# 2nd order diff equation!

• Oct 27th 2010, 02:48 PM
FunkyBudduh
2nd order diff equation!
Hi, I have this differential equation to solve its broken down in to 3 parts but im having trouble doing the 2nd part any ideas would be great!

Consider the intial value problem

(y x^2) - (4xy) + 6y = g(x) x>0 y(0)=y(0)=0

part a) Write this de in self adjoint form which i have done

(1/x^4 y) + 6y/x^6 = g(x)x^-6

part b) Find the influence function for this

i know the formula for this but i have to find y1 and y2 but i have no idea on how to do this can anyone help me?

thanks
• Oct 27th 2010, 03:05 PM
zzzoak

$
\displaystyle {y=x^r
}
$
• Oct 27th 2010, 04:27 PM
Krizalid
Quote:

Originally Posted by FunkyBudduh
(y x^2) - (4xy) + 6y = g(x) x>0 y(0)=y(0)=0

put $t=\ln x$ then $\displaystyle\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt},$ then $\displaystyle\frac{{{d}^{2}}y}{d{{t}^{2}}}-5\frac{dy}{dt}+6y(t)=g({{e}^{t}}).$

homogeneous solution is $y_h(t)=c_1e^{3t}+c_2e^{2t}$ now find the particular solution.
• Oct 28th 2010, 06:03 AM
FunkyBudduh
Hi Krizalid,

Why is it -5dy/dx?

thanks
• Oct 28th 2010, 06:24 AM
Krizalid
it's dt not dx.
• Oct 28th 2010, 07:03 AM
FunkyBudduh
Hi Krizalid,

Why are we using the function t=ln(x) instead of y=x^r seeming as the LHS of the equation is a polynomial and not a exponential. When i have used y= x^r i have got y=x^2 and y=x^3.

thanks
• Oct 28th 2010, 07:04 AM
Krizalid
it was a matter of choice, both ways work.
• Oct 29th 2010, 04:58 PM
FunkyBudduh
Hi Krizalid,

part 3 of the question is that $g(x) = (x^4)(e^x)$

i have used $y=x^2$ and $y=x^3$
to obtain a really nasty general solution, with exponentials and powers of xs

it looks like

$e^x(-x^7 + 8x^6 - 48x^5.....$

pretty sure this is wrong though, any idea on how to solve this i been tryin to do this for days now (Headbang)...

thanks

much appreciated
• Oct 30th 2010, 04:31 PM
FunkyBudduh
I HAVE SOLVED THIS!!!!!!!!!!!!!!! Thx For your help all