(I) Fundamentals and Terminology

A differential equation (D.E.) is an equation relating an unknown function and one or more of its derivatives.

Example 1

1. \(\frac{dx}{dt} = x^2 + t^2 + 2\)

  • Unknown function: \(x(t)\)
  • Independent variable: \(t\)

2. \(y'' - 2y' + 2xy = 0\)

  • Unknown function: \(y\)
  • Independent variable: \(x\)

The order of a differential equation is the order of the highest derivative that appears in it. The general form of an \(n\)-th order D.E. is:

$$F(x, y, y', y'', \dots, y^{(n)}) = 0$$

A continuous function \(u(x)\) is a solution of the D.E. on the interval \(I\) provided that its derivatives exist and satisfy the equation for all \(x\) in \(I\).

Types of D.E.

(II) Mathematical Modeling

Mathematical modeling is the process of translating a real-world situation into a mathematical model, performing analysis, and interpreting the results.

Example 2

In a city with a fixed population of \(P\) persons, the rate of change of the number \(N\) of those persons infected with a certain contagious disease is proportional to the product of the number who have the disease and the number who don't:

$$\frac{dN}{dt} = kN(P - N)$$
Example 3

1. Verify that the time rate of change of a population \(P(t) = Ce^{kt}\) satisfies \(\frac{dP}{dt} = kP\):

\(L.H.S. = \frac{d}{dt}(Ce^{kt}) = kCe^{kt} = kP = R.H.S.\)

2. Assume the population at \(t = 0\) was 1000, and the population doubled after 1 hour:

  • \(P(0) = 1000 \implies C = 1000\)
  • \(P(1) = 2000 \implies 1000e^k = 2000 \implies e^k = 2 \)

\(\implies k = \ln 2 \approx 0.693\)

3. Find the population after 1.5 hrs:

\(P(1.5) = 1000 \cdot 2^{1.5} \approx 2828\)

Example 4: Geometric D.E. Construction

Find a D.E. where the tangent line to the graph at \((x, y)\) passes through the point \((0, x)\):

\(y' = \frac{y - x}{x - 0} = \frac{y - x}{x}\)

(III) Verification of Solutions

Example 5

Verify that \(y(x) = (x + C)\cos x\) satisfies the D.E. \(y' + y \tan x = \cos x\). Then determine \(C\) such that \(y(\pi) = 0\).

1. Verification via differentiation:

\(\begin{aligned} \text{L.H.S.} &= [\cos x - (x+C)\sin x] + [(x+C)\cos x]\tan x \\ &= \cos x - (x+C)\sin x + (x+C)\sin x \\ &= \cos x = \text{R.H.S.} \end{aligned}\)

2. Solving the Initial Value Problem:

\(y(\pi) = (\pi + C)\cos \pi = -(\pi + C) = 0 \implies C = -\pi\)

Particular solution: \(y(x) = (x - \pi)\cos x\)

Example 6

Substitute \(y = e^{rx}\) into the D.E. \(3y'' + 4y' - 4y = 0\) to determine all values of the constant \(r\).

1. Verification via differentiation:

\(y = e^{rx}, \quad y' = re^{rx}, \quad y'' = r^2e^{rx}\)

2. Substitution:

\(3(r^2e^{rx}) + 4(re^{rx}) - 4(e^{rx}) = 0 \implies (3r^2 + 4r - 4)e^{rx} = 0\)

Since \(e^{rx} \neq 0\), we solve: \(3r^2 + 4r - 4 = 0 \implies (r+2)(3r-2) = 0\).

Values of constant: \(r = -2, \frac{2}{3}\).