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Applications First-order Linear ODE's

There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else - but persistent, Raoul Bott

Arithmetic is where numbers fly like pigeons in and out of your head, Carl Sandburg

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., $\frac{dy}{dx} = 3x +5y$):

Solving differential equations

To solve ordinary differential equations (ODEs), several methods can be employed, each suited to different types of equations. Here are some of the key methods:

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:

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.

Homogeneous Equations

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.

Existence and Uniqueness Theorem

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

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.

First-order Linear ODE

Standard Form

The general first-order linear ODE can be rearranged into the standard linear form: y’ + p(x)y = q(x) where:

Expressing the equation in this standard form makes it easier to apply systematic methods for finding solutions, such as using an integrating factor.

Solving First-Order Linear ODEs Using Integrating Factors

The method of integrating factors is a common technique for solving first-order linear ODEs in standard form.

Given the equation: y′ +p(x)y = q(x) we can find an integrating factor μ(x) defined by: $μ(x)=e^{\int p(x)dx}$

Multiplying both sides of the differential equation by μ(x) transforms the left-hand side into the derivative of μ(x)y, allowing us to integrate both sides more easily.

Applications of First-Order Linear ODEs

First-order linear ODEs are widely used to model various physical phenomena, including:

Newton’s Law of Cooling

Newton’s law of cooling describes the rate at which an object changes temperature when in contact with an environment with a different temperature.

The law states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient or external temperature Te of its surroundings.

Mathematically, this law is expressed by the differential equation: $\frac{dT}{dt}= -k(T -T_e)$ or equivalently $\frac{dT}{dt}=k(T_e-T)$ where:

Interpretation. This differential equation implies that:

In both cases, the object approaches the ambient temperature over time.

Solving Newton’s Law of Cooling Using Integrating Factors

  1. Rewrite in Standard Form. To solve the differential equation, we first rewrite it in standard linear form: $\frac{dT}{dt} + kT = kT_e$. This matches the standard form y’ + p(t)y = q(t), where y = T(t), p(t) = k and q(t) = kTe.
  2. Find the integrating factor: $μ(t)=e^{\int p(t)dt} = e^{\int kdt} = e^{kt}$
  3. Multiply both sides by the integrating factor: $e^{kt}\frac{dT}{dt} + e^{kt}kT = e^{kt}·kT_e ↭[\text{Simplify the left-hand side using the product rule:}] \frac{d}{dt}(e^{kt}T) = e^{kt}·kT_e$
  4. Integrate both sides with respect to t: $e^{kt}T = \int e^{kt}kT_edt =[\text{Since k and Tₑ are constants}] T_ee^{kt} ↭ e^{kt}T = T_ee^{kt}+C$, where C is the constant of integration.
  5. Solve for T: $T = T_e + Ce^{-kt}$.
  6. Apply initial conditions. If T(0) = T0, then $T_0 = T_e + Ce^{-0·t} = T_e + C ⇒ C = T_0 - T_e$
  7. Write the Final Solution: $T(t) = T_e + (T_0 - T_e)e^{-kt}$. This is the general solution to Newton’s Law of Cooling, showing how the temperature of the object changes over time.
  8. Interpretation of the Solution. The object's temperature T(t) exponentially approaches the external temperature Te over time. Mathematically, as t → ∞, $e^{-kt}→ 0⇒ \lim_{t \to \infty} T(t) = T_e$. The term $e^{-kt}$ represents exponential decay.

The rate at which T(t) approaches the ambient temperature Te depends on the cooling constant k:

Example: Determining the Cooling Constant k for a Pot of Coffee

A brewing pot of coffee is initially at a temperature of 180°F. It is placed in a room where the ambient (external) temperature is constant at 76°F. After 5 minutes, the temperature of the coffee drops to 168°F. Determine the cooling constant k using Newton’s Law of Cooling.

Refer to Figure B for a visual representation and aid in understanding it.

First order Linear The general solution to Newton’s Law of Cooling is: $T = T_e + (T_0-T_e)e^{-kt}$

  1. At time t = 0, the temperature is T(0) = 180°F, Te = 76°F. $T = T_e + (T_0-T_e)e^{-kt} ↭ T = 76 + (180-76)e^{-kt} = 76 +104e^{-kt}$
  2. Apply the temperature at t = 5 minutes: $ T(t) = 76 +104e^{-5·k} = 168 ↭ 104e^{-5·k} = 168 -76 = 92 ↭ 104e^{-5·k} = 92$
  3. Solve for $e^{-5·k}$: $e^{-5·k} = \frac{92}{104} = 0.8846.$
  4. Take the natural logarithm of both sides: $-5k = ln(0.8846)$.
  5. Solve for k: $k = \frac{-ln(0.8846)}{5}≈ \frac{0.122}{5} ≈ 0.0245$ per minute. The cooling constant k is approximately 0.0244 per minute. This value quantifies the rate at which the coffee cools down in the given environment.

Diffusion Model: Salt Concentration

Another application of first-order linear ODEs is modeling diffusion processes, such as the diffusion of salt across a semi-permeable membrane.

Consider a chamber containing a solution with salt concentration C(t) at time t. The chamber is surrounded by a water bath with a constant salt concentration Ce. A semi-permeable membrane allows salt to diffuse between the chamber and the water bath. The rate of diffusion is proportional to the concentration difference across the membrane.

The diffusion process is modeled by the first-order differential equation: $\frac{dC}{dt} = k(C_e-C)$ where:

This equation is similar to the cooling equation. It states that the rate at which the internal salt concentration changes is proportional to the difference between the external and internal concentrations:

First order Linear

Rewriting the equation in standard linear form: $\frac{dC}{dt}+kC = kC_e$. Again, this matches the previously studied standard form y’ + p(t)y = q(t), where y = C(t), p(t) = k, and q(t) = kCe.

Solution Using Integrating Factor

  1. Find the integrating factor: $μ(t)=e^{\int p(t)dt} = e^{\int kdt} = e^{kt}$
  2. Multiply both sides by the integrating factor: $e^{kt}\frac{dC}{dt} + e^{kt}kC = e^{kt}kC_e ↭ \frac{d}{dt}(e^{kt}C) = e^{kt}kC_e$
  3. Integrate both sides: $e^{kt}C = \int e^{kt}kC_edt = C_ee^{kt} ↭ e^{kt}C = C_ee^{kt}+C’$, where C’ is the constant of integration.
  4. Solve for C: $C = C_e + C’e^{-kt}$.
  5. Apply initial conditions. If C(0) = C0, then $C_0 = C_e + C’e^{-0·t} = C_e + C’ ⇒ C’ = C_0 - C_e$
  6. Final Solution: $C(t) = C_e + (C_0 - C_e)e^{-kt}$.
  7. Interpretation of the Solution. The internal concentration C(t) exponentially approaches the external concentration Ce over time. As t → ∞, $e^{-kt}→0, \lim_{t \to \infty} C(t) = C_e$.

    The term $e^{-kt}$ represents exponential decay.

The rate of diffusion k determines how quickly the internal concentration C(t) approaches the external concentration Ce:

In both applications, the solutions exhibit exponential behavior, reflecting how the quantity of interest approaches equilibrium over time.

Solving First-Order Linear Ordinary Differential Equations (ODEs)

In this section, we will solve several first-order linear ODEs using the method of integrating factors.

  1. Rewrite the differential equation in standard form (Since we are considering x > 0 to avoid division by zero): $y’ -\frac{1}{x}y = x^2$, where $p(x) = \frac{-1}{x}, q(x) = x^2.$
  2. Calculate the integrating factor: $μ = e^{\int p(x)dx} = e^{\int \frac{-1}{x}dx}$ =[The absolute value ln|x| ensures the logarithm is defined for x < 0 as well, but for simplicity, we assume x > 0] $e^{-ln(x)} = e^{ln(x^{-1})} = x^{-1} = \frac{1}{x}$.

    By the power rule of logarithms log(ab) = b*log(a).

  3. Multiplying both sides of the differential equation by the integrating factor μ(x): $\frac{1}{x}y’-\frac{1}{x^2}y = x$
  4. Recognize the left-hand side as a product derivative μ(x)y, $\frac{d}{dx}(\frac{1}{x}y) = \frac{1}{x}y’-\frac{1}{x^2}y$.
  5. Integrate both sides of the equation with respect to x: $\frac{1}{x}y = \int xdx = \frac{x^2}{2} + C$ where C is the constant of integration.
  6. Solve for y, $y(x) = \frac{x^3}{2}+Cx$ where C is an arbitrary constant determined by initial conditions (if any are provided).

This equation is not immediately in the standard linear form, so we will manipulate it to fit that form and then apply the method of integrating factors.

To bring the equation into the standard linear form y′ +P(x)y=Q(x), we divide both sides by x2: $\frac{1}{x^2}(x^2y’ + xy) = \frac{1}{x^2}(3x) ↭ y’ + \frac{1}{x}y = \frac{3}{x}$

Compute the integrating factor, μ(x). It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int \frac{1}{x}dx} = e^{ln|x|} = |x|$ Since x is in the denominator and we need x ≠ 0, we’ll consider x > 0 for simplicity, so μ(x) = x.

We write the differential equation in the form $\frac{d}{dx}(μ·y)=μ·Q(x)$ by multiplying both sides of the differential equation by the integrating factor, $xy’ + y = 3 ↭ \frac{d}{dx}(x·y) = 3$.

Next, we can solve the differential equation integrating with respect to x, $xy = \int 3dx = 3x + C ⇒[\text{Solve for y}] y = 3 + \frac{C}{x}$ where C is the constant of integration determined by initial conditions.

  1. Rewrite the differential equation in standard form: $y’ -\frac{sin(x)}{1+cos(x)}y = \frac{2x}{1+cos(x)}$ where $p(x) = -\frac{sin(x)}{1+cos(x)}, q(x) = \frac{2x}{1+cos(x)}$
  2. Compute the integrating factor: $μ = e^{\int p(x)dx} = e^{\int -\frac{sin(x)}{1+cos(x)}dx} = e^{ln(1+cos(x))} = 1 + cos(x)$.

    $\int -\frac{sin(x)}{1+cos(x)}dx$ =[Let u = 1+cos(x), then du = −sin(x)dx] $\int \frac{du}{u} = ln|u| = ln|1+cos(x)|=$[Since 1 + cos(x) ≥ 0 for all real x, we can drop the absolute value:] ln(1 + cos(x))

  3. Multiplying both sides of the differential equation by the integrating factor μ(x): $(1+cos(x))y’ -sin(x)y = 2x.$
  4. By construction, the left-hand side becomes the derivative of the product μ(x)y, i.e., $(1+cos(x)y)’ = 2x$
  5. Integrate both sides of the equation with respect to x: $1+cos(x)y = \int 2xdx = x^2 + C$ where C is the constant of integration.
  6. Solve for y, $y = \frac{x^2+C}{1+cos(x)}$
  7. Apply the initial condition $y(0) = 1, 1 = \frac{C}{1+cos(0)} ↭ 1 = \frac{C}{2} ↭[\text{Solving for C}] C = 2$.
  8. Final solution. Substitute C = 2 back into the general solution. The particular solution to the differential equation, satisfying y(0) = 1, is: $y = \frac{x^2+2}{1+cos(x)}$
  1. Write the differential equation in standard form. It is already in standard form where $p(x) = \frac{3}{x}, q(x) = \frac{e^x}{x^3}$
  2. Compute the integrating factor: $μ = e^{\int p(x)dx} = e^{\int \frac{3}{x}} = e^{3ln(x)} = e^{ln(x^3)} = x^3$ where x > 0 (to avoid division by zero and ensure μ(x) is defined).
  3. Multiplying both sides of the differential equation by the integrating factor μ(x): $x^3\frac{dy}{dx} +3x^2y = e^x.$
  4. By construction, the left-hand side becomes the derivative of the product μ(x)y, i.e., $(x^3y)’ = e^x$
  5. Integrate both sides of the equation with respect to x: $x^3y = \int e^xdx = e^x + C$ where C is the constant of integration.
  6. Solve for y, $y = \frac{e^x+C}{x^3}$ where C is an arbitrary constant determined by initial conditions (if any are provided).

Bibliography

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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  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
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  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
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