Here's my working for this ODE problem - could someone venture an opinion on the result please??
Before I start 'topsquark' from this forum was kind enough to offer the suggestion in reply to my earlier post to make a substitution x = e^t which confused me a little since he called 'x' the dependent variable, so perhaps someone may have a thought on that advice too??
Anyways, here's my attempt
x^2*y" + 4x*y' - 4y = 0 need to find 'p' where y = x^p satisfies the ODE. My analysis:
This is a 2nd order ODE of form f(x, y', y") so I'm using reduction of order process. I've found one solution by trial and error (y = 4x)
Form of 2nd solution is:
y(x) = xv , y'(x) = v + xv' , y"(x) = 2v' + xv"
x^2(2v' + xv") + 4x(v + xv') - 4x(v) = 0
x^3*v" + 6x^2*v' = 0
Let p = v' and p' = v" ----> x^3p' + 6x^2*p = 0
(1st order ODE) separate the variables
x^3*dp/dx = -6x^2*p ----> 1/p dp = -6/x dx ---> integral 1/p dp = imtegral -6/x dx
lnp = -6lnx + C = -6lnx + lnK
(lnp - lnK)/ lne^-6 = lnx = lne^t = t
t = ln(p/K)/ lne^(-6) ---> -6t = ln(p/K) ---> e^(-6t) = p/K
p = K*e^(-6t) = K(e^t)^(-6) = K*x^(-6)
p = K*x(-6)
Correct?????? How about the x = e^t advice?????