Reduction of Order for variable coefficients?

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January 9th 2010, 02:03 PM
Kaitosan

Reduction of Order for variable coefficients?

4x^2 y′′ + 8 x y′ − 3 y = 0 , x > 0

y(4) = 8, y′ (4) = 0

So far I've only seen characteristic solutions being solved via reduction of order but I'm not so sure about equations that include variable coefficients and in which Y1 and Y2 subsets of the general solution Y are not given.....a quick clarification on this would be much appreciated! Thanks!
January 9th 2010, 03:32 PM
Chris L T521

Quote:

Originally Posted by Kaitosan

4x^2 y′′ + 8 x y′ − 3 y = 0 , x > 0

y(4) = 8, y′ (4) = 0

So far I've only seen characteristic solutions being solved via reduction of order but I'm not so sure about equations that include variable coefficients and in which Y1 and Y2 subsets of the general solution Y are not given.....a quick clarification on this would be much appreciated! Thanks!

This is a Cauchy-Euler Equation of the form $ax^2\frac{\,d^2y}{\,dx^2}+bx\frac{\,dy}{\,dx}+cy=0$ . Supposing a solution of $y=x^r$ (where $x\neq0$ ), you can come up with a characteristic equation as seen here.

You can find out more about these types of equations and how to solve them in that link I gave you. Try to see if you can get anywhere with this information, and post back if you get stuck! :)
January 9th 2010, 03:35 PM
scienceismylady

A slightly more general version of Chris's start is:

The differential equation you gave is an equidimensional equation(a power of x matches the order of the derivative on each term).

The first step towards solving is assuming $x=e^{\xi}$ and $y(x)=Y(\xi)$ . You can then calculate what $y'(x)$ and $y''(x)$ are in terms of $Y(\xi)$ (by chain rule). It will reduce to a simple constant-coefficient equation for $Y(\xi)$ . You can then plug in $\xi=\ln(x)$ and get your solution, $y(x)$ .

(This is more general since it solves the equation when the roots $r$ are repeated or imaginary)