Differential Equations Notes

Copyright Notice

  • Do not copy for others, including friends or family.
  • Do not distribute online in any form, including social media.
  • Do not sell to tutoring companies or other organizations.
  • These files are strictly for personal use only.
  • Agreement to all conditions is required before using the files.
  • If you do not agree, please inform the instructor to be removed from D2L.

Introduction to Differential Equations

Definitions

  • Ordinary Differential Equation (ODE): An equation containing an unknown function of a single variable together with its derivatives with respect to that variable.
  • Partial Differential Equation (PDE): An equation containing an unknown function of two or more variables together with its partial derivatives with respect to these variables.
  • Order of a Differential Equation: The order of an ODE or PDE is the order of the highest derivative present.

Examples

  • ODE Example: \frac{d^2x}{dt^2} = t

    • Unknown function: x = x(t) (dependent variable).
    • Independent variable: t.
    • Order: Two (second-order derivative present).

  • PDE Example: K \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, K > 0

    • Unknown function: u = u(x, t) (dependent variable).
    • Independent variables: x and t.
    • Order: Two (second-order PDE).
    • Known as the Heat Conduction Equation.

Mathematical Modeling

  • Differential equations (ordinary or partial) occur in connection with numerous applications in various branches of science and engineering.

  • Examples of Problems:

    1. Conduction of heat in a rod, plate, or slab.
    2. Vibration in a string or a membrane.
    3. Determining gravitational or electrical potential.
    4. Determining the motion of a projectile, planet, or satellite.
    5. Reaction of chemicals.
    6. Growth and decay problems.
    7. Determining curves having specific geometric properties (e.g., in robotics).
    8. Mixture problems.
    9. Hanging cables.

  • These problems involve not only a differential equation but also a set of supplementary conditions, such as:

    • Initial Conditions
    • Boundary Conditions

Solution to an ODE

  • Definition: A solution to an ODE in an interval I is a function having sufficient derivatives in I and which identically satisfies the ODE.

  • Example: Show that x(t) = e^{5t} is a solution to the ODE \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} - 35x = 0 on \mathbb{R} = (-\infty, \infty) .

  • Solution:

    • Given ODE: x'' + 2x' - 35x = 0 (*)
    • x = e^{5t} (1)
    • x' = 5e^{5t} (2)
    • x'' = 25e^{5t} (3)
    • x, x', and x'' exist on \mathbb{R}.

    Substitute (1), (2), and (3) into the left-hand side (L.H.S) of (*):

L.H.S = 25e^{5t} + 2(5e^{5t}) - 35e^{5t} = 25e^{5t} + 10e^{5t} - 35e^{5t} = 0 = R.H.S

Verified. Therefore, x(t) = e^{5t} is a solution to the ODE in \mathbb{R}.

  • Example: Given the DE t^2 x'' - t x' - 8x = 0 , find all values of the constant real number K so that x(t) = t^K is a solution to the DE on (0, \infty).

  • Solution:

    • Given DE: t^2 x'' - t x' - 8x = 0 (*.I)
    • x = t^K (1)
    • x' = K t^{K-1} (2)
    • x'' = K(K-1) t^{K-2} (3)

Substitute (1), (2), and (3) in DE (*.I):

t^2 [K(K-1)t^{K-2}] - t [Kt^{K-1}] - 8[t^K] = 0
K(K-1)t^K - Kt^K - 8t^K = 0
[K(K-1) - K - 8]t^K = 0

Since t^K \neq 0 (for t \in (0, \infty)), then:

K(K-1) - K - 8 = 0
K^2 - K - K - 8 = 0
K^2 - 2K - 8 = 0
(K-4)(K+2) = 0
K = 4, -2

There are two solutions: x1(t) = t^4 and x2(t) = t^{-2}.

Initial Value Problem (IVP) vs Boundary Value Problem (BVP)

  • Initial Value Problem (IVP): An ODE together with the prescribed initial conditions. The conditions are prescribed at exactly one value of t.

    • Example: x''(t) + x(t) = 0 with supplementary conditions x(0) = 5 and x'(0) = -7 is an IVP.

  • Boundary Value Problem (BVP): An ODE together with the prescribed boundary conditions. The conditions are prescribed at more than one value of t.

    • Example: x''(t) + x(t) = 0 together with the supplementary conditions x(0) = 5 and x'(\pi) = -2 is a BVP.

  • Note: The condition x(t0) = x0 means x = x0 at t = t0 .

General Solution to an ODE

  • Definition: The general solution to an ODE is a formula that describes all possible solutions to the DE.

  • The general solution to an nth-order ODE must contain n arbitrary constants C1, C2, …, C_n . These arbitrary constants may also be referred to as parameters, and accordingly, the general solution may be referred to as an n-parameter family of solutions.

  • Example: Given the DE \frac{dx}{dt} = 2t
    (a) Find the general solution to the DE.
    (b) Sketch few members of the one-parameter family of solutions.

  • Solution:
    (a) \frac{dx}{dt} = 2t
    Integrate both sides:
    x = \int 2t dt = t^2 + C , where C is an arbitrary constant.
    Therefore, x = t^2 + C is the general solution to the DE.
    (b) The general solution is a one-parameter family, namely x = t^2 + C, C \in \mathbb{R} . x - C = (t - 0)^2 . This is a family of parabolas with a vertex at (0, C) and which opens upward.

First-Order Equations

  • Consider first-order equations of the form \frac{dy}{dx} = f(x, y) .

Separable Equations

  • A separable equation is one in which f can be expressed as a function of x times a function of y, i.e., \frac{dy}{dx} = g(x)h(y) .

Method of Solution

  • Rewrite h(y) \neq 0 as \frac{dy}{h(y)} = g(x) dx .

  • Integrate both sides: \int \frac{dy}{h(y)} = \int g(x) dx .

  • The solution is given implicitly by H(y) = G(x) + C , where H'(y) = \frac{1}{h(y)} and G'(x) = g(x) .

  • Note: If h(y0) = 0 , then y(x) = y0 is a solution (called a singular solution).

Example of Separable Equations

  • Example: Solve the differential equation \frac{dy}{dx} = \frac{x^2}{1 - y^2} .

  • Solution: Separate variables:
    (1 - y^2) dy = x^2 dx .

Integrate both sides:
\int (1 - y^2) dy = \int x^2 dx
y - \frac{y^3}{3} = \frac{x^3}{3} + C .

So the solutions are given implicitly by y - \frac{y^3}{3} = \frac{x^3}{3} + C .

Homogeneous Equations

  • A homogeneous equation can be written in the form \frac{dy}{dx} = F(\frac{y}{x}) .

Method of Solution

  • Let v = \frac{y}{x} . Then y = vx . So \frac{dy}{dx} = v + x \frac{dv}{dx} .
  • Substitute into the original equation: v + x \frac{dv}{dx} = F(v) .
  • Separate variables and integrate: \frac{dv}{F(v) - v} = \frac{dx}{x} .

Note that the method for solving homogeneous equation may not work at v = F(v), hence it produces singular solutions.

Exact Differential Equations

  • An exact differential equation is an equation of the form M(x, y) dx + N(x, y) dy = 0 , where \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} .

Method of Solution

  • Find a function \psi(x, y) such that \frac{\partial \psi}{\partial x} = M(x, y) and \frac{\partial \psi}{\partial y} = N(x, y) .
  • Then the solution is given implicitly by \psi(x, y) = C , where C is a constant.

-Note: The reason this approach works is that if \psi(x, y) = C , then by the chain rule, \frac{\partial \psi}{\partial x} dx + \frac{\partial \psi}{\partial y} dy = 0 , which is the original differential equation.

Integrating Factors

  • If the equation M(x, y) dx + N(x, y) dy = 0 is not exact, it may be possible to find a function \mu(x, y) such that the equation \mu(x, y) M(x, y) dx + \mu(x, y) N(x, y) dy = 0 is exact. Then \mu(x, y) is called an integrating factor.

Linear First-Order Equations

  • A linear first-order equation is one that can be written in the form
    \frac{dy}{dx} + p(x)y = q(x)

Method of Solution

  1. Calculate the integrating factor: \mu(x) = e^{\int p(x) dx}
  2. Multiply both sides of the equation by \mu(x) :
    \mu(x) \frac{dy}{dx} + \mu(x) p(x) y = \mu(x) q(x)
  3. Observe that the left-hand side is the derivative of \mu(x) y :
    \frac{d}{dx} (\mu(x) y) = \mu(x) q(x)
  4. Integrate both sides with respect to x:
    \mu(x) y = \int \mu(x) q(x) dx + C
  5. Solve for y:
    y = \frac{1}{\mu(x)} \int \mu(x) q(x) dx + \frac{C}{\mu(x)}