Wronskian

The Wronskian of $n$ functions $y_1(x),y_2(x),\dots ,y_n(x)$ is the determinant of the $n\times n$ matrix whose $k$-th row (for $k=0,1,\dots ,n-1$) consists of the $k$-th derivatives:

$W[y_1,\dots ,y_n](x)=\det \left[\begin{array}{cccc}{y}_1& y_2& \cdots & y_n\\ y_1^{\prime }& y_2^{\prime }& \cdots & y_n^{\prime }\\ ⋮& & & ⋮\\ y_1^{(n-1)}& y_2^{(n-1)}& \cdots & y_n^{(n-1)}\end{array}\right]$

It’s a test for linear independence of solutions of an $n$-th order linear ODE: if $W\ne 0$ at some point, the $n$ solutions are linearly independent on that interval.

In this course we mostly work with second-order linear ODEs, so the two-function form

$W[y_1,y_2](x)=\det \left[\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right]=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$

shows up everywhere. The remainder of the worked examples below are for the second-order case.

Why it matters

A second-order linear ODE has a 2-dimensional solution space. To write the general solution

$y(x)=c_1{y}_1(x)+c_2{y}_2(x)$

we need $y_1$ and $y_2$ to be linearly independent. If they’re proportional ($y_2=\alpha y_1$), they span only a 1D subspace and miss half the solutions.

The Wronskian tells you whether they’re independent. If $W[y_1,y_2](x_0)\ne 0$ for some $x_0$, they’re linearly independent on the interval. (For homogeneous linear ODEs, the Wronskian is either always zero or never zero on the interval — this is Abel’s theorem.)

Worked example: complex roots

For the ODE $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ with characteristic roots $\alpha \pm i\beta$ (where $\beta \ne 0$), the real-valued solutions are

$y_1(x)=e^{\alpha x}\cos (\beta x),y_2(x)=e^{\alpha x}\sin (\beta x)$

Compute the Wronskian:

Step 1: derivatives.

$y_1^{\prime }(x)=e^{\alpha x}[\alpha \cos (\beta x)-\beta \sin (\beta x)]$

$y_2^{\prime }(x)=e^{\alpha x}[\alpha \sin (\beta x)+\beta \cos (\beta x)]$

Step 2: determinant.

$W=e^{\alpha x}\cos (\beta x)\cdot e^{\alpha x}[\alpha \sin (\beta x)+\beta \cos (\beta x)]-e^{\alpha x}\sin (\beta x)\cdot e^{\alpha x}[\alpha \cos (\beta x)-\beta \sin (\beta x)]$

Factor $e^{2\alpha x}$:

$W=e^{2\alpha x}[\cos (\beta x)[\alpha \sin (\beta x)+\beta \cos (\beta x)]-\sin (\beta x)[\alpha \cos (\beta x)-\beta \sin (\beta x)]]$

Expand:

$W=e^{2\alpha x}[\alpha \cos \sin +\beta {\cos }^2-\alpha \cos \sin +\beta {\sin }^2]=e^{2\alpha x}\cdot \beta ({\cos }^2+{\sin }^2)=\beta e^{2\alpha x}$

Since $\beta \ne 0$ (otherwise the roots wouldn’t be complex), $W\ne 0$. So $y_1,y_2$ are linearly independent.

When the Wronskian vanishes

If $W[y_1,y_2](x_0)=0$ at some specific $x_0$, you can’t conclude anything immediately — for general functions, the Wronskian might vanish at isolated points without implying linear dependence. (Counterexamples exist.)

But for solutions of a linear homogeneous ODE, Abel’s theorem says the Wronskian is either identically zero or never zero on the interval. So for ODE solutions, if $W=0$ anywhere, $W=0$ everywhere, and the solutions are linearly dependent.

Abel’s theorem also gives an explicit formula:

$W(x)=W(x_0)\exp \left(-{\int }_{x_0}^{x}p(s)ds\right)$

for an ODE in standard form $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$. Notice this is never zero unless $W(x_0)$ is zero, confirming the all-or-nothing pattern.

In context

The Wronskian is the key tool for:

• Verifying linear independence of solutions you’ve found.
• The Representation theorem: the general solution of a homogeneous linear ODE is $c_1{y}_1+c_2{y}_2$ exactly when $W[y_1,y_2]\ne 0$.
• The Method of variation of parameters: the formula uses $W$ in the denominator.

For the higher-dimensional generalization to systems of ODEs, see Linear independence of vector functions.