I have yet to see any problem, however complicated, which, when you looked at it in the right way, did not become still more complicated, Poul Anderson.

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

A power series is a type of infinite series where each term is a power of a variable, typically written in the form: $\sum_{n=0}^\infty a_nx^n = A(x)$.

In this notation, the a_{n}’s represent the coefficients of the power series, and A(x) is the resulting function that the series converges to.

We can rewrite the series by treating the coefficients a_{n} as the values of a discrete function, say a(n). This gives us a new way to represent the power series as a discrete sum: $\sum_{n=0}^\infty a(n)x^n = A(x)$.

Here, a(n) is a function defined on the discrete set of non-negative integers. Now, the key idea is that we can relate this discrete function a(n) to a real, continuous function A(x), through known series expansions, for example:

- If a(n) = 1, then $A(x) = \frac{1}{1-x}$, for |x| < 1, $a(n) = 1 \leadsto A(x) = \frac{1}{1-x}$
- If a(n) =
^{1}⁄_{n!}, then A(x) = e^{x}, a(n) = $\frac{1}{n!} \leadsto A(x) = e^x$

These examples illustrate that discrete functions a(n) can correspond to well-known continuous functions A(x).

The Laplace Transform is based on a similar concept but extends this relationship to integrals instead of sums. The continuous analog of the power series is an integral involving a continuous variable t, rather than a discrete sum over n. The goal is to construct an integral of the form: $\int_{0}^{∞} a(t)x^tdt = A(x)$ where a(t) is a continuous function of t.

However, integrals involving terms like x^{t} can be challenging to compute directly because they are not always convenient to work with. To simplify this expression, we make a transformation that will result in a more manageable form. Let’s proceed step-by-step.

We begin by rewriting x^{t} in terms of the exponential function: x = e^{ln(x)} ⇒ $x^t = (e^{ln(x)})^t= [\text{Exponents Laws}] e^{t·ln(x)}$.

To make the resulting expression easier to handle, we now make a substitution for x. Assume that 0 < x <1, which implies that ln(x) < 0. Let: s = -ln(x). Since ln(x) is negative for 0 < x <1, this substitution guarantees that s > 0. Now, we have: $x^t = e^{t·ln(x)} = e^{-st}$ where the new variable s replaces the logarithm.

For clarity (aka cosmetic purposes or consistency with standard notation), we rename the function a(t) as f(t), a more common notation when discussing transforms. Thus, our integral now becomes: $\int_{0}^{∞} f(t)e^{-st}dt = F(s)$. This is the Laplace Transform. In this transformation, the input is a function of t (i.e., f(t)), and the output is a function of s (i.e., F(s)).

We denote the Laplace Transform of f(t) as: $\mathcal{L}(f(t)) = F(s)$ or in an alternative notation, simply: $f(t) \leadsto F(s)$

It’s important to recognize that **a transform takes a function of one variable** (in this case, t) **and transforms it into a function of a different variable** (in this case, s). This is different from an operator, which typically maps a function of a given variable into another function of the same variable. For example, a differentiation operator acts on f(t) and produces another function of t, such as f’(t).

In contrast, the Laplace Transform maps a function of t into a new function of a different variable s.

One of the most useful properties of the Laplace Transform is its linearity. If we have two functions f(t) and g(t), then: $\mathcal{L}(f+g) = \mathcal{L}(f)+\mathcal{L}(g), \mathcal{L}(af+bg) = a\mathcal{L}(f)+b\mathcal{L}(g)$ for any constants a and b. In other words, the Laplace Transform of a sum is the sum of the individual Laplace Transforms.

$\mathcal{L}(af+bg)(t) = \mathcal{L}(af(t) +bg(t)) = \int_{0}^{∞} e^{-st}[af(t) + bg(t)]dt = a\int_{0}^{∞} e^{-st}f(t)dt + b\int_{0}^{∞} e^{-st}g(t)dt = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$

The Laplace Transform is a powerful tool in mathematics and engineering, particularly for solving differential equations.

- Laplace Transform of the Constant Function 1

By definition, the Laplace Transform is given by: $\mathcal{L}(1) = \int_{0}^{∞} 1e^{-st}dt = \int_{0}^{∞} e^{-st}dt =[\text{To evaluate this improper integral, we consider the limit as the upper bound approaches infinity:}] \lim_{R \to ∞}\int_{0}^{R} e^{-st}dt$

Compute the definite integral over the finite interval [0, R]: $\int_{0}^{R} e^{-st}dt = \frac{e^{-st}}{-s}\bigg|_{0}^{R} = \frac{e^{-sR}-1}{-s}$

$\int_{0}^{∞} e^{-st}dt = \lim_{R \to ∞}\int_{0}^{R} e^{-st}dt = \lim_{R \to ∞} \frac{e^{-sR}-1}{-s} =[\text{Since s > 0, as R → ∞, } e^{-sR}→ 0] \frac{1}{s}$, this is true only for s > 0.

$\mathcal{L}(1) = \frac{1}{s}$ or $1 \leadsto \frac{1}{s}$. This means that the function f(t) = 1 maps to F(s) = ^{1}⁄_{s} under the Laplace Transform.

- Laplace Transform of the exponential function e
^{at}

First, we are going to calculate the Laplace Transform of e^{at}f(t), where f(t) is a function whose Laplace Transform is F(s).

By definition, the Laplace Transform of e^{at}f(t) is: $\mathcal{L}(e^{at}f(t)) = \int_{0}^{∞} e^{at}f(t)e^{-st}dt =[\text{Simplify the exponentials:}] \int_{0}^{∞} e^{-(s-a)t}f(t)dt = F(s-a)$ provided that s-a > 0, which implies s > a.

$e^{at}f(t) \leadsto F(s-a)$ for s > a, assuming $\int_{0}^{∞} f(t)e^{-st}dt = F(s)$. This result is known as the **exponential shift formula**. This formula shows that multiplying f(t) by an exponential term e^{at} shift the Laplace Transform F(s) to F(s -a).

Consider the specific case where f(t) = 1. Then, $e^{at}·1 = e^{at}\leadsto F(s−a) = \frac{1}{s-a}$ for s > a, since $1 \leadsto \frac{1}{s}$.

- Laplace Transform of u(t -a)

The unit step function u(t) is defined as: u(t) = $\begin{cases} 0, t < 0 \\ 1, t ≥ 0 \end{cases}$

The shifted unit step function u(t−a) is defined as: u(t-a) = $\begin{cases} 0, t < a \\ 1, t ≥ a \end{cases}$

$\mathcal{L}(u(t-a)) = \int_{0}^{∞} e^{-st}u(t-a)dt =[\text{Since u(t−a) = 0 for t < a, we can change the limits of integration}] \int_{a}^{∞} e^{-st}·1dt = \frac{e^{-st}}{-s}\bigg|_{a}^{∞} = 0-\frac{e^{-sa}}{-s} = \frac{e^{-sa}}{s}$

- Laplace Transform of Window Function

The window function $\Pi_{a, b}(t)$, also known as the rectangular function or boxcar function is defined as: $\Pi_{a, b}(t) = \begin{cases} 1, &a < t < b \\ 0, &otherwise \end{cases} =[\text{It can be expressed as the difference of two unit step functions}] u(t-a)-u(t-b)$ where u(t) is the unit step function (also known as the Heaviside step function).

$\mathcal{L}(\Pi_{a, b}(t)) = \mathcal{L}(u(t-a)-u(t-b)) =[\text{Using the linearity property of the Laplace transform}] \mathcal{L}(u(t-a)) -\mathcal{L}(u(t-b)) = \frac{e^{-sa}}{s}-\frac{e^{-sb}}{s} = \frac{e^{-sa}-e^{-sb}}{s}$. This result is valid for $s > 0$, which ensures the convergence of the Laplace transform.

- Laplace Transform of Dirac Delta function

The Dirac delta function δ(t) is a distribution that is defined as: δ(t) = $\begin{cases} 0, t ≠ a \\ ∞, t = 0 \end{cases}$

However, it is not a function in the traditional sense but rather a distribution that satisfies the following properties:

**Normalization**. $\int_{-∞}^{∞} δ(t-a)dt = 1. $This means that the total area under the delta function is 1.**Sifting Property**: $\int_{-∞}^{∞} δ(t-a)·f(t)dt = f(a)$

To find the Laplace transform of δ(t−a)δ(t−a), we compute: $\mathcal{L}(δ(t-a)) = \int_{0}^{∞} e^{-st}δ(t-a)dt =[\text{Since the delta function is zero everywhere except at t=a, we can also express this integral over the full range:}] \int_{-∞}^{∞} e^{-st}δ(t-a)dt =[\text{Applying the Sifting Property}] e^{-st}\bigg|_{a} = e^{-sa}$

- $\mathcal{L}(f(t-a)u(t-a)), \mathcal{L}(f(t)u(t-a))$

$\mathcal{L}(f(t-a)u(t-a)) = \int_{0}^{∞} e^{-st}f(t-a)u(t-a)dt =[\text{The integral starts from a because u(t-a) = 0 for t < a.}] \int_{a}^{∞} e^{-st}f(t-a)dt = $

Substitution of variables v = t -a, dv = dt, t = a ⇒v = 0, t = ∞ ⇒v = ∞

$= \int_{0}^{∞} e^{-s(v+a)}f(v)dv = \int_{0}^{∞} e^{-sv}e^{-sa}f(v)dv = e^{-sa}\int_{0}^{∞} e^{-sv}f(v)dv = e^{-sa}\mathcal{L}(f(t)) = e^{-sa}F(s)$ where F(s) is the Laplace transform of f(t).

$\mathcal{L}(f(t)u(t-a)) =$[The trick here is to let g(t-a) = f(t), which allows us to use the result from the first part.] =$\mathcal{L}(g(t-a)u(t-a)) = e^{-sa}\mathcal{L}(g(t))$= [g(t) = g((t+a)-a) = f(t+a)] $e^{-sa}\mathcal{L}(f(t+a))$

- $\mathcal{L}(f(t-2)^2u(t-2)), \mathcal{L}(t^2u(t-2))$

Recall the previous result $\mathcal{L}(f(t-a)u(t-a)) = e^{-sa}F(s)$ where F(s) is the Laplace transform of f(t).

$\mathcal{L}(f(t-2)^2u(t-2)) = e^{-s·2}\mathcal{L}(t^2) =[\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}] e^{-2s}\frac{2}{s^3} = \frac{2e^{-2s}}{s^3}$

$\mathcal{L}(t^2u(t-2))$ = [Recall the previous result $\mathcal{L}(f(t)u(t-a)) = e^{-sa}\mathcal{L}(f(t+a))$] $e^{-s·2}\mathcal{L}((t+2)^2) = e^{-2s}\mathcal{L}(t^2+4t+4) =[\text{Using linearity and } \mathcal{L}(t^n) = \frac{n!}{s^{n+1}}] e^{-2s}(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s})$

- Laplace Transform of Complex exponential functions.

We can extend the exponential shift formula to complex exponents.

Let b ∈ ℝ, consider $e^{(a+bi)t}, \mathcal{L}(e^{(a+bi)t}) = \int_{0}^{∞} e^{(a+bi)t}e^{-st}dt =[\text{Combine the exponents:}] \int_{0}^{∞} e^{-(s-a-ib)t}dt$

The integrand is now in the form e^{-ct} where c = s -a -ib. Evaluate the integral:

$\int_{0}^{∞} e^{-ct}dt = -\frac{1}{c}e^{-ct}\bigg|_{0}^{∞}$ =

$-\frac{1}{s -a -ib}e^{-(s -a -ib)t}\bigg|_{0}^{∞}$

Apply the limits: As t → ∞, $e^{-(s -a -ib)t} = e^{-(s-a)t} \cdot e^{ibt} → 0$ 💡 assuming s > a. t = 0, $e^{-(s -a -ib)·0} = 1.$

💡 The $e^{ibt}$ part is just oscillating between -1 and 1 (it’s cos(bt) + i·sin(bt)), so it doesn’t affect whether the overall expression approaches zero or not. The key part is $e^{-(s-a)t}$. For this to approach zero as t approaches infinity, we need the exponent to be negative, i.e., -(s-a) < 0 ↭ s > a

$\mathcal{L}(e^{(a+bi)t}) = -\frac{1}{s -a -ib}·0 +\frac{1}{s -a -ib}·1 = \frac{1}{s -a -ib}$

$\mathcal{L}(e^{(a+bi)t}) = \frac{1}{s-(a+ib)}$ valid for s > a, or alternatively, $e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}$.

Thus, the complex exponential function e^{(a + bi)t} transform into $\frac{1}{s-(a+bi)}$, demonstrating a direct application of the exponential shift formula with a complex shift.

- Laplace Transform of cos(at) and sin(at)

Euler’s formula states that: $e^{iat} = cos(at) + isin(at), e^{-iat} = cos(at) - isin(at)$

From these, we can express the trigonometric functions in terms of exponentials: $cos(at) = \frac{e^{iat}+e^{-iat}}{2}, sin(at) = \frac{e^{iat}-e^{-iat}}{2i}$

$cos(at) =[\text{Euler’s formula}] \frac{e^{iat}+e^{-iat}}{2}, \mathcal{L}(cos(at)) =[\text{By linearity of the Laplace Transform}, e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}] \frac{1}{2}(\frac{1}{s-ia}+\frac{1}{s+ia}) =$

If we have a complex expression and replacing i with -i doesn’t change the result, it means the expression is real, that’s the case with the previous formula.

$\frac{1}{2}·\frac{(s+ia)+(s-ia)}{(s-ia)(s+ia)} = \frac{1}{2}\frac{s+ia+s-ia}{s^2+a^2} = \frac{1}{2}\frac{2s}{s^2+a^2} = \frac{s}{s^2+a^2}, cos(at) \leadsto \frac{s}{s^2+a^2}$ where s > 0.

$sin(at) =[\text{Euler’s formula}] \frac{e^{iat}-e^{-iat}}{2i}, \mathcal{L}(sin(at)) =[\text{By linearity of the Laplace Transform}, e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}] \frac{1}{2i}(\frac{1}{s-ia}-\frac{1}{s+ia}) = \frac{1}{2i}\frac{(s+ia)-(s-ia)}{s^2+a^2} = \frac{1}{2i}\frac{2ia}{s^2+a^2} = \frac{a}{s^2+a^2}$. Thus, the Laplace Transform of sin(at) is: $sin(at) \leadsto \frac{a}{s^2+a^2}$. This result is also valid for s > 0.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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