Homogeneous Equations, the Wronskian, and Real Roots

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

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 $\varphi$ by

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

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.

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.¹

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 $\varphi$ is available in general, so the proof proceeds by abstract methods and is not reproduced here.² For the constant-coefficient equation we can exhibit solutions directly, which settles existence; uniqueness still leans on the theorem.

Solving the constant-coefficient equation

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

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

With two distinct real roots $r_1\ne r_2$, the functions $y_1=e^{r_{1}t}$ and $y_2=e^{r_{2}t}$ are both solutions, and by superposition so is

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

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.

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.

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.

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 $y^{\prime }(t_0)=y_0^{\prime }$ on $c_1{y}_1+c_2{y}_2$ produces the linear system

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

If $W[y_1,y_2](t_0)\ne 0$, the initial-value system above is solvable for any $y_0,y_0^{\prime }$, 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.

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 $\varphi$, choose $c_1,c_2$ so that $c_1{y}_1+c_2{y}_2$ matches $\varphi$ and ${\varphi }^{\prime }$ at $t_0$; the two functions then solve the same initial value problem and so coincide by the uniqueness theorem.

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

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 $y^{\prime \prime }-y=0$ both $\{e^t,e^{-t}\}$ and $\{\cosh t,\sinh t\}$ are fundamental sets, related by an invertible change of constants.

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 $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ gives $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$, and substituting $y_i^{\prime \prime }=-p{y}_i^{\prime }-q{y}_i$ from (H) collapses the $q$-terms and leaves

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

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 $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ 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.

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 $0$.