Find The Wronskian Of The Following ODE Let $y_1$ and $y_2$ be two solutions of the problem,
$$\begin{align}
y''(t)+ay'(t)+by(t)=0,t\in \Bbb R\\
y(0)=0\;\;\;\;\;\;\;\;\;
\end{align}$$
where $a$ and $b$ are real constants.
Let $W$ be the Wronskian of $y_1$ and $y_2$. Then
(A) $W(t)=0\;, \forall t\in \Bbb R$
(B) $W(t)=c\;, \forall t\in \Bbb R$, for some  positive constant $c$
(C) $W$ is a non-constant positive function.
(D) $\exists \;\;t_1,t_2\in \Bbb R $ such that $ W(t_1)<0<W(t_2)\;$
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My Attempt:
Option (D) Can be eliminated straight away as the Wronskian will always be strictly positive,strictly negative or zero.
I'm pretty confused with the first three options. Since $y_1$ and $y_2$ need not be $2$ independent solutions we cannot directly say the Wronskian vanishes.
Hints Please!
 A: If $y_1(0)=y_1'(0)=0$, then $y_1 \equiv 0$ by uniqueness of solutions, which forces the Wronskian $W(y_1,y_2)$ to be identically $0$.
Assuming that $y_1'(0)\ne 0$, $y_2'(0)\ne 0$ gives non-zero constants $A$ and $B$ such that $y=Ay_1+By_2$ is a solution of your equation that satisfies $y(0)=y'(0)=0$; hence, $y\equiv 0$, which proves that $y_1,y_2$ are linearly dependent, and that forces $W(y_1,y_2)\equiv 0$.
So, in all cases, the Wronskian vanishes identically.
A: The following approach works for any continuous $a(t)$, $b(t)$;  they needn't be constant.
If $y_1$ and $y_2$ are any solutions of
$y'' + a(t)y' + b(t)y = 0, \tag{1}$
where we allow for the moment $a(t)$ and $b(t)$ to be time dependent, the the Wronskian of these solutions is
$W = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix};  \tag{2}$
we thus see that
$W = y_1 y_2' - y_2 y_1', \tag{2}$
whence
$W' = y_1' y_2' + y_1 y_2'' - y_2' y_1' - y_2 y_1'' = y_1 y_2'' - y_2 y_1''. \tag{3}$
In the event that $y_1$, $y_2$ satisfy (1), we can elininate the second derivatives from (3) in favor of the $y_i'$, $i = 1,2$:
$y_i'' = -a(t) y_i' - b(t) y_i, \tag{4}$
so that
$W = y_1 (-a(t) y_2' - b(t) y_2) - y_2 (-a(t) y_1' - b(t) y_1) = -a(t) y_1 y_2' - b(t) y_1 y_2 + a(t) y_2 y_1' + b(t) y_1 y_2$
$= a(t)(y_2 y_1' - y_2' y_1) = -a(t)W. \tag{5}$
$W$ thus satisfies the simple differential equation
$W'= -a(t)W,  \tag{6}$
the solution of which is readily seen to be
$W(t) = W(0) \exp \left (\displaystyle -\int_0^t a(s) ds \right).  \tag{7}$
Now since,in this specific case, we have $y(0) = 0$, it follows that $W(0) _= 0$, and hence from (7) that
$W(t) = 0 \tag{8}$
for all $t \in \Bbb R$.  The correct answer is thus (A).
A: Since $y_1$ and $y_2$ are the solutions of the ODE they must satisfy the initial condition. 
$y_1(0)=0$ and $y_2(0)=0$
$\implies W=0$
