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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!

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  • $\begingroup$ Well it does say THE solutions to the problem I would have assumed they would be the two unique solutions in the form of exponentials $\endgroup$ – Triatticus Jun 11 '17 at 14:24
  • $\begingroup$ @Triatticus Oops! The question says $2$ solutions. I miss typed it. $\endgroup$ – Naive Jun 11 '17 at 14:26
  • $\begingroup$ Ah well, that might change things, though you can still know the general form of the answer but without Independence I can see why it is ambiguous. I think you can at least eliminate one more option using the general solution of the ODE $\endgroup$ – Triatticus Jun 11 '17 at 14:35
  • $\begingroup$ @Triatticus I guess the initial conditions will play a role here to determine the Wronskian. But I cannot come up with anything as of now. $\endgroup$ – Naive Jun 11 '17 at 14:38
  • $\begingroup$ Well recall that the solutions to this equation are of the form $y(t) = Ae^{r_1 t} + B e^{r_2 t}$ where the r's are the possibly complex roots of the characteristic equation $r^2 + a r + b = 0$. So you can write down any solution to this equation satisfying the initial condition which puts a condition on A and B. I feel like some info is missing like conditions on a and b. $\endgroup$ – Triatticus Jun 11 '17 at 15:25
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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

$W(t) = W(0) e^{-\int_0^t a(s) ds}. \tag{7}$

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).

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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.

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Since $y_1$ and $y_2$ are the solutions of the ODE they must satisfy the initial condition.

$y_1(0)=0$ and $y_2(0)=0$

$\implies W=0$

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