Differential Equations Notes

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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: $x<em>1(t) = t^4$ and $x</em>2(t) = t^{-2}$.

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 $x(t<em>0) = x</em>0$ means $x = x<em>0$ at $t = t</em>0$.

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 $C<em>1, C</em>2, …, C_n$. 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 $h(y<em>0) = 0$, then $y(x) = y</em>0$ 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:
   $y = \frac{1}{\mu(x)} \int \mu(x) q(x) dx + \frac{C}{\mu(x)}$