Linear ODE, Wronskian, variation of parameters, systems

In one sentence

The solution set of a linear $n$-th order equation is an $n$-dimensional vector space, the Wronskian detects a basis — but only for solutions of a common equation, and that qualifier is the whole trap.

Why the exam asks it

Because the Wronskian test is stated in most courses without its hypothesis, and $x^2$ with $x|x|$ is the small pair that shows why the hypothesis is there. And because systems reduce everything to matrix exponentials, which ties this topic to the canonical forms chapters.

The idea

For $y^{(n)} + p_{n-1}(x)y^{(n-1)} + \cdots + p_0(x)y = 0$ with continuous coefficients on an interval $I$:

• Solutions form a vector space of dimension exactly $n$.
• Every solution exists on all of $I$ — no blow-up, as the previous chapter noted.
• The general solution of the inhomogeneous equation is one particular solution plus the general homogeneous one.

The Wronskian

$$W(y_1,\dots,y_n) = \det\begin{pmatrix} y_1 & \cdots & y_n \\ y_1' & \cdots & y_n' \\ \vdots & & \vdots \\ y_1^{(n-1)} & \cdots & y_n^{(n-1)}\end{pmatrix}.$$

For solutions of the same linear equation: $W \ne 0$ at one point $\iff$ $W$ is never zero $\iff$ the solutions are independent.

Abel's formula is why the two ends of that are equivalent: $W' = -p_{n-1}W$, so

$$W(x) = W(x_0)\exp\left(-\int_{x_0}^x p_{n-1}\right),$$

and an exponential is never zero. The Wronskian of solutions is therefore either identically zero or nowhere zero — never zero at isolated points.

For arbitrary functions the test fails. Take $f(x) = x^2$ and $g(x) = x|x|$ on $\mathbb{R}$. They agree for $x \ge 0$ and differ by a sign for $x < 0$, so neither is a multiple of the other — independent. Yet on each half-line one is $\pm$ the other, so $W \equiv 0$ everywhere.

No contradiction: they are not both solutions of a single second-order linear equation with continuous coefficients. That is exactly the hypothesis the option omits.

Variation of parameters

For $y'' + py' + qy = g$ with homogeneous solutions $y_1, y_2$, a particular solution is

$$y_p = -y_1\int\frac{y_2 g}{W} + y_2\int\frac{y_1 g}{W}.$$

It works for any continuous $g$, unlike undetermined coefficients, which needs $g$ to be a polynomial, exponential, sine or cosine — or a product of those. Choosing between the two methods is usually the first decision.

Systems

$\mathbf{y}' = A\mathbf{y}$ has solution $\mathbf{y}(t) = e^{At}\mathbf{y}(0)$, where

$$e^{At} = \sum_{k\ge0}\frac{(At)^k}{k!}.$$

Computing it uses the canonical forms from the algebra unit:

• $A$ diagonalisable with eigenpairs $(\lambda_i, v_i)$: solutions are $\sum c_i e^{\lambda_i t}v_i$.
• Repeated eigenvalue with too few eigenvectors: generalised eigenvectors appear and solutions carry polynomial factors $t e^{\lambda t}$ — which is exactly a Jordan block of size $2$.
• Complex eigenvalues $\alpha \pm i\beta$: real solutions $e^{\alpha t}(\cos\beta t, \sin\beta t)$ combinations.

An $n$-th order scalar equation becomes a first-order system with a companion matrix, whose characteristic polynomial is the auxiliary equation — the same companion matrix as in the rational canonical form chapter.

A fundamental matrix $\Phi$ has independent solutions as columns, and $\det\Phi$ is the Wronskian. Abel's formula becomes $\frac{d}{dt}\det\Phi = \operatorname{tr}(A)\det\Phi$.

Where intuition breaks

"$W \equiv 0$, so the functions are dependent." $x^2$ and $x|x|$. True only for solutions of a common linear equation.

"$W(x_0) = 0$ for one $x_0$, so they are dependent there." For solutions, Abel makes $W$ either always or never zero. A Wronskian vanishing at a single point means it vanishes everywhere, so the functions are not both solutions of one such equation.

"The solution space has dimension $n$ for any $n$-th order equation." Linear ones, with continuous coefficients. Nonlinear equations have no vector space of solutions at all.

"Undetermined coefficients always works." Only for the special right-hand sides. Variation of parameters handles any continuous $g$ and is the safe choice.

"A repeated eigenvalue gives $e^{\lambda t}$ twice." Only if there are two independent eigenvectors. Otherwise a $te^{\lambda t}$ appears — the diagonalisability question from the algebra unit, in another costume.

"$e^{At}e^{Bt} = e^{(A+B)t}$." Only when $A$ and $B$ commute. The same failure as for matrix exponentials generally.

"The Wronskian of a fundamental matrix can vanish somewhere." $\det\Phi$ satisfies Abel's equation, so it is nowhere zero.

The exam's angle

1. Check whether the functions are solutions of a common equation before applying any Wronskian claim. If they are just functions, the test is one-directional.
2. Use Abel to answer "can $W$ vanish here" — for solutions, never at isolated points.
3. For systems, find eigenvalues and count eigenvectors. Too few means polynomial factors.
4. Match the method to the right-hand side: undetermined coefficients for the standard forms, variation of parameters otherwise.
5. Convert order-$n$ to a system when the question is about structure rather than an explicit solution; the companion matrix makes the algebra visible.

The night before

• Linear $n$-th order, continuous coefficients: solution space has dimension $n$, defined on the whole interval.
• Abel: $W' = -p_{n-1}W$, so $W$ is nowhere zero or identically zero.
• $W \ne 0$ $\iff$ independent — for solutions of a common linear equation only.
• $x^2$ and $x|x|$: $W \equiv 0$, independent. The standard counterexample.
• Variation of parameters works for any continuous $g$; undetermined coefficients only for polynomial, exponential and trigonometric forms.
• $\mathbf{y}' = A\mathbf{y}$ gives $e^{At}$. Repeated eigenvalue with too few eigenvectors $\Rightarrow$ $te^{\lambda t}$ terms.
• $\frac{d}{dt}\det\Phi = \operatorname{tr}(A)\det\Phi$.
• $e^{At}e^{Bt} = e^{(A+B)t}$ needs $AB = BA$.