Table of Contents
Differential Equations
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Courses

  1. Linear Algebra
  2. Multivariable Calculus
  3. Differential Equations
  4. Miscellaneous Topics

Differential Equations

  1. Introduction
    1. Notation and Definitions
    2. Verifying Solutions
    3. Initial Values Problems
  2. First Order Differential Equations
    1. Direction Fields
    2. Equilibria, Stability, and Phase Lines
    3. Separable Equations
    4. Integrating Factor
    5. Bernoulli Equations and Substitutions
    6. Exact Equations
    7. Exact Equations with Integrating Factor
    8. Euler's Method
    9. Existence and Uniqueness
    10. Interval of Validity
    11. Applications
      1. Radioactive Decay and Population Growth
      2. Mixing Tank Problem
      3. Terminal Velocity
      4. Continuous Compound Interest
  3. Wrońskian
  4. Second Order Differential Equations
    1. $ay''+by'+cy = 0$
      1. Distinct Real Roots
      2. Repeated Real Root
      3. Imaginary Roots
    2. Spring-Mass-Damper
      1. Equivalent RLC Circuit
      2. Underdamped, Overdamped, Critically Damped
      3. Undamped Oscillations and Resonance
    3. Method of Undetermined Coefficients
    4. Reduction of Order
    5. Variation of Parameters
    6. Cauchy-Euler Equations
  5. Laplace Transform
    1. Lookup Table and WolframAlpha
  6. Systems of ODEs
    1. Simplest Case $\begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$
      1. Poincaré Diagram
  7. Preview of Dynamical Systems
    1. Self-Study
    2. Rössler Attractor

Additional Resources

  1. WolframAlpha Examples
  2. Direction Field Plotter by Ariel Barton
  3. Phase Plane Plotter by Ariel Barton


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Initial Value Problems (IVPs) | youtube icon Topic Playlist

Number of Initial Conditions | youtube icon Explanation Video

An IVP is an ODE paired with an initial condition specifying the value of the unknown function at a point. For a first order ODE, the initial condition is $y(x_{0})=y_{0}$ where $x_{0}$ and $y_{0}$ are some specified values. For a second order ODE, the initial condition is specified as $y(x_{0})=y_{0}$ and $y'(x_{0})=y'_{0}$ where $x_{0}$, $y_{0}$, and $y'_{0}$ are specified values. For an nth order ODE, the initial condition is specified as $y(x_{0})=y_{0}$, $y'(x_{0})=y'_{0}$, . . ., and $y^{(n-1)}(x_{0})=y^{(n-1)}_{0}$ where $x_{0}$, $y_{0}$, $y_{1}$, . . , and $y^{(n-1)}_{0}$ are specified values.

Initial conditions for an ODE have an intuitive meaning in physics, like the velocity of a falling object or the movement of a mass attached to a spring with friction.

Example 1 | youtube icon Solution Video

The following ODE models the velocity of a falling object with air resistance, with $m$ the mass of the object, $k$ the drag coefficient, and $v(t)$ the velocity at time $t$. The second equation is after dividing by $m$.

\begin{equation} m\frac{\text{d}v}{\text{d}t} = -mg - kv \qquad \frac{\text{d}v}{\text{d}t} = -g - \frac{k}{m}v \end{equation}

For a concrete example $g = 9.8\dfrac{\text{m}}{\text{s}^{2}}$, $m = 4\text{kg}$, $k = 0.5\dfrac{\text{kg}}{\text{s}}$, and $v(0s) = 10\dfrac{\text{m}}{\text{s}}$. Using the general solution to the ODE and the initial condition, find the particular solution to the IVP.

\begin{equation} \frac{\text{d}v}{\text{d}t} = -9.8 - 0.125v \qquad v(0) = 10 \end{equation} \begin{equation} v(t) = Ce^{-0.125t} - 78.4 \end{equation}

Solution

\begin{equation} v(t) = 88.4e^{-0.125t} - 78.4 \end{equation}

Most methods to solve ODEs apply an indefinite integral in the method, resulting in a constant of integration $C$ appearing. The reason the constant of integration $C$ is necessary in the general solution is so any initial condition can be solved for. From the physics points of view, we couldn't have figured out the velocity at any point in the future if we didn't know the velocity to begin with.

Example 2 | youtube icon Solution Video

The following ODE models a mass on a spring with friction, with $m$ the mass of the object, $b$ the friction coefficient, $k$ the spring constant, and $z(t)$ the position of the object at time $t$ relative to the rest position of the spring.

\begin{equation} mz'' + bz' + kz = 0 \end{equation}

For a concrete example $m = 1\text{kg}$, $b = 0.1\dfrac{\text{kg}}{\text{s}}$, $k = 10\dfrac{\text{kg}}{\text{s}^{2}}$, $z(0s) = -0.3\text{m}$, and $z'(0s) = 1\dfrac{\text{m}}{\text{s}}$. Using the general solution to the ODE and the initial condition, find the particular solution to the IVP.

\begin{equation} z'' + 0.1z' + 10z = 0 \qquad z(0) = -0.3 \qquad z'(0) = 1 \end{equation} \begin{equation} z(t) = c_{1} e^{-0.05 t} \sin(3.16188 t) + c_{2} e^{-0.05 t} \cos(3.16188 t) \end{equation}

Solution

\begin{equation} z(t) = e^{-0.05 t} \big(0.311523 \sin(3.16188 t) - 0.3 \cos(3.16188 t)\big) \end{equation}

Two constants of integration $c_{1}$ and $c_{2}$ are in the general solution for a 2nd order ODE, which are needed to solve for the initial condition. From a physics point of view, we needed to know both the starting position of the object and the starting velocity of the object to be able to predict the position of the object in the future.