Question

I have this bessel equation
x^2*y"+xy'+(3x^2-2)y=0
-in solution i have (Jv)and(Yv) how can i solve it?
-If you have any EX attach it please

Answer 1

mm.. i think we should change variable
p = sqrt(3)x
so
dy/dx = dy/dp * dp/dx = dy/dp * sqrt(3)
and hence
d2y/dx2 = 3d2y/dp2

so your equation will become: now y' is derivative with respect to p.
p^2*y'' + py' + (p^2-2)y=0

Answer 2

now you should be able to solve it