Bessell function of the first kind $J_v$ of the Bessel equation $x^2y''+xy'+(\lambda^2 x^2-v^2)y=0$

Question

If we have an equation $x^2y''+xy'+(x^2-v^2)y=0$

then the solution of the first kind $J_v(x)=x^v\sum_{m=0}^{\infty}\frac{(-1)^mx^{2m}}{2^{2m+v}m!(m+v)!}$.

Then how would you find the solution of the first kind $J_v$ of the equation $x^2y''+xy'+(\lambda^2 x^2-v^2)y=0$ ?

I read that it is $J_v(\lambda x)$ but I don't know how to get that.

Answer 1

$x^2 y''(x) + x y'(x) + (\lambda^2 x^2 - \nu^2 ) y(x) = 0$

Change of variable : $X=\lambda x$

$y'(x) = \lambda y'(X)$ and $y''(x) = \lambda^2 y''(X)$

$X^2 y''(X) + X y'(X) + (X^2 - \nu^2 ) y(X) = 0$

$J_\nu(X) = J_\nu(\lambda x)$ is a solution.