Second-Order Differential Equations — Mathematics Knowledge System

Overview

A linear differential equation is one that is linear in the unknown function y and its derivatives — it can be written a₀(x)y + a₁(x)y′ + a₂(x)y″ + ⋯ + aₙ(x)y⁽ⁿ⁾ = b(x), where the coefficients a₀,…,aₙ and the forcing term b(x) are arbitrary functions of x (they need not themselves be linear). A second-order differential equation is the n=2 case, a₂(x)y″ + a₁(x)y′ + a₀(x)y = b(x), and is the most common order encountered in physics because Newton's second law relates acceleration (a second derivative) to force.

Intuition

A first-order equation needs one piece of information (an initial value) to pin down a unique solution; a second-order equation needs two, typically an initial position and an initial velocity, exactly like classical mechanics needs both a starting point and a starting speed to predict a trajectory. When the equation is linear with constant coefficients, exponential functions eˣ are natural building blocks because differentiation just multiplies eʳˣ by r — so plugging in y = eʳˣ turns the differential equation into an ordinary polynomial equation in r.

Formal Definition

Definition

The general linear second-order ODE, and the constant-coefficient homogeneous case whose solution structure is governed by a quadratic 'characteristic equation':

a_2(x)y'' + a_1(x)y' + a_0(x)y = b(x)

General linear 2nd-order ODE

ay'' + by' + cy = 0 \quad (a,b,c \text{ constant}, a \ne 0)

Constant-coefficient homogeneous case

ar^2 + br + c = 0

Characteristic equation, from trial solution y=e^{rx}

Notation

Notation	Meaning
$y''$	Second derivative of y with respect to x
$r$	Root of the characteristic equation ar² + br + c = 0
$\Delta = b^2 - 4ac$	Discriminant of the characteristic equation; its sign determines the solution type

Derivation

Substitute the trial solution y = eʳˣ into ay″ + by′ + cy = 0. Since y′ = re^{rx} and y″ = r²e^{rx}, dividing through by the never-zero factor e^{rx} yields the characteristic equation ar² + br + c = 0. Its roots determine the solution's form, in three cases governed by the discriminant Δ = b² - 4ac:

ar^2e^{rx} + bre^{rx} + ce^{rx} = 0 \implies ar^2+br+c=0

Divide out e^{rx} (never zero) to get a quadratic in r

y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \quad (\Delta > 0,\ r_1 \ne r_2 \text{ real})

Case 1: two distinct real roots — solution is a combination of two exponentials

y = (C_1 + C_2 x)e^{rx} \quad (\Delta = 0,\ r_1=r_2=r)

Case 2: repeated real root — second independent solution is xe^{rx}

y = e^{\alpha x}(C_1\cos\beta x + C_2 \sin\beta x) \quad (\Delta<0,\ r=\alpha\pm i\beta)

Case 3: complex conjugate roots α±iβ — solution is an oscillation with exponential envelope

Applications

Simple harmonic motion (mx″ + kx = 0), damped oscillators (mx″ + cx′ + kx = 0), and RLC electrical circuits are all governed by constant-coefficient linear second-order ODEs.

Worked Examples

Form the characteristic equation from y = e^{rx}.
$r^2 - 5r + 6 = 0$
Factor to find the two distinct real roots.
$(r-2)(r-3) = 0 \implies r_1 = 2,\ r_2 = 3$
Write the general solution as a combination of the two exponential solutions.
$y = C_1 e^{2x} + C_2 e^{3x}$

Answer: y = C₁e^(2x) + C₂e^(3x)

Practice Problems

Difficulty 6/10

Solve y″ + 2y′ + 5y = 0 (complex roots case).

Difficulty 5/10

Solve the simple harmonic oscillator equation x″ + 9x = 0 with x(0)=2, x′(0)=0.

Common Mistakes

Common Mistake

Using y = C₁e^{rx} + C₂e^{rx} (the same exponential twice) when the characteristic equation has a repeated root.

A repeated root only supplies one independent solution, e^{rx}; the second independent solution is xe^{rx}, giving y = (C₁+C₂x)e^{rx}.

Common Mistake

Writing the complex-root solution with imaginary exponentials, e.g. y = C₁e^{(\alpha+i\beta)x} + C₂e^{(\alpha-i\beta)x}, and stopping there.

That form is correct but not real-valued unless C₁, C₂ are complex conjugates; using Euler's formula it should be rewritten in real form as y = e^{αx}(C₁cos βx + C₂sin βx) for physical solutions.

Quiz

Summary

A second-order linear ODE with constant coefficients, ay″+by′+cy=0, is solved via the characteristic equation ar²+br+c=0.
Two distinct real roots r₁≠r₂ give y = C₁e^{r₁x} + C₂e^{r₂x}.
A repeated real root r gives y = (C₁+C₂x)e^{rx} — note the extra factor of x.
Complex conjugate roots α±iβ give an oscillating solution y = e^{αx}(C₁cos βx + C₂sin βx).
Two initial conditions (position and velocity/derivative) are needed to fix the two arbitrary constants.

References

WebsiteWikipedia — Linear differential equation
BookBoyce, W. & DiPrima, R. Elementary Differential Equations, 11th ed., Ch. 3.