An expression for the Wronskian 
Consider a general $n$th order linear equation
  $$x^{n}(t)+a_{n-1}x^{n-1}(t)+ \dots + a_{1}x'(t) + a_{0}x(t).$$
  Let $x_1, x_2 , \dots , x_n$ be a fundamental set of solutions of above and set $W(t)=W(x_1, x_2 , \dots , x_n ; t).$

How to show that $W(t)=W(t_0) e^{-\int_{t_0}^{t} a_{n-1}(s)~ds}.$ 
So we know that since $x_1, x_2 , \dots , x_n$ form a fundamental set of solutions, the set of vectors $\{ x_1, x_2 , \dots , x_n \}$ are linearly independent. This implies $W(t)=W(x_1, x_2 , \dots , x_n ; t) \neq 0$ for any $t \in (-\infty, \infty).$ This is all I've got. I need help in solving this problem. Any help is much appreciated. 
 A: Hint:
You need to show that 
$$\frac{dW(t)}{dt}=-a_{n-1}(t)W(t),$$
using the expression of 
$$W(t)=\det\begin{pmatrix}x_1&\cdots&x_n\\&\cdots&\\x_1^{(n-1)}&\cdots&x_n^{(n-1)}\end{pmatrix}$$
The key step is to show that 
$$W'(t)=\det\begin{pmatrix}x_1&\cdots&x_n\\&\cdots&\\x_1^{(n-2)}&\cdots&x_n^{(n-2)}\\x_1^{(n)}&\cdots&x_n^{(n)}\end{pmatrix}.$$
To see this, you need to express $W(t)$ as
$$\sum_{\sigma} \text{sign}(\sigma)x_{\sigma(1)}x_{\sigma(2)}'\cdots x_{\sigma(n)}^{(n-1)},$$
where $\sigma$ is a permutation of $1,\dots, n$ and this sum is over all permutations. 
See http://hobbes.la.asu.edu/courses/site/442/dets.pdf for some explanations of writing determinants using permutations.
Now using product rule to find the derivative. 
$$W'(t)=\sum_{\sigma} \text{sign}(\sigma)x'_{\sigma(1)}x_{\sigma(2)}'\cdots x_{\sigma(n)}^{(n-1)}+\sum_{\sigma} \text{sign}(\sigma)x_{\sigma(1)}x_{\sigma(2)}''x''_{\sigma(3)}\cdots x_{\sigma(n)}^{(n-1)}+\sum_{\sigma} \text{sign}(\sigma)x_{\sigma(1)}x_{\sigma(2)}'\cdots x_{\sigma(n)}^{(n)}$$
In the above formula, you can see that each term is again the determinant of a matrix, all of which except the last one contain two identical rows. For example, the first term is the determinant of the matrix with the first row of $W(t)$ replaced by $(x_1'x_2'\cdots x_n')$, hence the first two rows are identical. So all terms are zero except the one containing the $n$-th derivative. 
By the original ODE, we have
$$W'(t)=W'(t)=\det\begin{pmatrix}x_1&\cdots&x_n\\&\cdots&\\x_1^{(n-2)}&\cdots&x_n^{(n-2)}\\-a_{n-1}x_1^{(n-1)}-\cdots-a_0x_1&\cdots&-a_{n-1}x_n^{(n-1)}-\cdots-a_0x_n\end{pmatrix}.$$
By the properties of determinants, adding a multiple of one row to another does not change the determinant, so we perform
$$a_0R_1+R_n\rightarrow R_n, \dots, a_{n-2}R_{n-1}+R_n\rightarrow R_n,$$
where $R_i$ is the $i$-th row of $W'(t)$. This gives us
$$W'(t)=\det\begin{pmatrix}x_1&\cdots&x_n\\&\cdots&\\x_1^{(n-2)}&\cdots&x_n^{(n-2)}\\-a_{n-1}x_1^{(n-1)}&\cdots&-a_{n-1}x_n^{(n-1)}\end{pmatrix}.$$
Hence 
$$W'(t)=-a_{n-1}W(t).$$
Now solving this ODE will give you the result.
