Basic linear ODE

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Recall

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, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

• y = y(x) is the unknown function of the independent variable x.
• y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
• a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

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 the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

Basic linear ODE

In mathematics, a linear ordinary differential equation (ODE) is an equation involving an unknown function and its derivatives, where both the function and its derivatives appear linearly. This means that the unknown function and its derivatives are raised only to the first power and are not multiplied together or composed within nonlinear functions.

This property of linearity makes linear ODEs fundamental in modeling various physical systems due to their simplicity and the powerful mathematical tools available for solving them. They often arise in fields such as:

• Fluid Dynamics: Describing the flow of fluids under various conditions where the flow can be approximated as linear.
• 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, where Hooke’s law (a linear relationship) applies.
• Radioactive Decay: Describing the rate at which unstable nuclei decay over time, which follows a linear differential equation leading to exponential decay.

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.

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.

Understanding Linearity in ODEs

An ODE is linear if it can be expressed in the form: $a_n(t)\frac{d^ny}{dt^n} +a_{n-1}(t)\frac{d^{n-1}y}{dt^{n-1}} + ··· + a_1(t)\frac{dy}{dt} + a_0(t)y = q(t)$ where

• y(t) is the unknown function.
• $\frac{d^ky}{dt^y}$ are the derivatives of y with respect to t.
• a_i(t) are known coefficient functions of t. They can be constants or functions of t, but they do not depend on y.
• q(t) is a known forcing function.

The function y and its derivatives appear to the first power and are not multiplied together.

Common Forms of First-Order Linear ODEs

First-order linear ODEs involve only the first derivative of the unknown function. They are foundational in understanding more complex differential equations. 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 the forcing function q(t) is proportional to k, q(t) = kq_e(t) with q_e(t) being the external input function, e.g., $\frac{dy}{dt}+2y=2e^{2t}$ where k = 2 and q_e(t) = e^2t.
• 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, e.g., $\frac{dy}{dt}+3y=cos(t)$ where k = 3 and q(t) = cos(t).
• Standard Linear Form. y’ + p(t)y = q(t), where y is the unknown function of time, p(t) is a function of time, and q(t) is a known function representing external input or forcing. This is the most general form of a first-order linear ODE, encompassing the previous two cases, e.g., $\frac{dy}{dt} + sin(t)y = e^t$ where p(t) = sin(t) and q(t) = e^t.

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

Flow Rate of Salt in a Tank

Understanding how substances mix and change their concentrations over time in a tank is a classic problem in differential equations. This scenario is commonly used to model real-world systems in chemistry, engineering, and environmental science.

Consider a tank that contains a well-mixed solution of salt and water. Saltwater flows into and out of the tank at a constant rate. Our goal is to model how the concentration of salt in the tank changes over time. Let’s define the variables involved:

• r represents the rate of inflow and outflow (volume per unit time.g., liters or gallons per minute or hour). The rate at which the solution flows into and out of the tank is constant.
• V is the volume of the solution (water) in the tank. The volume V of the solution in the tank remains constant because the inflow and outflow rates are equal.
• x(t) is the amount of salt in the tank at time t (mass, e.g., kilograms or grams).
• C_e(t) is the concentration of incoming salt as a function of time (mass per unit volume, e.g., kg/L, refer to Figure E for a visual representation and aid in understanding it).
• C(t): Concentration of salt in the tank at time t (mass per unit volume).

The tank is well-mixed, so the concentration of salt throughout the tank is uniform at all times, meaning the concentration of salt is uniform throughout the tank.

Deriving the Differential Equation

The fundamental principle governing the change in the amount of salt in the tank is that 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 rate of salt inflow (the product of the inflow rate r and the concentration of the incoming solution C_e(t)) r·C_e(t) minus the rate of salt outflow (the product of the outflow rate r and the concentration of the solution in the tank C(t):) $r·\frac{x}{V}$, where C(t) = ^x⁄_V is the concentration of salt in the tank at time t (since the tank is well-mixed).

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

Expressing the Equation in Terms of Concentration

It’s often more convenient and useful to work with concentration C(t) instead of the total amount x(t). Recall that C(t) = ^x⁄_V, leading to: $\frac{dx}{dt} + rC = r·C_e(t)$. Besides, x = V·C(t) and differentiating both sides with respect to time gives: $V\frac{dC}{dt} + rC = r·C_e(t)$

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 ordinary differential equationODE in terms of C(t), the concentration of salt in the tank.

Intuitive Understanding of the Model

• Large k: the flow rate is high or the volume is small, so the concentration of salt in the tank C(t) rapidly matches or adjusts to the incoming concentration C_e(t). 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) changes over time, e.g., it is a sinusoidal function (e.g., C_e(t) = cos(wt)), the concentration C(t) will also oscillate but may lag behind (with a phase lag Φ) and reduced amplitude depending on k.

Solving the Differential Equation

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

1. Compute the Integrating Factor μ(t): $μ(t) = e^{\int kdt} = e^{kt}$
2. 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: $\frac{d}{dt}(e^{kt}C)$
3. 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 C₁ is the constant of integration.
4. Solve for C(t): $C(t) = e^{-kt}[k\int e^{kt}C_e(t)dt + C_1]$

Special Case: Constant Input Concentration

Assume C_e(t) = C₀, a constant concentration.

1. Integrate the Right-Hand side: $\int e^{kt}C_e(t)dt = C_0\int e^{kt}dt = \frac{C_0e^{kt}}{k} + C_2$
2. Substitute back into the solution: $C(t) = e^{-kt}[k\int e^{kt}C_e(t)dt + C_1] = e^{-kt}[k(\frac{C_0e^{kt}}{k}) + kC_2 + C_1] = C_0 + kC_2e^{-kt} + C_1e^{-kt} = C_0 + (kC_2+C_1)e^{-kt} = C_0 + C’e^{-kt}$. $C(t) = C_0 + C’e^{-kt}$ where $C’ = kC_2+C_1$
3. 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: $C(t) = C_0 + (C_{initial} -C_0)e^{-kt}$

Interpretation

• Steady-State Solution: As t → ∞, the exponential term $e^{-kt}$ approaches zero, so C(t) → C₀. This means that as time goes on the concentration in the tank approaches the constant input concentration C₀.
• Transient Solution: The term $(C_{initial} -C_0)e^{-kt}$ represents the transient behavior, which decays exponentially over time.
• Rate of Convergence: The constant k determines how quickly the system approaches steady state. A larger k results in a faster decay of the transient term, meaning a faster convergence.

Example Problem

Scenario:

A vat contains 500 gallons of beer with an alcohol concentration of 4% (by volume). Beer with an alcohol concentration of 7% is pumped into the vat at a rate of 5 gallons per minute. The mixture is simultaneously pumped out at the same rate. Questions: (a) What is the amount of alcohol in the vat after an hour? (b) What is the percentage of alcohol after an hour?

Step 1: Define Variables and Initial Conditions: The vat contains V = 500 gallons of beer with C_initial = 4% = 0.04 alcohol, so initially, the amount of alcohol is: A(0) = V·C_initial = 500 × 0.04 = 20 gallons of alcohol. Inflow and Outflow Rate: r = 5 gal/min. Incoming Alcohol Concentration: C_e(t) = 7% = 0.07.

Step 2: Set Up the Differential Equation

Let A(t) be the amount of alcohol in the vat at time t (measured in gallons). The concentration of alcohol in the vat at time t is: $C(t) = \frac{A(t)}{V} = \frac{A(t)}{500}$ because concentration = ^{amount of alcohol}⁄_volume.

The rate of alcohol inflow and outflow can be written as: $\frac{dA}{dt} = \text{Rate of Alcohol Inflow - Rate of Alcohol Outflow} = rC_e(t) -r\frac{A(t)}{V} = 5\text{gal/min}·0.07 -5\text{gal/min}·\frac{A}{500}$

• 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: $\frac{dA}{dt}=0.35 -0.01A, A(0) = 20, A(60)?$

Step 3: Solve the Differential Equation. 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 C₂ = -0.01·C₁ 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 C₃ = e^C₂

$0.35 -0.01A = ± C_3e^{-0.01t} ↭ -0.01A = -0.35 + C_4e^{-0.01t}$ where C₄ = ± C₃.

Solving for A, $A = \frac{-0.35}{-0.01}+\frac{C_4}{-0.01}e^{-0.01t} ↭ A = 35 + C_5e^{-0.01t}$ where $C_5 = \frac{C_4}{-0.01}⇒[\text{Let C = C₅. Applying the initial conditions}] A(0) = 20 = 35 + Ce^{-0.01·0} ↭ C = 20 -35 = -15 ⇒[\text{Thus, the general solution is:}] A = 35 + Ce^{-0.01t} = 35 -15e^{-0.01t}$

Step 4: Calculate Amount of Alcohol After One Hour (60 minutes). $A(60) = 35 -15e^{-0.01·60} = 35 -15e^{-0.6}≈ 26.77$ gallons of alcohol. So, after an hour (60 minutes), the vat contains approximately 26.7 gallons of alcohol.

Step 5: Calculate the Percentage of Alcohol After One Hour. Alcohol Concentration = $\frac{A(60)}{500} x 100$% $= \frac{26.77}{500} x 100 = 0.0535 x 100 = 5.35$% is the percentage of alcohol after an hour.

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