All things are difficult before they are easy, Thomas Fuller.

When you have eliminated the impossible, whatever remains, however improbable, must be the truth,Sherlock Holmes.

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 first-order system of ordinary differential equations (ODEs) involves multiple dependent variables, such as x(t) and y(t), that change with respect to one independent variable, typically time t. Unlike a single ODE, a system comprises several equations that must be solved simultaneously because the variables are interdependent. The interdependence of x(t) and y(t) means that the solution of one equation depends on the solution of the other.

In such systems, each equation describes the rate of change of one dependent variable in terms of the other variables and possibly the independent variable t. These systems are foundational in modeling complex phenomena in physics, engineering, biology, and other fields where multiple quantities interact dynamically.

**A system is linear if the dependent variables and their derivatives appear to the first power and are not multiplied together**. In other words, no terms involve products or powers greater than one of the dependent variables (e.g., no x^{2}, xy, etc.).

A first-order system with two dependent variables x (t) and y(t) can be written as:

$\begin{cases} x’ = f(x, y, t) \\ y’ = g(x, y, t) \end{cases}$ where

- x’ and y’ represent the time derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ respectively.
- f( x, y, t) and g(x, y, t) are functions that describe how x and y change over time based on the current values of x, y, and possibly the independent variable t t.

In a linear system, the dependent variables x(t) and y(t) occur linearly, meaning no terms involve powers or products of x or y (e.g., no x^{2}, xy, etc.).

A linear first-order system has the form:

$\begin{cases} x’ = a(t)x + b(t)y + r_1(t) \\ y’ = c(t)x + d(t)y + r_2(t) \end{cases}$ where:

- a(t), b(t), c(t), and d(t) are functions of time t (they can be constants in simpler cases).
- r
_{1}(t) and r_{2}(t) represent external forcing functions or inputs that depend on t. - x(t
_{0}) = x_{0}and y(t_{0}) = y_{0}are the initial conditions.

**Linearity**: The dependent variables x(t) and y(t) occur linearly.**Homogeneous vs. Nonhomogeneous:**If both r_{1}(t) = 0 and r_{2}(t) = 0, the system is a homogeneous linear system. If either r_{1}(t) or r_{2}(t) is non-zero, the system is nonhomogeneous.

When r_{1}(t) = r_{2}(t) = 0, the system simplifies to:

$\begin{cases} x’ = a(t)x + b(t)y \\ y’ = c(t)x + d(t)y \end{cases}$

Homogeneous systems are fundamental because their solutions can often be used to construct solutions to nonhomogeneous systems.

To illustrate how we can apply a first-order system of ODEs, consider the following physical scenario:

Consider an egg consisting of two regions:

**Yolk**: Inner part with temperature T_{1}(t).**Egg White**: Surround part with temperature T_{2}(t).

The egg is placed in a pot of water (the external environment) with temperature T_{e}(t). Initially, the water is boiling, but it cools over time, especially when the egg is transferred to an ice bath. Refer to Figure iv for a visual representation and aid in understanding it.

We aim to model the temperature dynamics of the yolk and egg white over time using a first-order system of ODEs.

The temperature of the water is modeled as: T_{e}(t) = 100·e^{-kt} for some positive constant k, meaning the water cools over time, decays exponentially. For simplicity, we assume T_{e}(t) = 0°C when the egg is transferred to an ice bath.

Initial conditions: T_{1}(0) = 40, T_{2}(0) = 45

To model the temperature dynamics of the egg, we use Newton’s Law of Cooling, which states that the rate of change of temperature is proportional to the temperature difference between its temperature and the temperature of the surrounding environment.

Using Newton’s Law of Cooling, we can model the temperature changes as:

**Yolk Temperature Change:**The rate of temperature change of the yolk depends on the temperature difference between the egg white and the yolk: $\frac{dt_1}{dt} = a(T_2-T_1)$**Egg White Temperature Change.**The rate of temperature change of the egg white depends on:

- The temperature difference between the yolk and the egg white.
- The temperature difference between the external environment and the egg white.

$\frac{dT_2}{dt} = a(T_1-T_2) + b(T_e-T_2)$ where:

- T
_{1}(t) is the temperature of the yolk - T
_{2}(t) is the temperature of the egg white. - a and b are positive constants that describe the rates of heat transfer between the yolk and egg white, and between the egg white and the external environment (water).
- T
_{e}(t) is the external temperature (the pot of water or ice bath).

For this example, we assume a = 2 (heat transfer between the yolk and white), b = 3 (heat transfer between the egg white and the external environment). Initial conditions: T_{1}(0) = 40 (yolk) and T_{2}(0) = 45 (egg white). The external temperature T_{e}(t) = 0 when the egg is place into an ice bath.

**Rewrite the system**. Plugging in the values of a, b, and T_{e}, we get the system:

$\begin{cases} T_1’ = -aT_1 + aT_2 \\ T’_2 = aT_1 -(a+b)T_2 + bT_e \end{cases}$

$\begin{cases} T_1’ = -2T_1 + 2T_2 (i) \\ T’_2 = 2T_1 -5T_2 (ii)\end{cases}$

**Eliminate one variable**: To reduce this system to a single equation, **we can solve for one variable in terms of the other**. From the first equation (i), solve for T_{2}: $T_2 = \frac{T_1’+2T_1}{2}$.

Now substitute this into the second equation (ii. $\frac{dT_2}{dt} = 2T_1-5T_2$): $(\frac{T_1’+2T_1}{2})’ = 2T_1 -5(\frac{T_1’+2T_1}{2}) ↭ (\frac{1}{2})(T_1’+2T_1)’ = 2T_1 -5(\frac{T_1’+2T_1}{2})↭[\text{Simplify both sides}] (T_1’+2T_1)’ = 4T_1-5(T_1’+2T_1) ↭ T_1’’ + 2 T_1’’ = 4T_1 -5T_1’ -10T_1 ↭ T_1’’ +7 T_1’ + 6T_1 = 0$. This is now a second-order linear differential equation. In general, a system of two first-order differential equations, will produce a second-order differential equation by substitution. A system of three first-order differential equations, will produce a third-order differential equation, and so on.

The equation $T_1’’+7T_1’ + 6T_1 = 0$ is a standard linear homogeneous ODE. The characteristic equation associated with the ODE is: r^{2} +7r + 6 = 0. Factor this equation: r^{2} +7r + 6 = (r + 1)(r + 6) = 0. The roots or solutions of this characteristic equation are -1 and -6. Hence, the general solution for T_{1}(t) is: $T_1 = C_1e^{-t} + C_2e^{-6t}$ where C_{1} and C_{2} are constants to be determined from initial conditions.

Solve for $T_2 = \frac{T_1’+2T_1}{2} = \frac{1}{2}(T_1’+2T_1)$

So, the solution is $T_2 = \frac{1}{2}(-C_1e^{-t} + -6C_2e^{-6t} + 2C_1e^{-t} + 2C_2e^{-6t}) = \frac{1}{2}C_1e^{-t} - 2C_2e^{-6t}$

Therefore, the solution for T_{2}(t) is:

$\begin{cases} T_1 = C_1e^{-t} + C_2e^{-6t} \\ T_2 = \frac{1}{2}C_1e^{-t} - 2C_2e^{-6t} \end{cases}$

Apply the Initial conditions T_{1}(0) = 40 and T_{2}(0) = 45:

$\begin{cases} 40 = C_1 + C_2 \\ 45 = \frac{1}{2}C_1 - 2C_2\end{cases}$

Multiply equation (ii) by 2 to eliminate the fraction:

$\begin{cases} 40 = C_1 + C_2 \\ 90 = C_1 - 4C_2\end{cases}$

(ii)-(i): 50 = -5C_{2} ⇒ C_{2} = -10, C_{1} = 40 -C_{2} = 40 + 10 = 50.

The final solution for the temperatures of the yolk and egg white is:

$\begin{cases} T_1 = 50e^{-t} -10e^{-6t} \\ T_2 = 25e^{-t} +20e^{-6t} \end{cases}$

The Yolk Temperature decreases over time due to heat transfer to the egg white and the external environment. The Egg White Temperature decreases over time. It cools both by transferring heat to the yolk and by losing heat to the surrounding ice bath.

- Solving a System of First-Order Linear Ordinary Differential Equations: $\begin{cases} x’ + y’ + 2x = 0 \\ x’ + y’ -x -y = sin(t) \end{cases}$

To apply systematic methods for solving the system, it’s beneficial to rewrite the equations in a standardized manner:

$\begin{cases} x’ + 2x + y’ = 0 \\ x’ -x + y’ -y = sin(t) \end{cases}$

Let’s denote the differential operator by D, where D = $\frac{d}{dt}$. Applying this operator to functions simplifies the notation for derivatives.

Applying the operator to each term in the system: $\begin{cases} (D+ 2)(x) + D(y) = 0 \\ (D -1)x + (D -1)y = sin(t) \end{cases}$

To solve the system, we aim to eliminate one of the variables, allowing us to find an expression for the other. Here’s how we can proceed.

- Multiply Equation 1 by (D−1)
- 2: Multiply Equation 2 by −D

$\begin{cases} (D-1) ((D+ 2)(x) + D(y)) = 0 \\ -D((D -1)x + (D -1)y = sin(t)) \end{cases}$

$\begin{cases} (D-1)(D+ 2)(x) + (D-1)D(y) = 0 \\ -D(D -1)x -D(D -1)y = -D(sin(t)) \end{cases}$

Combining the Equations to Eliminate y(t): $(D^2+D-2)(x)-(D^2-D)(x) = -D(sin(t))$

x’’ +x’ -2x -x’’ + x’= -cos(t)↭[Simplifying] 2x’-2x = -cos(t).

To simplify, divide both sides by 2: $x’-x = \frac{-1}{2}cos(t)$. It is in Standard Linear First-Order ODE: x’ + P(t) = Q(t)

Solving for x(t) Using an Integrating Factor

- Identify the Integrating Factor: $μ(t) = e^{\int P(t)dt}= e^{\int -1dt} = e^{-t}$
- Multiply Both Sides by the Integrating Factor: $e^{-t}x’ -e^{-t}x = \frac{-1}{2}e^{-t}cos(t) ↭ \frac{d}{dx}(e^{-t}x) = \frac{-1}{2}e^{-t}cos(t)$
- Integrate both sides: $e^{-t}x = \int \frac{-1}{2}e^{-t}cos(t)dt + C_1$

Use the standard integral $\int e^{at}cos(bt)dt = \frac{e^{at}(acos(bt)+bsin(bt))}{a^2+b^2} + C$. Here a = -1, b = 1.

$\int e^{-t}cos(t)dt =\frac{e^{-t}(-1·cos(t)+1·sin(t))}{(-1)^2+1^2} = \frac{e^{-t}(-cos(t) +sin(t))}{2}+ C$

Finally, $e^{-t}x = \frac{-1}{2}(\frac{e^{-t}(-cos(t) +sin(t))}{2}) + C_1 = \frac{1}{4}e^{-t}(cos(t) -sin(t)) + C_1$

Solve for x(t): $x(t) = C_1e^t + \frac{1}{4}cos(t) -\frac{1}{4}sin(t)$

Compute x’(t) = $C_1e^t - \frac{1}{4}sin(t) -\frac{1}{4}cost(t)$.

Having found x(t), we can now find y(t) using one of the original equations. Let’s use Equation 1: x’ + y’ + 2x = 0, we can rewrite it as y’ = -x’ -2x = $-(C_1e^t -\frac{1}{4}cos(t) -\frac{1}{4}sin(t)) -2(C_1e^t - \frac{1}{4}sin(t) +\frac{1}{4}cos(t)) ↭ y’ = -3C_1e^t+\frac{3}{4}sin(t)-\frac{1}{4}cos(t)$

$\int y’dt = \int(-3C_1e^t+\frac{3}{4}sin(t)-\frac{1}{4}cos(t)) ↭ y = -3C_1e^t-\frac{3}{4}cos(t)-\frac{1}{4}sin(t) + C_2$

The general solution to the system of ODEs is:

$\begin{cases} x(t) = C_1e^t + \frac{1}{4}cos(t) -\frac{1}{4}sin(t) \\ y(t) = -3C_1e^t-\frac{3}{4}cos(t)-\frac{1}{4}sin(t) + C_2 \end{cases}$ where C_{1} and C_{2} are constants determined by the initial conditions of the system.

An autonomous system is a system of differential equations where the independent variable, typically denoted by t (often representing time), does not explicitly appear on the right-hand side of the equations. In other words, the rate of change of the dependent variables depends only on the current state of the system, not directly on time.

A system of two first-order autonomous ODEs can be written as::

$\begin{cases} x’ = f(x, y) \\ y’ = g(x, y) \end{cases}$ where

- x = x(t) and y = y(t) are the dependent variables, functions of the independent variable t.
- x’ = $\frac{dx}{dt}$ and y’ = $\frac{dy}{dt}$ represent the time derivatives of the function x(t) and y(t), respectively.
- The functions f(x, y) and g(x, y) determine how x(t) and y(t) evolve over time based on their current values x and y.

**Key Characteristics of Autonomous Systems:**

**Time-Invariant Behavior**: Since t does not explicitly appear in f and g, the system’s behavior is the same at any time t, given the same state (x, y).**Dependence on State**: The evolution of the system depends solely on the current state (x, y), making the system’s behavior consistent across time.

The solutions to the autonomous system are functions x(t) and y(t), which together form a parametrized curve in the plane ℝ^{2}, meaning that for each value of t, there is a corresponding point (x(t), y(t)) in the plane. As t varies, the point
(x(t), y(t)) traces out a curve.

Each solution to the system corresponds to a different initial condition x(t_{0}) = x_{0}, y_{0} = y_{0} where t_{0} is the initial time (often t_{0} = 0) and (x_{0}, y_{0}) is the initial point in the plane (Refer to Figure v for a visual representation and aid in understanding it)

**Implications**

**Unique Solutions**. The curve depends on the initial conditions, which specify where the curve starts. For example, if x(0) = x_{0}and y(0) = y_{0}, the solution starts at the point (x_{0}, y_{0}) at t = 0. In other words, for a given set of initial conditions, there is a unique solution curve, provided that f and g satisfy certain conditions (e.g., Lipschitz continuity).**Family of Solutions**: Varying the initial conditions generates a family of solution curves, each representing a different trajectory in the plane.

The system of ODEs: $\begin{cases} x' = f(x, y) \\\ y' = g(x, y) \end{cases}$ defines a velocity field in the plane. At each point (x, y), the system assigns a vector $\vec{v}(x,y)=(f(x,y),g(x,y))$.

Explanation:

**Direction and Magnitude**: The vector $\vec{v}(x,y)=(f(x,y),g(x,y))$ indicates the direction and speed (magnitude) at which the system moves when it is at the point (x, y).**Tangent Vectors:**The vector $\vec{v}(x,y)=(f(x,y),g(x,y))$ is tangent (the velocity) to the solution curve passing through (x, y) at time t.- The plane ℝ
^{2}where x and y are plotted is called the**phase plane**. - The system of ODEs x′= f(x,y) and y′ = g(x,y) defines a velocity field. At each point (x,y) in the plane, the system assigns a vector (f(x,y), g(x,y)), which represents the velocity of the solution curve passing through that point. It is a vector that is always tangent to the solution curve (it describes the direction and speed at which the solution curve moves at each point).
**Solution Curves**: The solution curves are tangent to the velocity vectors at every point along their paths.At any point (x(t), y(t)) along a solution curve. The derivative (x’(t), y’(t)) represents the velocity of the system at time t. From the system equations, x’(t) = f(x(t), y(t)) and y’(t) = g(x(t), y(t)). Therefore, (x’(t), y’(t)) = (f(x(t), y(t)), g(x(t), y(t))).

Therefore, a system of two first order autonomous ODE is equal to a velocity field, and a solution is a parametrized curve that follows the velocity field, always moving in the direction specified by the field at each point.

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].

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn, and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
- MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
- Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.