Differential Equations and Applications

Overview of Differential Equations

  • Focuses on the applications of first-order Ordinary Differential Equations (ODEs).

Key Definitions and Types of ODEs

  • ODE: Ordinary Differential Equations, can be categorized into various types:
    • Separable Equations: Can be separated into two parts, one involving only dependent variables and the other only independent variables.
    • Homogeneous Equations: Equations that can be expressed in the form where all terms are of the same degree.
    • Linear Equations: Equations that can be written in the form of a linear polynomial.
    • Exact Equations: Equations where the total differential can be expressed as a differentiable function.

Applications of First-Order ODEs

  • Focus particularly on Separable and Linear types in real-world scenarios.
  • Examples include:
    • Newton's Law of Cooling: Used to model the cooling of objects.

Newton's Law of Cooling

  • Equation: dTdt=−k(T−Ts)\frac{dT}{dt} = -k(T - T_s)
    • Where:
    • $T$ = Temperature of the object.
    • $T_s$ = Surrounding (constant) temperature.
    • $k$ = Proportionality constant, a positive number.
    • Interpretation: The rate of temperature change of a cooling object is proportional to the difference in temperature between the object and its surroundings.

Solving the Equation

  • Separable Form: Rearrange to solve for $T$ by integrating both sides:
    • dTT−Ts=−kdt\frac{dT}{T - T_s} = -k dt
  • Integration:
    • Upon integrating, we get:
    • ln⁡∣T−Ts∣=−kt+C\ln |T - T_s| = -kt + C
    • Therefore, exponentiating gives:
    • T=Ts+Ae−ktT = T_s + A e^{-kt}
    • Where $A$ is determined by initial conditions.

Example Problem

  1. Object cools from 90°C to 30°C in a room at 10°C over 30 minutes:
    • Establish initial conditions:
      • $T(0) = 90°C$, $T(30) = 30°C$, $T_s = 10°C$.
    • Solve for $k$ and use derived equation to find temperatures at various time intervals.
    • Ultimately find out that the temperature after 20 minutes is around 41.75°C.

Important Notes on Constants

  • A general solution will typically contain two constants ($A$ and $k$), and conditions provided in the problem are crucial for solving the constants.
  • The temperature and time conditions must be translated into mathematical conditions for successful calculations.

Second Example of Application

  • Body Cooling Scenario:
    • Initial body temperature at discovery is 85°F, and then after 1 hour the temperature is 75°F in a room of 70°F.
    • Need to determine when it was at 98.6°F using derived equations.

Final Remarks

  • Reinforce understanding of how to extract initial and subsequent conditions properly for accurately resolving differential equations in real-world scenarios.
  • The law applies to various objects, not just human bodies (they might just serve as relatable examples).