I already understand that for second-order, linear, homogeneous ODE, if the Wronskian of two solutions to the ODE ( or functions that satisfy the ODE) is zero at a certain point, the functions are linearly dependent.
At another place (source at bottom), it has been written that if Wronskian is zero at a certain point, it is not necessary that they are linearly dependent. As an example,
for f(x) = x, g(x) = sin(x), we find W(f, g) = x cos(x) − sin(x) which is
nonzero, for example, at x = π. Hence, x and sin(x) are Linearly
Independent. Note that W(f, g) may be zero at some point such as x =
0.
The only way I can resolve these two statements is by saying that you can't construct a second-order, linear, homogeneous ODE which has these two functions, x and sin(x) as it's solutions (i.e. they satisfy the ODE).
I don't know how to formally prove this, except that I observe that sin(x) already is the solution of second-order, linear, homogeneous ODE, of which x is not a solution, so adding it might make a third-order or say non-linear ODE.
So, how do we actually resolve these statements and can we prove that we can't construct an ODE as required above?
EDIT: This question was solved earlier, but I have a doubt in the solution by mathcounterexamples.net now: why can we not change the coefficient of y'' to the function in the denominator (xcos(x) – sin(x)) ?
Source: https://home.iitk.ac.in/~sghorai/TEACHING/MTH203/ode7.pdf or
https://drive.google.com/file/d/1OGRE00YNB0kjVHam9ZSpsDQfba0PxkYG/view?usp=sharing
Relevant statement:
Theorem 3: Two solutions $y_1,y_2$ of
$$
y'' + p(x)y' + q(x)y = 0, \quad x \in \mathcal I
$$
are linearly dependent iff $W(y_1,y_2) = 0$ at a certain point $x_0 \in \mathcal I$.
Best Answer
Hint
Take $a,b$ real maps to define any required second-order, linear, homogeneous, ODE
$$y^{\prime \prime}(x) =a(x)y^\prime(x) +b(x)y(x)$$
and deduce a contradiction if $x, \sin x$ are both solutions.
Indeed, $x$ solution implies $a(x)=-xb(x)$. While if $\sin x $ is a solution
$$-\sin x =- x \cos x b(x)+ \sin x b(x)$$
and
$$b(x)=\frac{\sin x}{-\sin x + x \cos x}$$ which is not continuous at zero.