# ODEs- How to annihilate ln(x) ?

1. May 8, 2010

### Roni1985

1. The problem statement, all variables and given/known data

I need to annihilate ln(x)

2. Relevant equations

3. The attempt at a solution

my try was saying that this is a eular equation with
r1=r2=0
c1=0 and c2=1
so the annihilator should be D^2
but I don't think it works.

Any other suggestions ?

Thanks.

2. May 8, 2010

### gabbagabbahey

How about $(D+xD^2)$?

3. May 8, 2010

### Roni1985

Yep, it works :\
thanks.
But, how did you get it ?

Roni.

4. May 8, 2010

### gabbagabbahey

I got it by inspection, but if you are looking for a general method, I would say begin by finding $Dy(x)$ and $D^2y(x)$ and seeing if they are proportional to each other.

EDIT: Actually, I don't think you can claim that this is a valid annhilator, since it involves $x$ and is not just a polynomial (with constant coefficients) in $D$

5. May 8, 2010

### Roni1985

I see....
thanks very much for the help...

6. May 8, 2010

### The Chaz

Would it be accurate to say that we check if a finite amount of derivatives are a linear combination?

7. May 8, 2010

### LCKurtz

The method of annihilators is a method for solving constant coefficient DE's. An Euler DE is not that type, but it can be transformed into a constant coefficient DE by a change of variables. Then you can use the method of annihilators on the transformed equation and transform the answer back to solve your original Euler DE. The change of variables you need is t = ln(x). Using a dot for differentiation with respect to t you get, using t = ln(x):

$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx} = \dot y\frac 1 x$$

$$\frac{d^2y}{dx^2}= -\frac 1 {x^2}\dot y + \frac 1 x \frac{d\dot y}{dt}\frac{dt}{dx} =-\frac 1 {x^2}\dot y+\frac 1{x^2}\ddot y$$

When you substitute these into the left side of your Euler equation you will be left with a constant coefficient equation for y as a function of t. You will need to substitute x = et in the right side too. Solve that by annihilators and substitute t = ln(x) to get your final answer.

Last edited: May 8, 2010