Wronskian and Linear Independence of Solutions

What is the Wronskian?

The Wronskian of two differentiable functions $y_1(x)$ and $y_2(x)$ is the determinant $W(y_1, y_2)(x) \;=\; \det \begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix} \;=\; y_1 \, y_2' - y_2 \, y_1'.$

For $n$ functions, it's the $n \times n$ determinant of the matrix whose rows are $y_i$, $y_i'$, …, $y_i^{(n-1)}$.

It's one of the most useful tools in linear ODEs because it gives a single-number test for linear independence of solutions.

The linear-independence test

Theorem (Wronskian test). Let $y_1, \ldots, y_n$ be solutions of a linear homogeneous ODE on an interval $I$. Then the $y_i$ are linearly independent on $I$ if and only if their Wronskian is nonzero at some point of $I$ — equivalently, nonzero at every point (Abel's theorem guarantees $W$ is either identically zero or never zero on $I$).

So checking linear independence reduces to evaluating one determinant at one point.

Caution. For functions that are not simultaneously solutions of a single linear ODE, a nonzero Wronskian still implies linear independence, but zero Wronskian does not imply linear dependence. The theorem above is specifically for solutions of a linear ODE.

Abel's theorem (why it's all-or-nothing)

For solutions of $y'' + p(x) y' + q(x) y = 0$, the Wronskian satisfies $W(x) = W(x_0) \exp\!\left( -\int_{x_0}^{x} p(t) \, dt \right).$

The exponential factor never vanishes, so $W(x)$ is either identically zero (if $W(x_0) = 0$) or nonzero everywhere on $I$. Evaluating at one convenient point fully determines the answer.

For the ODE with $p(x) = 0$ (no $y'$ term), $W$ is constant.

The given problem

Compute $W(e^{x}, e^{2 x})$.

Step-by-step

$y_1 = e^{x}$, $y_2 = e^{2 x}$.

$y_1' = e^{x}$, $y_2' = 2 e^{2 x}$.

$W \;=\; \det \begin{pmatrix} e^{x} & e^{2 x} \\ e^{x} & 2 e^{2 x} \end{pmatrix} = e^{x} \cdot 2 e^{2 x} - e^{2 x} \cdot e^{x} = 2 e^{3 x} - e^{3 x} = e^{3 x}.$

$\boxed{\, W(e^{x}, e^{2 x}) = e^{3 x} \,}$

$e^{3 x}$ is never zero, so $e^{x}$ and $e^{2 x}$ are linearly independent on all of $\mathbb{R}$.

A sanity check — which ODE do these solve?

Both are solutions of $y'' - 3 y' + 2 y = 0$ (characteristic roots $r = 1, 2$; see #183). The Wronskian for this ODE, with $p(x) = -3$, satisfies by Abel $W(x) = W(x_0) e^{-\int -3 \, dt} = W(x_0) e^{3 (x - x_0)}.$

At $x_0 = 0$: $W(0) = 1$. So $W(x) = e^{3 x}$ — matching our direct computation exactly.

Examples — when functions are dependent

Compute $W(e^{x}, 3 e^{x})$: $W = \det \begin{pmatrix} e^{x} & 3 e^{x} \\ e^{x} & 3 e^{x} \end{pmatrix} = 3 e^{2 x} - 3 e^{2 x} = 0.$

Zero everywhere — these are linearly dependent ($3 e^{x} = 3 \cdot e^{x}$). Confirmed.

Another: $W(\sin x, \cos x)$: $W = \det \begin{pmatrix} \sin x & \cos x \\ \cos x & -\sin x \end{pmatrix} = -\sin^{2} x - \cos^{2} x = -1 \ne 0.$

Linearly independent — forming the standard basis for solutions of $y'' + y = 0$.

Why "linearly independent" is crucial

A second-order linear homogeneous ODE has a two-dimensional solution space. If you pick two solutions $y_1, y_2$ and they're linearly independent (Wronskian nonzero), they form a basis — every solution is $C_1 y_1 + C_2 y_2$ for some unique $C_1, C_2$. If they're dependent, they span only a 1D subspace, and you can't express every solution that way.

So the Wronskian test is what you run before declaring "this is the general solution" — to make sure your two candidate solutions really are independent.

Wronskian in variation of parameters

In variation of parameters (#187), the Wronskian appears as the denominator of the coefficients: $u_1' = -\frac{y_2 \, g}{W}, \qquad u_2' = \frac{y_1 \, g}{W}.$

If the Wronskian is zero on an interval, variation of parameters breaks down — which only happens if $y_1, y_2$ were already dependent (so they don't form a valid basis to begin with).

Higher-order and general-n case

For three solutions of a 3rd-order linear ODE, compute $W(y_1, y_2, y_3) = \det \begin{pmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{pmatrix}.$

Same theorem: nonzero at a point ↔ linearly independent.

Abel's theorem generalises: for $y^{(n)} + p_{n-1}(x) y^{(n-1)} + \ldots + p_0(x) y = 0$, $W(x) = W(x_0) \exp\!\left( -\int_{x_0}^{x} p_{n-1}(t) \, dt \right).$

Only the coefficient of the $(n-1)$-th derivative matters for the Wronskian's evolution — the lower coefficients don't contribute to the exponential.

Common mistakes

• Confusing the Wronskian matrix with the Wronskian determinant. The matrix is $W(y_1, y_2)$-matrix; the Wronskian is its determinant.
• Concluding dependence from zero Wronskian for arbitrary functions. The zero-Wronskian-implies-dependence direction is true only for solutions of a linear ODE, not for arbitrary function pairs.
• Evaluating at the wrong sign. $y_1 y_2' - y_2 y_1'$, not $y_1' y_2 - y_2' y_1$.
• Using the Wronskian on a set of functions that don't solve the same ODE. Then the test only half-applies — nonzero ⇒ independent, but zero doesn't mean dependent.

Try it in the visualization

Watch the Wronskian $W(x)$ as a curve over $x$, with the two underlying solutions $y_1, y_2$ overlaid. Slide the initial conditions; see the Wronskian stay the same sign (never crossing zero) whenever the two solutions are independent. Swap in a multiple of $y_1$ as $y_2$ and watch $W$ collapse to zero everywhere.