.}\)
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Example 3.2.9.

Use the Laplace transform to solve the initial value problem

\begin{equation*} y'+3y=4e^t, \quad y(0)=1. \end{equation*}

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