Wronskian Problem 
My approach for this question,
I consider $a$ and $b$ as some specific real numbers and then, after solving the differential equation, I get $y_i$, $i=1,2$ and as a function of $t$ and go on solving Wronskian by using $y_1$ and $y_2$.
I also get $0$, constant and some function of $t$ where totally depends on the choice of $a$ and $b$ which I choose, so I don't understand how to move forward.
Is my approach  not correct?
What is that I am missing?
 A: Given the differential equation
$y'' + a(t) y' + b(t) y = 0, \tag 0$
where we have generalized to allow the coefficients $a$ and $b$ to be functions of $t$, the Wronskian of two solutions $y_1(t)$ and $y_2(t)$ is
$W(t) = \det \left ( \begin{bmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{bmatrix} \right ) = y_1(t) y_2'(t) - y_2(t) y_1'(t); \tag 1$
we thus have
$W'(t) = y_1'(t) y_2'(t) + y_1(t) y_2''(t) - y_2'(t) y_1'(t) - y_2(t) y_1''(t); \tag 2$
from (0) we have that each solution satisfies
$y'' = - a(t) y' - b(t) y(t); \tag 3$
we may then substitute $-a(t) y'(t) - b(t) y(t)$ for instances of $y''(t)$ in (2):
$W'(t)$
$= y_1'(t) y_2'(t) + y_1(t)( -a(t) y_2'(t) - b(t) y_2(t)) - y_2'(t) y_1'(t) - y_2(t) ( -a(t) y_1'(t) - b(t)y_1(t))$
$ =  y_1'(t) y_2'(t) -a(t) y_1(t) y_2'(t) - b(t)y_1(t)y_2(t)$
$ - y_2'(t) y_1'(t) + a(t) y_1'(t) y_2(t) + b(t) y_1(t) y_2(t)$
$ = a(t) (y_1'(t) y_2(t) - y_1(t) y_2'(t)) = -a(t) W(t); \tag 4$
we thus see that $W(t)$ obeys the equation
$W'(t) = - a(t) W(t); \tag 5$
the solution to (5) initialized at some $t_0 \in \Bbb R$ is readily seen to be
$W(t) = W(t_0) \exp \left (\displaystyle -\int_{t_0}^t a(s) \; ds \right ); \tag 6$
if now we take $t_0 = 0$ then, since we are given that
$y_1(0) = y_2(0) = 0, \tag 7$
it follows from (1) that
$W(0) = 0, \tag 8$
and therefore (6) implies
$W(t) = 0, \; \forall t \in \Bbb R. \tag 9$
The correct solution is thus answer (1).
A: The wronskian is either identically zero or never zero. In this case we have $y_1(0)=y_2(0) =0.$
Thus $$W(0) = \det \left ( \begin{bmatrix} y_1(0) & y_2(0) \\ y_1'(0) & y_2'(0) \end{bmatrix} \right ) = y_1(0) y_2'(0) - y_2(0) y_1'(0) =0$$
Therefore the wronskian is identically zero. 
