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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 $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

In mathematics, a linear ordinary differential equation (ODE) is an equation that involves an unknown function and its derivatives. This property of linearity makes linear ODEs fundamental in modeling various physical systems, including:

- Fluid Dynamics: Describing the flow of fluids under various conditions.
- Electrical Circuits: Analyzing the behavior of circuits involving resistors, capacitors, and inductors.
- Mechanical Vibrations: Modeling the motion of oscillating systems, such as springs and pendulums.
- Radioactive Decay: Describing the rate of decay of radioactive substances.

The linearity of these equations allows for more tractable analytical solutions, often serving as approximations to more complex nonlinear systems. This makes them essential tools in both theoretical and applied mathematics.

The linearity of these ODEs means that the unknown function and its derivatives appear to the first power and are not multiplied together. In this article, we will explore different forms of linear ODEs, their applications, and their solutions, with a focus on understanding their behavior in real-world contexts.

First-order linear ODEs involve the first derivative of the unknown function. They can be expressed in several forms:

**Proportional Input Form**: y’ + ky = kq_{e}(t), where y’ is the derivative of y, k is a constant, and q_{e}(t) represents an external input function, scaled by a constant k.**Constant Coefficient Form**: y’ + ky = q(t) where k > 0 is a constant coefficient and q(t) represent an external (forcing) input. This is a slightly more general form where the forcing term doesn’t need to be a multiple of k.**Standard Linear Form**: y’ + p(t)y = q(t), where y is the unknown function of time, p(t) is a function of time that can vary, and q(t) is still an external input. This is the most general form of a first-order linear ODE, encompassing the previous two cases.

Each of these forms arises in different real-world scenarios, which we will break down further.

Consider a tank containing a solution of salt and water. Saltwater flows into and out of the tank at a constant rate. The goal is to model the concentration of salt in the tank over time. Let:

- r represents the rate of inflow and outflow (volume per unit time).
- V is the volume of the solution (water) in the tank (assumed constant).
- x(t) is the amount of salt in the tank at time t (mass).
- C
_{e}(t) is the concentration of incoming salt as a function of time (mass per unit volume, Refer to Figure E for a visual representation and aid in understanding it).

The tank is well-mixed, so the concentration of salt throughout the tank is uniform. The volume V remains constant because the inflow and outflow rates are equal.

The rate of change of the amount of salt in the tank is determined by the difference between the rate of salt inflow and the rate of salt outflow. This gives us the following differential equation:

$\frac{dx}{dt} = \text{Rate of Salt Inflow − Rate of Salt Outflow} = r·C_e(t) - r·\frac{x}{V}$

This equation states that the rate of change of salt in the tank $\frac{dx}{dt}$ is equal to the inflow rate of salt r·C_{e}(t) minus the outflow rate $r·\frac{x}{V}$, where C(t) = ^{x}⁄_{V} is the concentration of salt in the tank at time t.

$\frac{dx}{dt} = r·C_e(t) - r·\frac{x}{V}⇒[\text{Rearranging terms gives:}] \frac{dx}{dt} + \frac{r·x}{V} = r·C_e(t)$.

It’s often useful to work with concentration C(t) rather than the total amount x(t). C(t) = ^{x}⁄_{V}, leading to: $\frac{dx}{dt} + rC = r·C_e(t)$ or [x = V·C(t)]: $V\frac{dC}{dt} + rC = r·C_e(t)$

This equation can be rewritten in standard form: $\frac{dC}{dt} + \frac{r}{V}C = \frac{r}{V}C_e(t)$. Let k = ^{r}⁄_{V} where k represents the fractional flow rate (flow rate relative to the volume), indicating how quickly the tank’s contents are replaced.

The differential equation becomes: $\frac{dC}{dt} + kC = kC_e(t)$. This is a first-order linear ODE in terms of C(t), the concentration of salt in the tank.

- Large k: the flow rate is high or the volume is small, so the concentration of salt in the tank rapidly matches or adjusts to the incoming concentration. The system responds swiftly to changes in C
_{e}(t). - Small k: the flow rate is low or the volume is large, the system has a delayed response to changes in C
_{e}(t). It takes longer for the concentration in the tank C(t) to catch up with the incoming concentration C_{e}(t). - If C
_{e}(t) is a sinusoidal function (e.g., C_{e}(t) = cos(wt)), the concentration C(t) will also oscillate but with a phase lag Φ and reduced amplitude depending on k.

$\frac{dC}{dt} + kC = kC_e(t)$

**Compute the Integrating Factor μ(t)**: $μ(t) = e^{\int kdt} = e^{kt}$**Multiply both sides by μ(t)**: $e^{kt}\frac{dC}{dt} + ke^{kt}C = ke^{kt}C_e(t)$. The left-hand side of the equation simplifies to: $\frac{d}{dt}(e^{kt}C)$**Integrate Both Sides**: $\int \frac{d}{dt}(e^{kt}C)dt = \int ke^{kt}C_e(t)dt ↭ e^{kt}C = k\int e^{kt}C_e(t)dt + C_1$ where C_{1}is the constant of integration.**Solve for C(t)**: $C(t) = e^{-kt}[k\int e^{kt}C_e(t)dt + C_1]$

Suppose C_{e}(t) = C_{0} a constant.

- Integrate: $\int e^{kt}C_e(t)dt = C_0\int e^{kt}dt = \frac{C_0e^{kt}}{k} + C_2$
- Substitute back: $C(t) = e^{-kt}[k\int e^{kt}C_e(t)dt + C_1] = e^{-kt}[k(\frac{C_0e^{kt}}{k}) + kC_2 + C_1] = C_0 + kC_2e^{-kt} + C_1e^{-kt} = C_0 + (kC_2+C_1)e^{-kt} = C_0 + C’e^{-kt}$ where $C’ = kC_2+C_1$
- Suppose we have an initial condition C(0) = C
_{initial}: $C_{initial} = C_0 + C’e^{-0·t} = C_0 + C’⇒ C’ = C_{initial} -C_0$. The general solution to the differential equation with a constant input concentration is: $C(t) = C_0 + (C_{initial} -C_0)e^{-kt}$

**Steady-State Solution**: As t → ∞, the exponential term $e^{-kt}$ approaches zero, so C(t) → C_{0}. This means the concentration in the tank approaches the constant input concentration C_{0}.**Transient Solution**: The term $C_{initial} -C_0)e^{-kt}$ represents the transient behavior, which decays exponentially over time.**Rate of Convergence**: The rate constant k determines how quickly the system reaches steady state. A larger k results in a faster decay of the transient term.

A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 7% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. (a) What is the amount of alcohol after an hour? (b) What is the percentage of alcohol after an hour?

Initial condition: The vat contains 500 gallons of beer with 4% alcohol, so initially, the amount of alcohol is: A(0)=500×0.04=20 gallons of alcohol.

The rate of alcohol inflow and outflow can be written as:

$\frac{dA}{dt} = \text{Rate of Alcohol Inflow - Rate of Alcohol Outflow} = 5\text{gal/min}·0.07-5\text{gal/min}·\frac{A}{500}$ because concentration = ^{amount of alcohol}⁄_{volume}.

**Rate of inflow**: Beer with 7% alcohol is being pumped in at 5 gal/min, so the rate of alcohol inflow is: 5 gal/min × 0.07 = 0.35 gal/min.**Rate of outflow**: The mixture is being pumped out at the same rate of 5 gal/min, and the concentration of alcohol in the vat at any time t is^{A(t)}⁄_{500}where A(t) is the amount of alcohol at time t. Thus, the rate of alcohol outflow is: 5 gal/min x^{A(t)}⁄_{500}= 0.01·A(t).

Thus, the differential equation becomes: $\frac{dA}{dt}=0.35 -0.01A, A(0) = 20, A(60)?$

This is a separable differential equation. Let’s separate the variables: $\frac{1}{0.35 -0.01A}dA = dt$.

Integrate both sides: $\int \frac{1}{0.35 -0.01A}dA = \int dt ⇒ \frac{1}{-0.01}ln|0.35 -0.01A| = t + C_1 ⇒[\text{Simplifying:}] ln|0.35 -0.01A| = -0.01(t + C_1) ↭ ln|0.35 -0.01A| = -0.01t + C_2$ where C_{2} = -0.01·C_{1} is a new constant of integration.

On the left-hand side, the integral requires a substitution u = 0.35 - 0.01A, so du = -0.01dA.

$ln|0.35 -0.01A| = -0.01t + C_2 ↭[\text{Take exponentials of both sides to solve for A:}] |0.35 -0.01A| = e^{-0.01t + C_2}↭ |0.35 -0.01A| = C_3e^{-0.01t}$ where C_{3} = e^{C2}

$0.35 -0.01A = ± C_3e^{-0.01t} ↭ -0.01A = -0.35 + C_4e^{-0.01t}$ where C_{4} = ± C_{3}.

Solving for A, $A = \frac{-0.35}{-0.01}+\frac{C_4}{-0.01}e^{-0.01t} ↭ A = 35 + C_5e^{-0.01t}$ where $C_5 = \frac{C_4}{-0.01}⇒[\text{Applying the initial conditions}] A(0) = 20 = 35 + C_5e^{-0.01·0} ↭ C_5 = 20 -35 = -15 ⇒[\text{Thus, the general solution is:}] A = 35 + C_5e^{-0.01t} = 35 -15e^{-0.01t}$

$A(60) = 35 -15e^{-0.01·60} = 35 -15e^{-0.6}≈ 26.77$ gallons of alcohol. So, after an hour (60 minutes), the vat contains approximately 26.7 gallons of alcohol.

$\frac{A(60)}{500} x 100 = \frac{26.77}{500} x 100 = 0.0535 x 100 = 5.35%$ is the percentage of alcohol after an hour.

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.