Linear differential equation and its Wronskian Let $a(t),b(t)$ be continuous functions and $x_1(t),x_2(t)$ two solutions of the differential equation $$x''(t)+a(t)x'(t)+b(t)x(t)=0$$
We define $w(t)=x_1(t)x_2'(t)-x_2(t)x_1'(t)$. Show that 
(i) $w'(t)=-a(t)w(t)$.
(ii) If $a(t) \geq M>0$, then $\lim_{t \to \infty} w(t)=0$.
I've derived $w'(t)=x_1x_2''-x_2x_1''$. I need to prove the equality $$x_1x_2''-x_2x_1''=-a(t)(x_1x_2'-x_2x_1')$$.
By the hypothesis, $-a(t)x'=x''+bx$. I don't know how to cleverly use this information, I would appreciate hints (for example, taking a conveniente linear combination of $x_1,x_2$ as a solution). I am pretty lost with (i) as well.
 A: Put $x_1$ and $x_2$ into the hypothesis (separately) and substitute into the equality you wish to prove. I have all terms disappearing.
$$-a(t)x_1'=x_1''+bx_1 \qquad -a(t)x_2'=x_2''+bx_2$$
For part ii) if $w(t)$ is positive, $w'(t)$ is negative and thus $w(t)$ is heading towards zero. If $w(t)$ is negative, $w'(t)$ is positive and thus $w(t)$ is heading towards zero.
A: My main interest here is contributing an answer to part (ii) of the question, but what the hey, in for a penny, in for a pound:
With
$w = x_1'x_2 - x_1 x_2', \tag{1}$
we see that
$w' = x_1'' x_2 + x_1' x_2' - x_1' x_2' - x_1 x_2' = x_1'' x_2 - x_1 x_2'', \tag{2}$
as our OP user156441 claimed.  Since
$x_i'' + ax_i' + bx_i = 0 \tag{3}$
for $i = 1, 2$, we can insert
$x_i'' = -a x_i' - bx_i \tag{4}$
into (2):
$w' = x_1'' x_2 - x_1 x_2'' = (-a x_1' - bx_1) x_2 - x_1 (-ax_2' - bx_2)$
$= -a(x_1' x_2 - x_1 x_2'') = -aw, \tag{5}$
since terms containing $b$ cancel out.  So we have (i).
As for part (ii), note that if $w(t_0) > 0$ for some $t_0$, then $w(t) > 0$ everywhere, since $w(t) = 0$ is the unique solution taking the value zero at any particular value of $t$; thus for $w(t) > 0$ we have from (5) that
$(\ln w(t))' = \dfrac{w'(t)}{w(t)} = -a(t), \tag{6}$
whence, integrating from some $t_0$ to $t$, 
$\ln (\dfrac{w(t)}{w(t_0)}) = \ln w(t) - \ln w(t_0) = -\int_{t_0}^t a(s)ds, \tag{7}$
or
$w(t) = w(t_0)e^{-\int_{t_0}^t a(s)ds}. \tag{8}$
If now $a(t) \ge M >0$,
$\int_{t_0}^t a(s)ds \ge \int_{t_0}^t M ds = M(t - t_0), \tag{9}$
and thus
$e^{-\int_{t_0}^t a(s)ds} \le e^{-M(t - t_0)} \to 0 \tag{10}$
as $t \to \infty$; thus by (8) $w(t) \to 0$ as well.  In the event that $w(t) < 0$, we apply the argument to $-w(t)$ and thereby reach the same conclusion.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
