Wronskian Problem

Question

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?

Answer 1

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.

Answer 2

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).