---
title: Homogeneous Equations, the Wronskian, and Real Roots
module: Second-Order Linear Equations
moduleNumber: 3
lessonNumber: 1
order: 301
summary: >
  A second-order linear homogeneous equation with constant coefficients is
  solved by guessing an exponential and reducing to the quadratic characteristic
  equation. Two solutions span every solution exactly when their Wronskian is
  nonzero; that condition, superposition, and Abel's formula give the full
  structure of the general solution for the case of two distinct real roots.
topics: [Second-Order Linear Equations]
draft: false
sources:
  - book: Boyce
    ref: "Ch. 3 — Second-Order Linear Equations; §3.1 Homogeneous Equations with Constant Coefficients; §3.2 Solutions of Linear Homogeneous Equations; the Wronskian"
  - book: Simmons
    ref: "Ch. 3 §15 General Solution of the Homogeneous Equation; §17 The Use of Operators"
---

A second-order equation involves the unknown function and its first two
derivatives. The linear ones are the tractable case, and among those the
constant-coefficient homogeneous equation

$$
a y'' + b y' + c y = 0, \qquad a \neq 0,
$$

is where the theory is cleanest. With two distinct real roots of the associated
quadratic, its general solution is a two-parameter family of exponentials. Three
structural facts justify that claim and carry over unchanged to the complex- and
repeated-root cases: superposition, the Wronskian test, and Abel's formula for
the Wronskian.

## The linear operator

Write $p$ and $q$ for continuous coefficient functions on an open interval $I$.
The **second-order linear differential operator** $L$ acts on any twice-differentiable
$\phi$ by

$$
L[\phi] = \phi'' + p\,\phi' + q\,\phi,
$$

producing another function. In operator shorthand $L = D^2 + pD + q$, where $D$
is differentiation. The equation we study is $L[y] = 0$, usually written

$$
L[y] = y'' + p(t)\, y' + q(t)\, y = 0. \tag{H}
$$

The constant-coefficient case is $p(t) = b/a$, $q(t) = c/a$ (after dividing by
$a$), but the general theory below needs only that $p$ and $q$ be
continuous.

> **Definition (Second-order linear homogeneous equation).** An equation
> $y'' + p(t)y' + q(t)y = 0$ with $p, q$ continuous on $I$. It is
> **homogeneous** because the right-hand side is zero, so $y \equiv 0$ is always
> a solution (the _trivial_ solution). "Linear" means $L$ satisfies
> $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]$.

Linearity has two immediate consequences.

- **Existence and uniqueness.** An initial value problem for a linear equation
  has exactly one solution, defined on the whole interval where the coefficients
  are continuous.
- **Superposition.** Sums and scalar multiples of solutions are again solutions,
  so the solution set is a vector space.

Nonlinear second-order equations have neither guarantee in such a clean form,
which is why the linear theory is developed first.[^boyce31]

> **Theorem (Existence and Uniqueness).** Consider
> $y'' + p(t)y' + q(t)y = g(t)$ with $y(t_0) = y_0$, $y'(t_0) = y_0'$, where
> $p, q, g$ are continuous on an open interval $I$ containing $t_0$. This problem
> has exactly one solution $y = \phi(t)$, and that solution exists throughout
> $I$.

The theorem covers the nonhomogeneous equation, which is needed later. It makes
three claims: a solution exists, it is unique, and it is defined on all of $I$
rather than only near $t_0$. Unlike
the first-order case, no formula for $\phi$ is available in general, so the proof
proceeds by abstract methods and is not reproduced here.[^boyce32a] For the
constant-coefficient equation we can exhibit solutions directly, which settles
existence; uniqueness still leans on the theorem.

$$
% caption: Applying $L$ to a solution returns the zero function; the operator
% is linear, so any combination of solutions is annihilated in the same way.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=20mm, minimum height=11mm, align=center}]
  \definecolor{acc}{HTML}{2A6F97}
  \node[box] (y1) at (0,1.1) {$y_1$};
  \node[box] (y2) at (0,-1.1) {$y_2$};
  \node[box, draw=acc, text=acc, thick] (comb) at (4,0) {$c_1 y_1 + c_2 y_2$};
  \node[box] (L) at (8,0) {$L[\,\cdot\,]$};
  \node[box] (zero) at (11.6,0) {$0$};
  \draw[->, acc, thick] (y1) -- (comb);
  \draw[->, acc, thick] (y2) -- (comb);
  \draw[->, thick] (comb) -- (L);
  \draw[->, thick] (L) -- (zero);
  \node[anchor=south, font=\scriptsize, text=black] at (2,1.15) {solutions of (H)};
  \node[anchor=north, font=\scriptsize, text=black] at (9.8,-0.15) {again a solution};
\end{tikzpicture}
$$

## Solving the constant-coefficient equation

For $a y'' + b y' + c y = 0$ the coefficients are constant, so try a solution
that reproduces itself under differentiation up to a scalar: $y = e^{rt}$. Then
$y' = re^{rt}$ and $y'' = r^2 e^{rt}$, and substitution gives

$$
(a r^2 + b r + c)\, e^{rt} = 0.
$$

Since $e^{rt}$ is never zero, $y = e^{rt}$ solves the equation exactly when $r$
is a root of the **characteristic equation**.

> **Definition (Characteristic equation).** For $a y'' + b y' + c y = 0$, the
> quadratic $a r^2 + b r + c = 0$ obtained by substituting $y = e^{rt}$. Its
> roots $r_1, r_2$ determine the form of the solution. The case of real,
> distinct roots is the discriminant condition $b^2 - 4ac > 0$.

With two distinct real roots $r_1 \neq r_2$, the functions $y_1 = e^{r_1 t}$ and
$y_2 = e^{r_2 t}$ are both solutions, and by superposition so is

$$
y = c_1 e^{r_1 t} + c_2 e^{r_2 t}. \tag{$\ast$}
$$

This two-parameter family is the general solution: below we verify that its
Wronskian is nonzero — the condition that guarantees it contains
every solution.

> **Worked example (Distinct real roots).** Solve $y'' + 5y' + 6y = 0$. The
> characteristic equation is $r^2 + 5r + 6 = (r+2)(r+3) = 0$, so $r_1 = -2$,
> $r_2 = -3$ and the general solution is $y = c_1 e^{-2t} + c_2 e^{-3t}$. To
> satisfy $y(0) = 2$, $y'(0) = 3$: the two conditions give $c_1 + c_2 = 2$ and
> $-2c_1 - 3c_2 = 3$, so $c_1 = 9$, $c_2 = -7$ and
> $y = 9e^{-2t} - 7e^{-3t}$.

Both exponents are negative, so this solution decays to zero; its positive
initial slope forces a single interior maximum before the decay dominates. When
the roots have opposite signs one exponential grows and the solution is
eventually unbounded, and when a root is zero the corresponding term is a
constant. The sign pattern of $(r_1, r_2)$ determines the long-run behavior.

$$
% caption: The two real roots as points on the $r$-axis; the sign pattern of
% $(r_1, r_2)$ fixes whether the solution decays, grows, or tends to a constant.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{2A6F97}
  \draw[->, thick] (-4.2,0) -- (4.4,0) node[right] {$r$};
  \draw[thick] (0,-0.12) -- (0,0.12);
  \node[anchor=south] at (0,0.14) {$0$};
  \fill[acc] (-2.6,0) circle (2.4pt);
  \fill[acc] (-1.1,0) circle (2.4pt);
  \node[acc, anchor=south] at (-2.6,0.12) {$r_2$};
  \node[acc, anchor=south] at (-1.1,0.12) {$r_1$};
  \node[draw, align=center, font=\scriptsize, minimum height=9mm] at (-2.5,-1.5)
    {both roots $<0$\\ solution decays to $0$};
  \node[draw, align=center, font=\scriptsize, minimum height=9mm] at (1.6,-1.5)
    {a root $>0$\\ solution grows without bound};
  \draw[->, black] (-2.0,-1.02) -- (-1.85,-0.15);
  \draw[->, black] (1.1,-1.02) -- (0.9,-0.15);
\end{tikzpicture}
$$

The exponentials themselves and their combinations are simple to picture: each
$e^{rt}$ is monotone, and a linear combination of two decaying exponentials with
opposite-sign coefficients can dip or peak once but cannot oscillate.

$$
% caption: Two decaying exponentials $e^{-2t}$, $e^{-3t}$ and the combination
% $9e^{-2t} - 7e^{-3t}$, which rises to a single maximum then decays.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{2A6F97}
  \draw[->, thick] (0,-0.3) -- (0,3.0) node[above] {$y$};
  \draw[->, thick] (-0.3,0) -- (5.4,0) node[right] {$t$};
  % e^{-2t} scaled
  \draw[black, thick] (0,2.4) .. controls (0.7,1.35) and (1.6,0.6) .. (2.6,0.35)
    .. controls (3.6,0.18) and (4.5,0.1) .. (5.2,0.07);
  \node[black, anchor=west, font=\scriptsize] at (2.7,0.55) {$e^{-2t}$};
  % e^{-3t} scaled
  \draw[black, thick, dashed] (0,1.7) .. controls (0.5,0.75) and (1.2,0.32) .. (2.0,0.18)
    .. controls (3.0,0.09) and (4.0,0.05) .. (5.2,0.03);
  \node[black, anchor=west, font=\scriptsize] at (1.05,0.95) {$e^{-3t}$};
  % combination 9e^{-2t} - 7e^{-3t}: starts at 2, peaks ~2.2 near t=0.15, decays
  \draw[acc, very thick] (0,0.9) .. controls (0.25,1.7) and (0.55,1.98) .. (0.9,1.9)
    .. controls (1.6,1.62) and (2.6,0.95) .. (3.6,0.5)
    .. controls (4.3,0.28) and (4.9,0.18) .. (5.2,0.14);
  \node[acc, anchor=west, font=\scriptsize] at (3.5,0.75) {$9e^{-2t} - 7e^{-3t}$};
\end{tikzpicture}
$$

## Superposition and the two-parameter family

Superposition is what turns two particular solutions into an infinite family.
The proof is a direct computation with the operator.

> **Theorem (Principle of Superposition).** If $y_1$ and $y_2$ solve (H), then
> $c_1 y_1 + c_2 y_2$ solves (H) for all constants $c_1, c_2$.

By linearity $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2] = c_1 \cdot 0 +
c_2 \cdot 0 = 0$. Taking $c_2 = 0$ recovers the special case that any scalar
multiple of a solution is a solution. The next question is whether the family
$c_1 y_1 + c_2 y_2$ captures _all_ solutions, or only some of them. That reduces
to whether the constants can always be tuned to meet a prescribed pair of
initial values.

Imposing $y(t_0) = y_0$ and $y'(t_0) = y_0'$ on $c_1 y_1 + c_2 y_2$ produces the
linear system

$$
\begin{aligned}
c_1 y_1(t_0) + c_2 y_2(t_0) &= y_0, \\
c_1 y_1'(t_0) + c_2 y_2'(t_0) &= y_0'.
\end{aligned}
$$

This $2 \times 2$ system has a unique solution $(c_1, c_2)$ for every right-hand
side precisely when its coefficient determinant is nonzero. That determinant is
the Wronskian.

## The Wronskian and fundamental sets

> **Definition (Wronskian).** For two differentiable functions $y_1, y_2$, the
> **Wronskian** is the determinant
> $$
> W[y_1, y_2](t) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}
>   = y_1 y_2' - y_1' y_2.
> $$

If $W[y_1, y_2](t_0) \neq 0$, the initial-value system above is solvable for any
$y_0, y_0'$, so every initial condition can be matched by some member of the
family. The converse also holds, and the combined statement is the central
structural theorem.

> **Theorem (General solution).** Let $y_1, y_2$ solve (H) on $I$. The family
> $y = c_1 y_1 + c_2 y_2$ includes every solution of (H) if and only if there is
> a point $t_0 \in I$ where $W[y_1, y_2](t_0) \neq 0$.

When such a point exists, $y_1$ and $y_2$ are said to form a **fundamental set
of solutions**, and $c_1 y_1 + c_2 y_2$ is the **general solution**. The proof of
the forward direction uses uniqueness: given any solution $\phi$, choose
$c_1, c_2$ so that $c_1 y_1 + c_2 y_2$ matches $\phi$ and $\phi'$ at $t_0$; the
two functions then solve the same initial value problem and so coincide by the
uniqueness theorem.

> **Definition (Fundamental set of solutions).** A pair $y_1, y_2$ of solutions
> of (H) whose Wronskian is nonzero at some (equivalently, every) point of $I$.
> Every solution is then a unique linear combination of $y_1$ and $y_2$, so the
> solution space is two-dimensional.

$$
% caption: A nonzero Wronskian makes the coefficient matrix of the
% initial-value system invertible, so $(c_1, c_2)$ is uniquely determined by
% any prescribed $(y_0, y_0')$.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=34mm, minimum height=12mm, align=center}]
  \definecolor{acc}{HTML}{2A6F97}
  \node[box] (data) at (0,0) {initial data\\ $y_0,\; y_0'$};
  \node[box, draw=acc, text=acc, thick] (W) at (4.7,0) {$W[y_1,y_2](t_0) \neq 0$};
  \node[box] (out) at (9.6,0) {unique\\ $c_1,\; c_2$};
  \draw[->, acc, thick] (data) -- (W);
  \draw[->, acc, thick] (W) -- (out);
  \node[anchor=north, font=\scriptsize, text=black] at (4.7,-0.75)
    {coefficient matrix invertible};
\end{tikzpicture}
$$

For the two distinct exponentials, the Wronskian is easy to compute and never
vanishes.

> **Worked example (Exponentials are a fundamental set).** For $y_1 = e^{r_1 t}$,
> $y_2 = e^{r_2 t}$,
> $$
> W[y_1, y_2](t) = \begin{vmatrix} e^{r_1 t} & e^{r_2 t} \\
>   r_1 e^{r_1 t} & r_2 e^{r_2 t} \end{vmatrix}
>   = (r_2 - r_1)\, e^{(r_1 + r_2)t}.
> $$
> Because the exponential is never zero and $r_1 \neq r_2$, this is nonzero for
> all $t$. Hence $e^{r_1 t}, e^{r_2 t}$ form a fundamental set, confirming that
> $(\ast)$ is the general solution.

Linear independence is the algebraic name for the same fact. Two solutions are
**linearly independent** on $I$ when the only constants with
$k_1 y_1 + k_2 y_2 \equiv 0$ are $k_1 = k_2 = 0$. For solutions of (H), linear
independence and a nonzero Wronskian are equivalent, and either one certifies a
fundamental set. A fundamental set is not unique: for $y'' - y = 0$ both
$\{e^t, e^{-t}\}$ and $\{\cosh t, \sinh t\}$ are fundamental sets, related by an
invertible change of constants.

$$
% caption: Independent solutions span a two-dimensional solution space; the
% dependent case collapses one solution onto a multiple of the other and the
% Wronskian vanishes.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{2A6F97}
  % independent: two arrows spanning
  \begin{scope}
    \draw[->, acc, very thick] (0,0) -- (1.9,0.3);
    \draw[->, acc, very thick] (0,0) -- (0.5,1.7);
    \node[acc, anchor=west, font=\scriptsize] at (1.9,0.3) {$y_1$};
    \node[acc, anchor=south, font=\scriptsize] at (0.5,1.7) {$y_2$};
    \node[align=center, font=\scriptsize] at (1.0,-0.7) {independent\\ $W \neq 0$};
  \end{scope}
  % dependent: two collinear arrows
  \begin{scope}[xshift=5.4cm]
    \draw[->, black, very thick] (0,0) -- (1.9,0.75);
    \draw[->, black, very thick] (0,0) -- (1.1,0.43);
    \node[black, anchor=west, font=\scriptsize] at (1.9,0.75) {$y_1$};
    \node[black, anchor=north, font=\scriptsize] at (1.0,0.28) {$y_2 = k y_1$};
    \node[align=center, font=\scriptsize] at (1.0,-0.7) {dependent\\ $W \equiv 0$};
  \end{scope}
\end{tikzpicture}
$$

## Abel's formula for the Wronskian

The Wronskian of two solutions is not an arbitrary function; it satisfies a
first-order equation of its own. Differentiating $W = y_1 y_2' - y_1' y_2$ gives
$W' = y_1 y_2'' - y_1'' y_2$, and substituting $y_i'' = -p y_i' - q y_i$ from (H)
collapses the $q$-terms and leaves

$$
W' + p(t)\, W = 0.
$$

This is separable and first-order, so it integrates immediately.

> **Theorem (Abel's formula).** If $y_1, y_2$ solve
> $y'' + p(t)y' + q(t)y = 0$ with $p$ continuous on $I$, then
> $$
> W[y_1, y_2](t) = C \exp\!\left(-\int p(t)\, \d t\right)
> $$
> for a constant $C$ depending on the pair. Consequently $W$ is either
> identically zero on $I$ (if $C = 0$) or never zero on $I$ (if $C \neq 0$).

Two facts follow at once. First, to decide whether a pair is a fundamental set,
evaluate $W$ at a single convenient point rather than everywhere. Second, the
Wronskian is known up to a multiplicative constant _without solving the
equation_, from $p$ alone. For a constant-coefficient equation $a y'' + b y' +
c y = 0$ in standard form $p = b/a$, so $W(t) = C e^{-(b/a)t}$, matching the
$(r_2 - r_1)e^{(r_1+r_2)t}$ computed above once $r_1 + r_2 = -b/a$ is used.

The dichotomy "always zero or never zero" is what makes the single-point test
legitimate: a Wronskian of solutions cannot vanish at one point and be nonzero
at another.

> **Worked example (Abel's formula).** The equation $2t^2 y'' + 3t y' - y = 0$
> on $t > 0$ has solutions $y_1 = t^{1/2}$ and $y_2 = t^{-1}$. Computing the
> Wronskian directly,
> $$
> W[y_1, y_2] = t^{1/2}\bigl(-t^{-2}\bigr) - \tfrac{1}{2}t^{-1/2}\cdot t^{-1}
>   = -t^{-3/2} - \tfrac{1}{2}t^{-3/2} = -\tfrac{3}{2}\,t^{-3/2}.
> $$
> Abel's formula reproduces the $t$-dependence from the coefficient alone. In
> standard form $y'' + \frac{3}{2t}\,y' - \frac{1}{2t^2}\,y = 0$, so
> $p(t) = 3/(2t)$ and
> $$
> W(t) = C\exp\!\left(-\int \frac{3}{2t}\,\d t\right)
>   = C\exp\!\left(-\tfrac{3}{2}\ln t\right) = C\,t^{-3/2},
> $$
> which matches the direct computation with $C = -\tfrac{3}{2}$.

| Quantity | Formula | Role |
| --- | --- | --- |
| Characteristic equation | $ar^2 + br + c = 0$ | roots set the solution form |
| Discriminant | $b^2 - 4ac$ | $> 0$ gives two distinct real roots |
| General solution | $c_1 e^{r_1 t} + c_2 e^{r_2 t}$ | all solutions, two parameters |
| Wronskian | $y_1 y_2' - y_1' y_2$ | nonzero $\Rightarrow$ fundamental set |
| Abel's formula | $C\exp(-\int p\,\d t)$ | $W$ from $p$ alone; zero-or-never |

## Root signs and long-run behavior

For two distinct real roots the qualitative behavior is dictated entirely by
their signs, since each term is a pure exponential with no oscillation.

- **Both roots negative:** every solution decays to $0$ as $t \to \infty$;
  the slower-decaying term (larger root) dominates the tail.
- **Roots of opposite sign:** a generic solution grows without bound, driven by
  the positive root; only the special initial data killing that term decays.
- **One root zero:** the corresponding term is a constant, so solutions tend to
  a constant rather than to $0$.

[^boyce31]: **Boyce**, _Elementary Differential Equations_, §3.1 — Homogeneous Equations with Constant Coefficients: the substitution $y = e^{rt}$, the characteristic equation, and the two-distinct-real-roots general solution $c_1 e^{r_1 t} + c_2 e^{r_2 t}$.
[^boyce32a]: **Boyce**, _Elementary Differential Equations_, §3.2 — Solutions of Linear Homogeneous Equations; the Wronskian: the existence-uniqueness theorem (3.2.1), superposition (3.2.2), the Wronskian test for a fundamental set (3.2.4), and Abel's theorem (3.2.7).