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For two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.