Repeated Roots - Differential Equations

SECTION 1

Historical Context & Motivation

Differential equations became essential tools for describing the natural world during the 1700s and 1800s, when mathematicians like Euler, d'Alembert, and Lagrange studied how physical systems — springs, pendulums, vibrating strings — change over time. A second-order linear ODE (ordinary differential equation) involves a function, its first derivative, and its second derivative, all combined in a linear way. Finding the general solution to such equations requires solving a polynomial called the characteristic equation. Most of the time, this polynomial yields two distinct roots, and the solution strategy is straightforward. But a fascinating special case arises when those two roots are exactly the same — the repeated-root case.

1739 • Euler's Exponential Solutions

Leonhard Euler showed that exponential functions e^rx naturally solve constant-coefficient linear ODEs, introducing the idea of a characteristic equation.

1762 • D'Alembert's Reduction of Order

Jean le Rond d'Alembert developed the reduction of order method, which provides a systematic way to find a second, independent solution when one solution is already known.

1788 • Lagrange Formalizes the Theory

Joseph-Louis Lagrange unified the treatment of second-order linear ODEs, clearly distinguishing the distinct-root, repeated-root, and complex-root cases.

1800s • Engineering Applications

The repeated-root case appeared in studies of critically damped mechanical and electrical systems, making it central to 19th-century engineering and physics.

Here's the core question that motivates this lesson: if the characteristic equation of a second-order ODE has two identical roots, the usual approach only produces one independent solution. But a second-order equation demands two independent solutions to form a complete general solution. How do we find the missing one? That's exactly what this lesson will teach you.

SECTION 2

Core Principles & Definitions

Before we tackle the repeated-root case directly, let's make sure the foundational ideas are clear. We are working with second-order, linear, homogeneous ODEs that have constant coefficients. The standard form looks like ay″ + by′ + cy = 0, where a, b, and c are constants and a ≠ 0. Solving these equations always begins with the same move: assume a trial solution of the form y = e^rx and substitute it into the ODE. This produces the characteristic equation ar² + br + c = 0, whose roots determine the nature of the general solution.

Characteristic Equation

Replacing y with e^rx in ay″ + by′ + cy = 0 yields the quadratic ar² + br + c = 0. Its roots tell you the form of the solution.

Discriminant Test

The discriminant Δ = b² − 4ac determines three cases: Δ > 0 gives two distinct real roots, Δ < 0 gives complex roots, and Δ = 0 gives a repeated root.

Linear Independence

A second-order ODE needs two linearly independent solutions. Two functions are independent if one is not a constant multiple of the other.

The Repeated Root Fix

When the discriminant is zero, the single root r gives only y₁ = e^rx. The second independent solution is y₂ = xe^rx.

General Solution

For repeated root r, the general solution is y = (C₁ + C₂x)e^rx, where C₁ and C₂ are arbitrary constants.

✦

KEY TAKEAWAY

Think of solving an ODE like needing two different keys to fully unlock a box. Normally, the characteristic equation hands you two different keys (two distinct roots). But when both roots are the same, it only gives you one key. The trick is to multiply by x — that's like cutting a slightly different groove into the same blank so it opens a different lock. The factor of x produces a genuinely new, independent solution.

SECTION 3

Visual Explanation

The diagram below shows the three possible cases for the roots of the characteristic equation, focusing on how the repeated-root case sits right on the boundary between two distinct real roots and two complex roots. Understanding this boundary position is key to seeing why repeated roots matter in both mathematics and physics.

The three panels correspond to the three cases of the discriminant. When Δ > 0 (left), two distinct exponential solutions exist. When Δ = 0 (center, thick gold border), we get a repeated root and must multiply by x to build the second solution. When Δ < 0 (right), the solution involves oscillation. Notice how the repeated-root case sits exactly on the boundary.

Notice the critical difference in the center panel. Because both roots are the same, you can't just write C₁e^rx + C₂e^rx — that would simplify to (C₁ + C₂)e^rx, which is really just one solution with a single constant. The extra factor of x in the dashed curve is what makes the second solution genuinely different from the first.

SECTION 4

Mathematical Framework

Let's build the solution step by step. We start with the general form of a second-order, linear, homogeneous ODE with constant coefficients.

STANDARD FORM

ay″ + by′ + cy = 0

Here a, b, and c are real constants with a ≠ 0. The primes denote derivatives with respect to x.

We guess a solution of the form y = e^rx. Substituting into the ODE gives y′ = re^rx and y″ = r²e^rx. Factoring out e^rx (which is never zero) produces the characteristic equation.

CHARACTERISTIC EQUATION

ar² + br + c = 0

This is a standard quadratic in r. Its solutions are r = (−b ± √(b² − 4ac)) / (2a).

The repeated-root case occurs exactly when the discriminant equals zero: b² − 4ac = 0. In that situation the quadratic formula simplifies to a single value r = −b / (2a). This one root gives us only one solution: y₁ = e^rx. We need a second independent solution.

REPEATED ROOT CONDITION

b² − 4ac = 0 ⟹ r = −b / (2a)

The discriminant is zero, so the quadratic has a double root. The characteristic polynomial factors as a(r − r₀)² = 0.

To find the missing second solution, we use a technique called reduction of order. We guess that the second solution has the form y₂ = v(x) × e^rx for some unknown function v(x). After substituting this into the ODE and using the fact that r is a double root, all terms involving v and v′ vanish, leaving only v″ = 0. Integrating twice gives v = C₁ + C₂x. The simplest non-trivial new piece is v(x) = x, so the second independent solution is y₂ = xe^rx.

GENERAL SOLUTION — REPEATED ROOTS

y = (C₁ + C₂x)e^(rx) where r = −b / (2a)

C₁ and C₂ are arbitrary constants determined by initial conditions. The factor of x in the second term is what makes the two parts linearly independent.

SECTION 5

Why Multiplying by x Works

You might wonder: why does multiplying by x produce a valid, independent solution? Let's break down the logic more carefully. When you substitute y₂ = v(x)e^rx into ay″ + by′ + cy = 0 and expand using the product rule, you get a new ODE in v(x). The key insight is that because r is a double root, the terms with v(x) itself and v′(x) both cancel out. You're left with av″e^rx = 0, which means v″ = 0. Integrating once gives v′ = constant, and integrating again gives v = C₁ + C₂x. The C₁ part just reproduces the first solution we already knew, so the genuinely new piece is the C₂x term. That's why y₂ = xe^rx is the correct second solution.

The left column shows the five-step reduction-of-order process. The right panel explains why the v and v′ terms cancel: the first vanishes because r satisfies the characteristic equation, and the second vanishes because r is specifically a double root (meaning 2ar + b = 0). Only v″ = 0 survives, giving v = x.

💡

Important Detail

The condition 2ar + b = 0 is exactly what you get by taking the derivative of the characteristic polynomial ar² + br + c. This is no coincidence — a repeated root of a polynomial is always a root of its derivative as well. This is why the v′ term drops out only in the repeated-root case, and not when the roots are distinct.

SECTION 6

Worked Example

Let's solve a complete initial value problem (IVP) involving a repeated root. We'll find the general solution and then use initial conditions to determine the specific constants.

Solve y″ − 6y′ + 9y = 0, with y(0) = 2 and y′(0) = 5

1
Step 1 — Write the Characteristic Equation

Assume y = e^rx. Substituting into y″ − 6y′ + 9y = 0 and dividing by e^rx gives the characteristic equation.

r² − 6r + 9 = 0
2
Step 2 — Solve for r

Factor the quadratic: r² − 6r + 9 = (r − 3)² = 0. This gives a repeated root.

r = 3 (double root)
3
Step 3 — Verify via Discriminant

Check: b² − 4ac = (−6)² − 4(1)(9) = 36 − 36 = 0. The discriminant is zero, confirming the repeated-root case.

Δ = 0 ✓
4
Step 4 — Write the General Solution

For a repeated root r = 3, the general solution is y = (C₁ + C₂x)e^3x. This includes both the standard solution e^3x and the "extra" solution xe^3x.

y = (C₁ + C₂x)e^3x
5
Step 5 — Apply Initial Condition y(0) = 2

Plug in x = 0: y(0) = (C₁ + C₂ × 0)e⁰ = C₁ × 1 = C₁. So C₁ = 2.

C₁ = 2
6
Step 6 — Find y′ and Apply y′(0) = 5

Using the product rule: y′ = C₂e^3x + 3(C₁ + C₂x)e^3x. At x = 0: y′(0) = C₂ + 3C₁ = C₂ + 6 = 5. Therefore C₂ = −1.

C₂ = −1
7
Step 7 — State the Final Answer

Substituting C₁ = 2 and C₂ = −1 into the general solution gives the particular solution to the IVP.

y = (2 − x)e^3x

SECTION 7

Comparing All Three Root Cases

To solidify your understanding of repeated roots, it helps to see all three cases side by side. The table below summarizes the discriminant condition, the nature of the roots, the form of the general solution, and a typical behavior you'd see in the graph. Each case is fundamentally determined by a single number: the discriminant Δ = b² − 4ac.

Feature	Distinct Real Roots	Repeated Root	Complex Roots
Discriminant	Δ > 0	Δ = 0	Δ < 0
Roots	r₁, r₂ (real, different)	r (one value, multiplicity 2)	α ± βi (complex pair)
General Solution	C₁ $e^{r₁x}$ + C₂ $e^{r₂x}$	(C₁ + C₂x) $e^{rx}$	$e^{αx}$ [C₁cos(βx) + C₂sin(βx)]
Typical Behavior	Sum of two different exponentials	Exponential × linear polynomial	Exponential envelope × oscillation
Physical Example	Overdamped spring	Critically damped spring	Underdamped spring (oscillation)

Summary of all three cases for ay″ + by′ + cy = 0

✦

KEY TAKEAWAY

Imagine pushing a door closed on a hydraulic hinge. If the hinge is too loose (overdamped/distinct real roots), the door creeps shut slowly. If the hinge is too tight (underdamped/complex roots), the door swings past the frame and bounces back and forth. The repeated-root case is like a perfectly tuned hinge — the door returns to the closed position as quickly as possible without any bouncing. Engineers call this critical damping, and it is exactly the boundary between the other two behaviors.

SECTION 8

Connection to Advanced Theory

The repeated-root idea appears again in several advanced settings. For higher-order ODEs (third-order, fourth-order, and beyond), a root of multiplicity k generates k independent solutions of the form e^rx, xe^rx, x²e^rx, …, x^k−1e^rx. The second-order case is just the simplest instance of this general pattern.

Concept	This Lesson (2nd-Order)	Advanced Extension
Order	Second-order (n = 2)	n-th order for any n
Multiplicity	Root repeated exactly twice	Root repeated k times (k ≤ n)
Extra solutions	Multiply by x once → $xe^{rx}$	Multiply by x⁰, x¹, …, $x^{k−1}$
Coefficients	Constant coefficients only	Variable coefficients (Cauchy–Euler, Frobenius method)
Nonhomogeneous	Not covered here	Variation of parameters or undetermined coefficients

How repeated roots generalize to higher-order and more complex ODEs

If you continue studying differential equations, you'll also encounter the method of undetermined coefficients for nonhomogeneous equations. In that setting, the repeated-root concept matters again: if the forcing term on the right-hand side happens to be a solution of the homogeneous equation, you must multiply your guess by an extra factor of x — the same kind of fix you learned here. The underlying reason is identical: you need linearly independent functions.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the general solution for the repeated-root case cannot simply be y = C₁e^rx + C₂e^rx. What goes wrong, and what must be done to fix it?

Show answer

If both roots are the same value r, then C₁

e^{rx}

+ C₂

e^{rx}

simplifies to (C₁ + C₂)

e^{rx}

, which is just a single constant times

e^{rx}

. This means you have only one independent solution instead of the two that a second-order ODE requires. The two functions

e^{rx}

and

e^{rx}

are not linearly independent — one is a constant multiple of the other (specifically, 1 × the other). To fix this, you multiply the second solution by x, giving y₂ =

xe^{rx}

. The functions

e^{rx}

and

xe^{rx}

are linearly independent because one cannot be obtained by multiplying the other by a constant.

Question 2: BASIC CALCULATION

Find the general solution of y″ + 4y′ + 4y = 0.

Show answer

The characteristic equation is r² + 4r + 4 = 0. Factoring gives (r + 2)² = 0, so r = −2 is a repeated root. You can verify by computing the discriminant: Δ = 4² − 4(1)(4) = 16 − 16 = 0. Since r = −2 is a double root, the general solution is y = (C₁ + C₂x)

e^{−2x}

. The two independent solutions that form this are y₁ =

e^{−2x}

and y₂ =

xe^{−2x}

Question 3: INTERMEDIATE

Solve the initial value problem: 4y″ − 12y′ + 9y = 0, with y(0) = 1 and y′(0) = 4.

Show answer

The characteristic equation is 4r² − 12r + 9 = 0. The discriminant is (−12)² − 4(4)(9) = 144 − 144 = 0, confirming a repeated root. Using the quadratic formula: r = 12 / (2 × 4) = 3/2. The general solution is y = (C₁ + C₂x)

e^{3x/2}

. Applying y(0) = 1: C₁ × e⁰ = C₁ = 1. Next, compute y′ = C₂

e^{3x/2}

+ (3/2)(C₁ + C₂x)

e^{3x/2}

. At x = 0: y′(0) = C₂ + (3/2)(1) = C₂ + 3/2 = 4, so C₂ = 5/2. The particular solution is y = (1 + (5/2)x)

e^{3x/2}

Question 4: APPLIED

A spring-dashpot system is modeled by mx″ + cx′ + kx = 0, where m = 2 kg, k = 50 N/m, and the system is critically damped. Find the value of the damping coefficient c and write the general solution for the displacement x(t).

Show answer

Critical damping corresponds to the repeated-root case, which requires the discriminant to be zero: c² − 4mk = 0. Substituting m = 2 and k = 50: c² − 4(2)(50) = 0, so c² = 400 and c = 20 N·s/m. The characteristic equation becomes 2r² + 20r + 50 = 0, or r² + 10r + 25 = 0, which factors as (r + 5)² = 0. The repeated root is r = −5. The general solution is x(t) = (C₁ + C₂t)

e^{−5t}

. The negative exponent ensures the displacement decays to zero, and the factor of t allows the system to return to equilibrium as fast as possible without oscillating — the hallmark of critical damping.

Question 5: CRITICAL THINKING

Consider the ODE y″ − 2αy′ + α²y = 0, where α is a real constant. Show that for every value of α, this equation has a repeated root. Then find a specific solution satisfying y(0) = 0 and y(1) = e^α, and describe how the solution's behavior changes as α goes from negative to positive.

Show answer

The characteristic equation is r² − 2αr + α² = 0, which factors as (r − α)² = 0 for every real α. The discriminant is (−2α)² − 4(1)(α²) = 4α² − 4α² = 0, confirming a repeated root r = α for all values of α. The general solution is y = (C₁ + C₂x)

e^{αx}

. Applying y(0) = 0: C₁ = 0, so y = C₂

xe^{αx}

. Applying y(1) = e^α: C₂ × 1 × e^α = e^α, so C₂ = 1 and the solution is y =

xe^{αx}

. When α < 0, the exponential decay eventually dominates the linear growth of x, so the solution rises briefly and then decays toward zero. When α = 0, the solution is simply y = x (a straight line). When α > 0, both the exponential and the linear factor grow, so the solution increases without bound as x → ∞. The parameter α thus controls whether the system decays, stays linear, or grows exponentially.

SUMMARY

Lesson Summary

Summary

When solving a second-order linear homogeneous ODE with constant coefficients (ay″ + by′ + cy = 0), the characteristic equation ar² + br + c = 0 determines the form of the solution. The repeated-root case occurs when the discriminant b² − 4ac equals zero, producing a single root r = −b/(2a) of multiplicity two. Because two identical exponential solutions are not linearly independent, we use reduction of order to discover the second solution y₂ = $xe^{rx}$ .

The general solution for the repeated-root case is y = (C₁ + C₂x) $e^{rx}$ , where C₁ and C₂ are constants fixed by initial conditions. Physically, this case corresponds to critical damping — the fastest return to equilibrium without oscillation. Mastering this special case prepares you for higher-order repeated roots, nonhomogeneous equations, and the broader theory of linear differential equations.