The language of change in the natural world. Differential equations connect rates of change to the quantities themselves, giving us the power to model everything from population growth and heat flow to the motion of planets and the behavior of electrical circuits.
A differential equation is an equation that relates a function to one or more of its derivatives. While algebraic equations involve unknown numbers, differential equations involve unknown functions. They are the primary mathematical tool for modeling systems that change over time or space.
At its simplest, a differential equation is any equation that contains derivatives. For example:
This tells us that the rate of change of y with respect to x is 3x². Solving it means finding a function y(x) whose derivative equals 3x². In this case, y = x³ + C, where C is an arbitrary constant.
A more complex example:
Here we must find a function y(x) such that its second derivative plus five times its first derivative plus six times the function itself equals zero.
The order of a differential equation is the highest derivative that appears in the equation.
A differential equation is linear if the unknown function y and all its derivatives appear only to the first power and are not multiplied together. Otherwise it is nonlinear.
A linear ODE is homogeneous if the right-hand side is zero, and non-homogeneous if it is not:
A solution to a differential equation is a function that satisfies the equation when substituted in. There are several types:
Show that y = e²ˣ is a solution of y″ − 4y = 0.
Step 1: Compute derivatives: y = e²ˣ, y′ = 2e²ˣ, y″ = 4e²ˣ.
Step 2: Substitute into the equation: y″ − 4y = 4e²ˣ − 4e²ˣ = 0. ✓
Since the left side equals zero, y = e²ˣ is indeed a solution.
Solve dy/dx = 6x² with y(0) = 5.
Step 1: Integrate both sides: y = ∫ 6x² dx = 2x³ + C.
Step 2: Apply the initial condition y(0) = 5: 5 = 2(0)³ + C → C = 5.
Solution: y = 2x³ + 5.
A first-order ODE is separable if it can be written so that all terms involving y are on one side and all terms involving x are on the other. Separable equations are among the most common and easiest to solve.
A separable equation has the form:
We separate the variables:
Then integrate both sides:
Step 1: Separate variables: dy/y = x dx (assuming y ≠ 0).
Step 2: Integrate both sides: ∫ dy/y = ∫ x dx → ln|y| = x²/2 + C₁.
Step 3: Solve for y: |y| = e^(x²/2 + C₁) = e^C₁ · e^(x²/2).
Step 4: Let A = ±e^C₁ (an arbitrary nonzero constant): y = A · e^(x²/2).
Note: y = 0 is also a solution (the equilibrium solution when g(y) = y = 0). It corresponds to A = 0, so the general solution is y = Ae^(x²/2) for any constant A.
Step 1: Separate: dy/(1 + y²) = dx.
Step 2: Integrate: arctan(y) = x + C.
Step 3: Solve for y: y = tan(x + C).
Step 4: Apply IC y(0) = 1: 1 = tan(C), so C = π/4.
Solution: y = tan(x + π/4).
This solution is valid for −3π/4 < x < π/4, where the tangent is defined.
Step 1: Separate: dy/y² = cos(x) dx (for y ≠ 0).
Step 2: Integrate: ∫ y⁻² dy = ∫ cos(x) dx → −1/y = sin(x) + C.
Step 3: Apply IC y(0) = 1: −1/1 = sin(0) + C → C = −1.
Step 4: Solve for y: −1/y = sin(x) − 1 → y = 1/(1 − sin(x)).
Solution: y = 1/(1 − sin(x)), valid where sin(x) ≠ 1.
A first-order linear ODE has the standard form:
This equation is linear in y (no y², y·y′, etc.). Even when it is not separable, there is a systematic method using an integrating factor.
The key idea is to multiply both sides by a carefully chosen function μ(x) — the integrating factor — that makes the left side collapse into the derivative of a product.
Step 1: Write the equation in standard form: dy/dx + P(x)·y = Q(x).
Step 2: Compute the integrating factor:
Step 3: Multiply both sides by μ(x). The left side becomes d/dx[μ(x)·y]:
Step 4: Integrate both sides:
Step 5: Solve for y:
Step 1: P(x) = 2, Q(x) = e⁻ˣ.
Step 2: Integrating factor: μ = e^(∫ 2 dx) = e²ˣ.
Step 3: Multiply: e²ˣ·dy/dx + 2e²ˣ·y = e²ˣ·e⁻ˣ = eˣ.
Left side is d/dx[e²ˣ·y], so: d/dx[e²ˣ·y] = eˣ.
Step 4: Integrate: e²ˣ·y = ∫ eˣ dx = eˣ + C.
Step 5: Solve for y: y = e⁻²ˣ·(eˣ + C) = e⁻ˣ + Ce⁻²ˣ.
General solution: y = e⁻ˣ + Ce⁻²ˣ.
Step 1: Divide by x to get standard form: dy/dx − (3/x)·y = x³.
Here P(x) = −3/x and Q(x) = x³.
Step 2: Integrating factor: μ = e^(∫ −3/x dx) = e^(−3 ln|x|) = x⁻³ = 1/x³.
Step 3: Multiply by 1/x³: (1/x³)·dy/dx − (3/x⁴)·y = 1.
Left side: d/dx[y/x³] = 1.
Step 4: Integrate: y/x³ = x + C.
Step 5: Solve for y: y = x⁴ + Cx³.
Step 6: Apply IC y(1) = 2: 2 = 1 + C → C = 1.
Solution: y = x⁴ + x³.
An equation of the form:
is called exact if there exists a function F(x, y) such that:
In that case, the solution is simply:
By Clairaut's theorem, if F has continuous second partial derivatives, then ∂²F/∂y∂x = ∂²F/∂x∂y. This means:
Step 1: Verify ∂M/∂y = ∂N/∂x.
Step 2: Integrate M with respect to x (holding y constant): F = ∫ M dx + g(y), where g(y) is an unknown function of y.
Step 3: Differentiate F with respect to y and set it equal to N: ∂F/∂y = ∂/∂y[∫ M dx] + g′(y) = N. Solve for g′(y) and integrate to find g(y).
Step 4: Write the solution F(x, y) = C.
Step 1: Identify M = 2xy + 3 and N = x² + 4y.
Check: ∂M/∂y = 2x and ∂N/∂x = 2x. Since ∂M/∂y = ∂N/∂x, the equation is exact. ✓
Step 2: Integrate M with respect to x: F = ∫ (2xy + 3) dx = x²y + 3x + g(y).
Step 3: Differentiate F with respect to y: ∂F/∂y = x² + g′(y). Set equal to N: x² + g′(y) = x² + 4y → g′(y) = 4y → g(y) = 2y².
Step 4: The solution is F(x, y) = C:
Solution: x²y + 3x + 2y² = C.
Step 1: M = 3x²y + eʸ, N = x³ + xeʸ − 2y.
∂M/∂y = 3x² + eʸ, ∂N/∂x = 3x² + eʸ. Exact ✓
Step 2: F = ∫ (3x²y + eʸ) dx = x³y + xeʸ + g(y).
Step 3: ∂F/∂y = x³ + xeʸ + g′(y) = x³ + xeʸ − 2y → g′(y) = −2y → g(y) = −y².
Solution: x³y + xeʸ − y² = C.
A second-order linear ODE with constant coefficients has the general form:
where a, b, c are real constants and a ≠ 0. These equations arise naturally in physics—especially in the study of vibrations, oscillations, and electrical circuits.
We guess a solution of the form y = eʳˣ. Substituting into ay″ + by′ + cy = 0 gives:
Since eʳˣ ≠ 0, we need:
The nature of the roots r₁ and r₂ determines the form of the general solution:
r₁ = r₂ = r = −b/(2a). A single exponential is not enough; we need a second linearly independent solution:
r = α ± βi, where α = −b/(2a) and β = √(4ac − b²)/(2a). Using Euler's formula:
This represents oscillatory behavior with exponential growth or decay (depending on the sign of α).
Step 1: Write the characteristic equation: r² − 5r + 6 = 0.
Step 2: Factor: (r − 2)(r − 3) = 0, so r₁ = 2, r₂ = 3.
Step 3: Two distinct real roots → y = C₁e²ˣ + C₂e³ˣ.
Step 1: Characteristic equation: r² + 4r + 4 = 0.
Step 2: Factor: (r + 2)² = 0, so r = −2 (repeated root).
Step 3: Repeated root → y = (C₁ + C₂x)e⁻²ˣ.
Step 1: Characteristic equation: r² + 2r + 5 = 0.
Step 2: Discriminant: 4 − 20 = −16 < 0. Use the quadratic formula:
r = (−2 ± √(−16))/2 = (−2 ± 4i)/2 = −1 ± 2i.
So α = −1, β = 2.
Step 3: Complex roots → y = e⁻ˣ(C₁ cos(2x) + C₂ sin(2x)).
This represents a decaying oscillation since α = −1 < 0.
Step 1: Characteristic equation: r² + 9 = 0 → r = ±3i. Pure imaginary roots (α = 0, β = 3).
Step 2: General solution: y = C₁ cos(3x) + C₂ sin(3x).
Step 3: Apply y(0) = 2: C₁ cos(0) + C₂ sin(0) = C₁ = 2.
Step 4: y′ = −3C₁ sin(3x) + 3C₂ cos(3x). Apply y′(0) = −3: 3C₂ = −3 → C₂ = −1.
Solution: y = 2 cos(3x) − sin(3x). (Simple harmonic motion.)
A non-homogeneous second-order linear ODE has the form:
where f(x) ≠ 0 is the forcing function (or input, or driving term).
The general solution is the sum of two parts:
This method works when f(x) is a polynomial, exponential, sine, cosine, or a combination of these. We guess the form of y_p based on f(x), plug it into the equation, and solve for the unknown coefficients.
| f(x) | Guess for y_p |
|---|---|
| aₙxⁿ + … + a₁x + a₀ | Aₙxⁿ + … + A₁x + A₀ |
| aeᵏˣ | Aeᵏˣ |
| a cos(kx) + b sin(kx) | A cos(kx) + B sin(kx) |
| Product/sum of above | Product/sum of corresponding guesses |
Step 1: Solve the homogeneous equation: r² − 3r + 2 = 0 → (r−1)(r−2) = 0 → r = 1, 2.
y_h = C₁eˣ + C₂e²ˣ.
Step 2: Guess y_p = Ax + B (a polynomial matching f(x) = 4x).
Step 3: y_p′ = A, y_p″ = 0. Substitute: 0 − 3A + 2(Ax + B) = 4x.
2Ax + (2B − 3A) = 4x + 0.
Matching coefficients: 2A = 4 → A = 2; 2B − 3(2) = 0 → B = 3.
Solution: y = C₁eˣ + C₂e²ˣ + 2x + 3.
Step 1: Homogeneous: r² + 1 = 0 → r = ±i. y_h = C₁ cos(x) + C₂ sin(x).
Step 2: Guess y_p = A cos(2x) + B sin(2x).
Step 3: y_p″ = −4A cos(2x) − 4B sin(2x). Substitute:
−4A cos(2x) − 4B sin(2x) + A cos(2x) + B sin(2x) = 3 sin(2x).
−3A cos(2x) − 3B sin(2x) = 3 sin(2x).
Coefficients: −3A = 0 → A = 0; −3B = 3 → B = −1.
Solution: y = C₁ cos(x) + C₂ sin(x) − sin(2x).
This method works for any forcing function f(x), not just the special types above. Given a homogeneous solution y_h = C₁y₁ + C₂y₂, we seek a particular solution of the form:
where u₁ and u₂ are unknown functions determined by:
and W is the Wronskian:
Step 1: Homogeneous solution: y₁ = cos(x), y₂ = sin(x).
Step 2: Wronskian: W = cos(x)·cos(x) − sin(x)·(−sin(x)) = cos²(x) + sin²(x) = 1.
Step 3: Compute u₁′ and u₂′ (with f(x) = sec(x), since the equation is already in standard form with a = 1):
u₁′ = −sin(x)·sec(x)/1 = −sin(x)/cos(x) = −tan(x).
u₂′ = cos(x)·sec(x)/1 = cos(x)/cos(x) = 1.
Step 4: Integrate: u₁ = ∫ −tan(x) dx = ln|cos(x)|. u₂ = ∫ 1 dx = x.
Step 5: y_p = cos(x)·ln|cos(x)| + x·sin(x).
General solution: y = C₁cos(x) + C₂sin(x) + cos(x)·ln|cos(x)| + x·sin(x).
The Laplace transform is a powerful integral transform that converts a differential equation in the time domain into an algebraic equation in the frequency domain. After solving the algebraic equation, we apply the inverse Laplace transform to obtain the solution in the original variable.
The Laplace transform of a function f(t), defined for t ≥ 0, is:
provided the integral converges.
| f(t) | F(s) = ℒ{f(t)} |
|---|---|
| 1 | 1/s (s > 0) |
| tⁿ | n!/s^(n+1) (s > 0) |
| eᵃᵗ | 1/(s − a) (s > a) |
| sin(bt) | b/(s² + b²) |
| cos(bt) | s/(s² + b²) |
| eᵃᵗ sin(bt) | b/((s−a)² + b²) |
| eᵃᵗ cos(bt) | (s−a)/((s−a)² + b²) |
| tⁿeᵃᵗ | n!/(s−a)^(n+1) |
The procedure is:
Step 1: Take ℒ of both sides. Using the derivative property:
ℒ{y″} = s²Y − sy(0) − y′(0) = s²Y − s
ℒ{y′} = sY − y(0) = sY − 1
So: (s²Y − s) + 3(sY − 1) + 2Y = 0.
Step 2: Simplify: s²Y − s + 3sY − 3 + 2Y = 0 → Y(s² + 3s + 2) = s + 3.
Step 3: Solve for Y: Y(s) = (s + 3)/((s + 1)(s + 2)).
Step 4: Partial fractions: (s + 3)/((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2).
s + 3 = A(s + 2) + B(s + 1). Setting s = −1: 2 = A. Setting s = −2: 1 = −B → B = −1.
Y(s) = 2/(s + 1) − 1/(s + 2).
Step 5: Inverse transform: y(t) = 2e⁻ᵗ − e⁻²ᵗ.
Verification: y(0) = 2 − 1 = 1 ✓. y′(t) = −2e⁻ᵗ + 2e⁻²ᵗ, y′(0) = −2 + 2 = 0 ✓.
Step 1: Transform: (sY − 3) + 2Y = 1/(s + 1).
Step 2: Y(s + 2) = 3 + 1/(s + 1) = (3s + 4)/(s + 1).
Y(s) = (3s + 4)/((s + 1)(s + 2)).
Step 3: Partial fractions: (3s + 4)/((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2).
3s + 4 = A(s + 2) + B(s + 1). s = −1: 1 = A. s = −2: −2 = −B → B = 2.
Y(s) = 1/(s + 1) + 2/(s + 2).
Step 4: y(t) = e⁻ᵗ + 2e⁻²ᵗ.
Many real-world problems involve multiple interacting quantities, each changing over time. These are modeled by systems of differential equations—a set of coupled ODEs.
A system of two first-order linear ODEs can be written as:
In matrix form:
where A is the coefficient matrix.
For the homogeneous system x′ = Ax, we look for solutions of the form x(t) = veᵏᵗ, where v is a constant vector and λ is a scalar. Substituting gives:
So λ must be an eigenvalue of A, and v its corresponding eigenvector.
The general solution is:
Step 1: Write in matrix form: A = [[3, 1], [1, 3]].
Step 2: Characteristic equation: det(A − λI) = (3−λ)² − 1 = 0 → λ² − 6λ + 8 = 0 → (λ−2)(λ−4) = 0.
Eigenvalues: λ₁ = 2, λ₂ = 4.
Step 3 (λ₁ = 2): (A − 2I)v = 0: [[1,1],[1,1]][v₁,v₂]ᵀ = 0 → v₁ + v₂ = 0 → v₁ = [1, −1]ᵀ.
Step 4 (λ₂ = 4): (A − 4I)v = 0: [[−1,1],[1,−1]][v₁,v₂]ᵀ = 0 → −v₁ + v₂ = 0 → v₂ = [1, 1]ᵀ.
General solution:
x(t) = C₁e²ᵗ + C₂e⁴ᵗ
y(t) = −C₁e²ᵗ + C₂e⁴ᵗ
Both eigenvalues are positive, so this is an unstable node.
Step 1: A = [[0, −1], [1, 0]]. Characteristic equation: λ² + 1 = 0 → λ = ±i.
Step 2: Complex eigenvalues ±i with eigenvectors. For λ = i: v = [1, −i]ᵀ.
Step 3: The real-valued general solution is:
x(t) = C₁ cos(t) + C₂ sin(t)
y(t) = C₁ sin(t) − C₂ cos(t)
Step 4: Apply ICs: x(0) = C₁ = 1. y(0) = −C₂ = 0 → C₂ = 0.
Solution: x(t) = cos(t), y(t) = sin(t). The trajectory is a circle — a center.
Any nth-order ODE can be converted to a system of n first-order ODEs. For y″ + py′ + qy = f(t), let x₁ = y and x₂ = y′. Then:
This technique is essential for numerical methods and theoretical analysis.
Differential equations are the backbone of mathematical modeling in science and engineering. Here are some of the most important applications.
The simplest model assumes the rate of growth is proportional to the current population:
This is separable: P(t) = P₀eᵏᵗ, where P₀ is the initial population and k is the growth rate constant.
Exponential growth is unrealistic for large populations due to limited resources. The logistic equation introduces a carrying capacity K:
When P is small compared to K, growth is approximately exponential. As P approaches K, growth slows and eventually stops.
A population of bacteria follows dP/dt = 0.5P(1 − P/1000) with P(0) = 100. Find P(t).
Step 1: This is the logistic equation with k = 0.5 and K = 1000.
Step 2: The general solution to the logistic equation is:
P(t) = K / (1 + ((K − P₀)/P₀)·e⁻ᵏᵗ)
Step 3: Substitute values: K = 1000, P₀ = 100, k = 0.5:
P(t) = 1000 / (1 + 9e⁻⁰·⁵ᵗ)
Behavior: P(0) = 1000/(1+9) = 100 ✓. As t → ∞, P → 1000/1 = 1000 (carrying capacity).
The rate of change of an object's temperature is proportional to the difference between its temperature and the ambient temperature T_a:
This is a first-order linear ODE with solution:
A cup of coffee at 90°C is placed in a room at 20°C. After 5 minutes it is 70°C. When will it be 40°C?
Step 1: T(t) = 20 + 70e⁻ᵏᵗ (since T₀ − T_a = 90 − 20 = 70).
Step 2: Use T(5) = 70: 70 = 20 + 70e⁻⁵ᵏ → 50 = 70e⁻⁵ᵏ → e⁻⁵ᵏ = 5/7.
k = −ln(5/7)/5 = ln(7/5)/5 ≈ 0.0673.
Step 3: Solve T(t) = 40: 40 = 20 + 70e⁻ᵏᵗ → 20 = 70e⁻ᵏᵗ → e⁻ᵏᵗ = 2/7.
t = −ln(2/7)/k = ln(7/2)/k ≈ 1.2528/0.0673 ≈ 18.6 minutes.
A mass m attached to a spring with spring constant k (and damping coefficient c) satisfies:
where y is the displacement from equilibrium and F(t) is any external force.
A 1 kg mass on a spring (k = 10 N/m) with damping c = 2 N·s/m. Find y(t) with y(0) = 0.5, y′(0) = 0.
Step 1: The ODE: y″ + 2y′ + 10y = 0.
Step 2: Characteristic equation: r² + 2r + 10 = 0 → r = (−2 ± √(4−40))/2 = −1 ± 3i.
Step 3: General solution: y = e⁻ᵗ(C₁ cos(3t) + C₂ sin(3t)).
Step 4: y(0) = C₁ = 0.5. y′ = e⁻ᵗ(−C₁ cos(3t) − C₂ sin(3t) − 3C₁ sin(3t) + 3C₂ cos(3t)).
y′(0) = −C₁ + 3C₂ = 0 → C₂ = C₁/3 = 1/6.
Solution: y(t) = e⁻ᵗ(0.5 cos(3t) + (1/6) sin(3t)).
The system oscillates at frequency 3 rad/s with exponentially decaying amplitude.
An RLC series circuit with resistance R, inductance L, and capacitance C, driven by a voltage E(t), satisfies:
Since I = dq/dt, this becomes a second-order ODE in the charge q:
This is mathematically identical to the spring-mass equation! The analogy is: L ↔ m, R ↔ c, 1/C ↔ k, E(t) ↔ F(t).
An RL circuit (no capacitor) with L = 2 H, R = 6 Ω, and E = 12 V. Find I(t) with I(0) = 0.
Step 1: The ODE: 2 dI/dt + 6I = 12 → dI/dt + 3I = 6.
Step 2: Integrating factor: μ = e³ᵗ. d/dt[e³ᵗ I] = 6e³ᵗ.
Step 3: Integrate: e³ᵗ I = 2e³ᵗ + C → I = 2 + Ce⁻³ᵗ.
Step 4: IC: I(0) = 0 → 0 = 2 + C → C = −2.
Solution: I(t) = 2(1 − e⁻³ᵗ) amperes.
The current rises from 0 and approaches the steady-state value I = E/R = 2 A.
Partial differential equations (PDEs) involve functions of several variables and their partial derivatives. PDEs are essential for describing phenomena that depend on multiple independent variables, such as position and time.
The heat equation models how temperature u(x, t) evolves in a rod over time:
where k > 0 is the thermal diffusivity. This is a parabolic PDE.
To solve the heat equation, we typically need:
We assume u(x, t) = X(x)·T(t). Substituting into the PDE:
Since the left side depends only on t and the right only on x, both must equal a constant −λ. This gives two ODEs:
With boundary conditions u(0,t) = u(L,t) = 0, we need X(0) = 0 and X(L) = 0. The eigenvalues and eigenfunctions are:
The corresponding time solutions are:
The general solution is a Fourier sine series:
where the coefficients Bₙ are determined by the initial condition via Bₙ = (2/L) ∫₀ᴸ f(x) sin(nπx/L) dx.
Solve ∂u/∂t = ∂²u/∂x² on 0 < x < π with u(0,t) = u(π,t) = 0 and u(x,0) = sin(x) + 3sin(2x).
Step 1: Here L = π, k = 1, so λₙ = n² and the general solution is u = Σ Bₙ sin(nx) e^(−n²t).
Step 2: Apply the initial condition: u(x,0) = Σ Bₙ sin(nx) = sin(x) + 3sin(2x).
By inspection: B₁ = 1, B₂ = 3, Bₙ = 0 for n ≥ 3.
Solution: u(x, t) = sin(x)·e⁻ᵗ + 3sin(2x)·e⁻⁴ᵗ.
The higher-frequency component (sin 2x) decays faster (e⁻⁴ᵗ vs. e⁻ᵗ), which is characteristic of diffusion.
The wave equation models vibrations (strings, membranes, sound):
where c is the wave speed. This is a hyperbolic PDE.
For a string of length L fixed at both ends (u(0,t) = u(L,t) = 0), with initial displacement u(x,0) = f(x) and initial velocity uₜ(x,0) = g(x), separation of variables gives:
where:
Solve ∂²u/∂t² = 4 · ∂²u/∂x² on 0 < x < 1 with u(0,t) = u(1,t) = 0, u(x,0) = sin(πx), uₜ(x,0) = 0.
Step 1: c = 2, L = 1. General solution: u = Σ [Aₙ cos(2nπt) + Bₙ sin(2nπt)] sin(nπx).
Step 2: uₜ(x,0) = 0 → all Bₙ = 0. u(x,0) = sin(πx) → A₁ = 1, Aₙ = 0 for n ≥ 2.
Solution: u(x, t) = sin(πx)·cos(2πt).
This is a standing wave — the string oscillates with period T = 1 (since 2π/(2π) = 1).
Laplace's equation models steady-state (time-independent) phenomena such as electrostatics, steady-state heat distribution, and incompressible fluid flow:
This is an elliptic PDE. Solutions are called harmonic functions and have many remarkable properties:
Solve ∇²u = 0 on the rectangle 0 < x < a, 0 < y < b with:
u(0,y) = 0, u(a,y) = 0, u(x,0) = 0, u(x,b) = f(x).
Step 1: Separation of variables u = X(x)Y(y) gives: X″/X = −Y″/Y = −λ.
Step 2: With X(0) = 0, X(a) = 0: Xₙ = sin(nπx/a), λₙ = (nπ/a)².
Step 3: Y equation: Y″ − (nπ/a)²Y = 0 with Y(0) = 0. Solution: Yₙ = sinh(nπy/a).
Step 4: u(x,y) = Σ Cₙ sin(nπx/a) sinh(nπy/a).
Step 5: Apply u(x,b) = f(x): Cₙ sinh(nπb/a) = (2/a)∫₀ᵃ f(x)sin(nπx/a)dx.
A general second-order PDE in two variables can be written as:
The classification depends on the discriminant B² − AC: