ODE: Transforms of Elementary Functions

Mastering the computation of Laplace transforms for common elementary functions is the indispensable first step toward harnessing the full power of the Laplace method for solving differential equations. This process translates time-domain functions into a complex frequency-domain representation, turning challenging calculus operations like differentiation into simpler algebra. Your fluency with these foundational transforms directly determines your efficiency and accuracy when tackling the initial value problems prevalent in engineering dynamics, circuit analysis, and control systems.

Defining the Laplace Transform and Its Core Property

The Laplace transform is an integral transform that converts a function of a real variable $t$ (typically representing time) into a function of a complex variable $s$ (representing complex frequency). For a function $f(t)$ defined for $t\ge 0$, its Laplace transform $F(s)$ is defined by the improper integral:

$$\mathcal{L}\{f(t)\}=F(s)={\int }_0^{\infty }{e}^{-st}f(t)dt$$

The transform's immense utility springs from its operational property concerning derivatives. It transforms the operation of differentiation with respect to $t$ into multiplication by $s$ in the frequency domain, subject to initial conditions. Specifically, for a function $f(t)$ with a well-behaved derivative:

$$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)$$

This property is what allows us to convert a differential equation in $t$ into an algebraic equation in $s$, which we can solve for $F(s)$ before transforming back to the time-domain solution. Before we can use this tool, however, we must build a reliable "dictionary" of basic transform pairs.

Core Transforms of Polynomial and Exponential Functions

We begin with the most elementary functions, which form building blocks for more complex ones. The transform of the constant function 1, or $t^0$, is derived directly from the definition. For $f(t)=1$:

$$\mathcal{L}\{1\}={\int }_0^{\infty }{e}^{-st}\cdot 1dt={\left[-\frac{e^{-st}}{s}\right]}_0^{\infty }=\frac{1}{s},\text{for }\text{Re}(s)>0$$

From this, we can derive the transform for $t^n$, where $n$ is a positive integer. Using integration by parts or the related Gamma function formula, the general result is:

$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}},\text{for }\text{Re}(s)>0$$

For example, $\mathcal{L}\{t^2\}=\frac{2!}{s^3}=\frac{2}{s^3}$.

Next, the exponential function $e^{at}$, where $a$ is any real or complex constant, is fundamental. Its transform is computed as:

$$\mathcal{L}\{e^{at}\}={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{-(s-a)t}dt=\frac{1}{s-a},\text{for }\text{Re}(s)>\text{Re}(a)$$

This result is a cornerstone. Notice that for $a=0$, it correctly reduces to $\mathcal{L}\{1\}=1/s$.

Transforms of Sine, Cosine, and Their Damped Variants

The transforms of sinusoidal functions $sin(bt)$ and $cos(bt)$ can be derived using Euler's formula, $e^{i\theta }=\cos \theta +i\sin \theta$, or by applying integration by parts twice. The standard results are:

$$\mathcal{L}\{\sin (bt)\}=\frac{b}{s^2+b^2}$$

$$\mathcal{L}\{\cos (bt)\}=\frac{s}{s^2+b^2}$$

Both are valid for $\text{Re}(s)>0$. A powerful application of the exponential transform rule is finding the transforms of damped sinusoids. This uses the frequency shift property (or the $s$-shifting theorem), which states: $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$, where $F(s)=\mathcal{L}\{f(t)\}$.

Applying this to our sine and cosine transforms gives us two immensely useful pairs for modeling oscillatory systems with decay or growth:

$$\mathcal{L}\{e^{at}\sin (bt)\}=\frac{b}{(s-a{)}^2+b^2}$$

$$\mathcal{L}\{e^{at}\cos (bt)\}=\frac{s-a}{(s-a{)}^2+b^2}$$

You can see the pattern: wherever $s$ appears in the undamped transform formula, it is replaced by $(s-a)$. For instance, to compute $\mathcal{L}\{e^{-3t}\cos (4t)\}$, you identify $a=-3$ and $b=4$, yielding $\frac{s-(-3)}{(s+3{)}^2+16}=\frac{s+3}{(s+3{)}^2+16}$.

Introducing the Unit Step Function

Many engineering systems involve inputs that are switched on at a specific time. This is modeled using the unit step function $u_c(t)$ (also called the Heaviside function), defined as:

$$u_c(t)=\{\begin{array}{ll}0& \text{if }t<c\\ 1& \text{if }t\ge c\end{array}$$

The step function "turns on" at $t=c$. Its Laplace transform is crucial for solving differential equations with discontinuous forcing functions:

$$\mathcal{L}\{u_c(t)\}=\frac{e^{-cs}}{s},\text{for }s>0$$

More importantly, the related time shift property governs how to transform a function that is multiplied by a step function, effectively delaying its start time: $\mathcal{L}\{u_c(t)f(t-c)\}=e^{-cs}F(s)$. For example, a pulse starting at $t=2$ defined as $u_2(t)\cdot (t-2{)}^2$ transforms to $e^{-2s}\cdot \mathcal{L}\{t^2\}=e^{-2s}\cdot \frac{2}{s^3}$.

Building Fluency Through Properties and Practice

True fluency comes from combining your knowledge of these basic transforms with the operational properties of the Laplace transform. Beyond the frequency and time shifts already discussed, the linearity property is your daily workhorse:

$$\mathcal{L}\{af(t)+bg(t)\}=aF(s)+bG(s)$$

This allows you to transform complex functions term-by-term. Consider a function like $f(t)=3-2e^{-4t}+5\sin (2t)$. You compute its transform by applying linearity to the transforms of its components:

$$\mathcal{L}\{f(t)\}=3\cdot \frac{1}{s}-2\cdot \frac{1}{s+4}+5\cdot \frac{2}{s^2+4}=\frac{3}{s}-\frac{2}{s+4}+\frac{10}{s^2+4}$$

Systematic practice in breaking down functions into these standard forms is the key to building the speed and accuracy required to use the Laplace transform as an effective problem-solving tool for differential equations.

Common Pitfalls

1. Misapplying the Frequency Shift Property: A frequent error is to incorrectly apply $e^{at}f(t)\to F(s-a)$. You must ensure that the exponential multiplies the entire function $f(t)$. For $\mathcal{L}\{e^{at}\cdot t\sin (bt)\}$, you first need the transform of $t\sin (bt)$ (found via the derivative property) before applying the $s$-shift.
2. Ignoring the Region of Convergence (ROC): While often implied in introductory engineering contexts, forgetting that transforms like $\mathcal{L}\{e^{at}\}=1/(s-a)$ are only valid for $\text{Re}(s)>\text{Re}(a)$ can lead to errors in more advanced applications involving inverse transforms and system stability analysis.
3. Incorrect Handling of Step Function Arguments: The time-shift property $\mathcal{L}\{u_c(t)f(t-c)\}=e^{-cs}F(s)$ requires the function to be written as $f(t-c)$. A mistake is to see $u_c(t)\cdot t$ and try to apply the property directly. You must rewrite it as $u_c(t)\cdot ((t-c)+c)$ to use the theorem correctly on the $(t-c)$ part.
4. Algebraic Errors in Partial Fraction Decomposition: The final step in solving an ODE often involves inverting a rational function $F(s)$. Errors in performing partial fraction expansion—such as mis-calculating residues or mis-handling repeated or complex roots—are the most common source of incorrect final answers, even with a correct transformed equation.

Summary

• The Laplace transform $\mathcal{L}\{f(t)\}={\int }_0^{\infty }{e}^{-st}f(t)dt$ converts time-domain functions into the complex frequency-domain, turning differential operations into algebraic ones.
• A core set of elementary transform pairs must be memorized: $t^n\to n!/s^{n+1}$, $e^{at}\to 1/(s-a)$, $\sin (bt)\to b/(s^2+b^2)$, $\cos (bt)\to s/(s^2+b^2)$.
• The frequency shift property, $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$, generates transforms of damped sinusoids: $e^{at}\sin (bt)\to b/((s-a{)}^2+b^2)$ and $e^{at}\cos (bt)\to (s-a)/((s-a{)}^2+b^2)$.
• The unit step function $u_c(t)$ models switching events, with $\mathcal{L}\{u_c(t)\}=e^{-cs}/s$, and the time-shift property $\mathcal{L}\{u_c(t)f(t-c)\}=e^{-cs}F(s)$ handles delayed inputs.
• Achieving computational fluency requires consistent practice using linearity to combine basic transforms and properties, forming the essential foundation for solving differential equations via the Laplace method.