> with(linalg):
Let us consider several examples of applications of the theorem about Wronskian.

Example 1. Prove that the functions

f(x)=x, g(x)=e^x, h(x)=e^2x, are linearly independent.

Notice that these functions have first and second (and all other) derivatives, so the Theorem about the Wronskian is applicable. Consider the matrix of derivatives of these functions:
> f:=x: g:=exp(x): h:=exp(2*x): f1:=1: f2:=0: g1:=exp(x): g2:=exp(x): h1:=2*exp(2*x): h2:=4*exp(2*x): A:=matrix([[f, g, h], [f1,g1,h1],[f2,g2,h2]]);

A :=	[ x	ex	e2x ]
	[ 1	ex	2e2x ]
	[ 0	ex	4e2x ]

> W:=simplify(det(A));

W := 2 x e^3x - 3 e^3x

Notice that if x=0 then W(x)=-3, W is not identically 0. Thus by the theorem about Wronskian the functions f, g, h are linearly independent.

Example 2. Prove that the functions e^ax, e^bx, e^cx, e^dx are linearly independent provided a, b, c, d are pairwise different numbers. Indeed, the matrix of derivatives of these functions is:

> A:=matrix([[exp(a*x), exp(b*x), exp(c*x), exp(d*x)], [a*exp(a*x), b*exp(b*x), c*exp(c*x), d*exp(d*x)], [a²*exp(a*x), b²*exp(b*x), c²*exp(c*x), d²*exp(d*x)],\ [a³*exp(a*x), b³*exp(b*x), c³*exp(c*x), d³*exp(d*x)]]);

A :=	[ eax	ebx	ecx	edx ]
	[ a eax	b ebx	c ecx	d edx ]
	[ a2eax	b2ebx	c2ecx	d2edx ]
	[ a3eax	b3ebx	c3ecx	d3edx ]

Using Maple, it is easy to check that the determinant of this matrix is equal to

(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)*e^(a+b+c+d)*x

Indeed,


> simplify(det(A)-(a-b)*(a-c)*(a-d)*(b-c)*(b-d)*(c-d)*exp((a+b+c+d)*x));

0

Thus the Wronskian is identically equal to 0 if and only if one of the differences a-b, a-c, a-d, b-c, b-d, c-d is 0. But by our assumption all these differences are non-zero, so the Wronskian is not 0 and we can apply the theorem about Wronskians.