Show that $J_1 (ax)$ is the finite solution of $x^2 y'' + x y' + (a^2 x^2 - 1) y = 0$.
I've been asked this question with almost not preface to Bessel functions so I am really stuck. I am assuming I solve the Bessel equation with $(ax)$ instead of $x$ and then check to see if that solution also fits the given equation. Another method I considered was graphically, where $J_1(ax)$ would just be the graph of $J_1(x)$ but dilated in the horizontal direction, so it's zeros locations would be changed by the factor of $a$.
I am then confused as to how to locate the solutions of the given equation, so any help would be appreciated, even just a kick in the right direction. Apologies for poor formatting.
You must start with the definition of the Bessel function, which you didn't provide.
$$x^2y''+xy'+(x^2-1)y=0.$$
Then perform the substitution $t=ax$, giving by the chain rule
$$y'=\frac{dy}{dx}=a\frac{dy}{dt}=a\dot y,\\y''=\frac{d^2y}{dx^2}=a^2\frac{d^2y}{dt^2}=a\ddot y.$$
Then plugged in the equation,
$$t^2\ddot y+t\dot y+(a^2t^2-1)y=0.$$