1. ## Tricky ODE problem

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

2. Originally Posted by dave d
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
Try changing the dependent variable to $\displaystyle x = e^t$

-Dan

3. ## Applying the x = e^t substitution

Thanks 'topsquark' for the suggestion to substitute e^t for x in my partial working. Here's my attempt at the rest of the problem - I get a very neat result which makes me think perhaps it is correct. Here's the rest from where I got stuck:

lnp = lne^(-6) * lnx + lnK
(lnp - lnK)/(lne^(-6)) = lnx = lne^t = t (if x = e^t)

t = (ln(p/K))/ (lne^6) = (ln(p/K))/6
6t =ln(p/K)
e^(6t) = p/ K
p = Ke^(6t) = K(e^t)^6 = Kx^6

p = Kx^6 (K is a constant)

Could someone venture a verdict please??