Differential Equations · Lesson 08

The Wronskian and independence of solutions

Two solutions give every solution only if they are genuinely different — linearly independent, in the language of the linear algebra course. The test is a determinant: the Wronskian, built from the two solutions and their derivatives, which is the determinant of the matrix of the last lesson. If it is not zero at one point, the two solutions form a basis of the solution space, which therefore has dimension two.

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W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}

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Intuition

Two solutions give every solution only if they are genuinely different — linearly independent, in the language of the linear algebra course. The test is a determinant: the Wronskian, built from the two solutions and their derivatives, which is the determinant of the matrix of the last lesson. If it is not zero at one point, the two solutions form a basis of the solution space, which therefore has dimension two.

Two directions on a map reach every point by combination only if they are not along the same line. The Wronskian is the area of the parallelogram the two solutions span at a point, measured in value and slope, and zero area means they point the same way.

A determinant that decides independence

For two solutions of $y''+p(x)y'+q(x)y=0$ on an interval, the Wronskian is the determinant below. Abel's formula shows it is either never zero or always zero. If it is never zero the solutions are independent and every solution is a combination of them: they are a fundamental set, a basis of the two-dimensional solution space.

W (t_{0}) \neq = 0

y_{1} y_{1}^{'} y_{2} y_{2}^{'}

y

''

+

p (x) y^{'} +

q (x) y =

0

W(x)=W(x_{0})\,e^{-\int\limits_{x_{0}}^{x}p(s)\,ds}

W=\det\begin{pmatrix}y_{1}&y_{2}\\y_{1}'&y_{2}'\end{pmatrix}=y_{1}y_{2}'-y_{1}'y_{2}

Computing it, and what it shows

$e^{x}$ and $e^{2x}$ : $W=e^{x}\cdot2e^{2x}-e^{x}\cdot e^{2x}=e^{3x}\ne0$ .
$\cos x$ and $\sin x$ : $W=\cos^{2}x+\sin^{2}x=1$ .
$e^{rx}$ and $xe^{rx}$ : $W=e^{2rx}\ne0$ , so the repeated-root pair is a basis too.
For two solutions of one equation, $W$ is zero everywhere or nowhere. For arbitrary functions it can vanish at some points and not others, so the dichotomy belongs to solutions.
The solution space has dimension exactly two: two independent solutions span it, and any three solutions are dependent.

Two solutions with a nonzero Wronskian give every solution

W(x_{0})\ne0\ \Longrightarrow\qquad y=C_{1}y_{1}+C_{2}y_{2}\ \text{for every solution}\ y

Take any solution and record its value and slope at the point. Constants that make a combination of the two solutions match them solve a two-by-two linear system, and its determinant is the Wronskian at that point, which is not zero — so the constants exist. The combination is a solution by superposition, with the same value and slope at the point as the given one. By uniqueness the two are the same function.

Proof steps

y(x_{0})=a,\qquad y'(x_{0})=b

Take any solution and record its value and slope at the point.

\begin{gathered}C_{1}y_{1}(x_{0})+C_{2}y_{2}(x_{0})=a\\[0.35em] C_{1}y_{1}'(x_{0})+C_{2}y_{2}'(x_{0})=b\end{gathered}

Ask for a combination of the two solutions with the same value and slope.

\det\begin{pmatrix}y_{1}(x_{0})&y_{2}(x_{0})\\y_{1}'(x_{0})&y_{2}'(x_{0})\end{pmatrix}=W(x_{0})\ne0

The determinant of that system is the Wronskian, so the constants exist.

z=C_{1}y_{1}+C_{2}y_{2},\qquad z(x_{0})=a,\ z'(x_{0})=b

The combination solves the equation by superposition, with the same value and slope.

z=y

By uniqueness the two solutions coincide: every solution is a combination of the two.

Applications

Practice

The Wronskian

Put the two solutions in the top row and their derivatives below, and take the determinant.

W=\det\begin{pmatrix}y_{1}&y_{2}\\y_{1}'&y_{2}'\end{pmatrix}=y_{1}y_{2}'-y_{1}'y_{2}

Try it

What is the Wronskian of $e^{x}$ and $e^{3x}$ ?