Hi - having trouble with this ODE question:
x^2y" + 2xy' - 4y = 0 I'm looking for a value of 'p' where y = x^p satisfies the ODE. My analysis -
This is a 2nd order ODE of the form f(x, y', y") so I'm using a 'reduction of order' process (found one solution by trial and error).
My 2nd solution is of the form y2(x) = v(x)y1(x) this gives me
x^3v" + 6x^2v' = 0 ---> x^3p' + 6x^2p = 0 with the subs p = v'
Separate the variables 1/p dp = -6/x dx --> integration gives
lnp = -6lnx + C or lnp = -6lnx + lnK --> lnp = -6lnKx -->
lnp = ln(Kx)^-6 --> now drop the logs ----> p = (Kx)^-6
After that things get hazy - can someone please resue me. Dave D