# Solution of the Cauchy-Euler equation

1. Oct 19, 2008

### Schmoozer

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

The Cauchy-Euler equation is:
x2y'+axy'+by=0
And has a solution of the form:
y1(x)=xm
Use the method of Variation of Parameters to show that the second independent solution
is:
y2(x)=xmln(x)

So that the overall solution is:
y(x)=[A+Bln(x)]xm
Hint: This equation suggests that the characteristic equation has two identical roots.

Hint: Use the following transformation:
x=et

2. Relevant equations

None

3. The attempt at a solution
I'm not really sure how to start this. Any help would be very much appreciated.

2. Oct 19, 2008

### Dick

You should start it by putting y=x^m into the original equation and then realizing you meant to say x^2*y''+a*x*y'+b*y=0. Then think about it again using the hints.

3. Oct 19, 2008

### Schmoozer

Ok, so:
m2+(a-1)m+b=0

So using variation of parameters the equation can be represented as:

$${\lambda}^{2}+ \left( a-1 \right) \lambda+b=0$$

and

$$\lambda=\mbox {{\tt \&+-}} \left( 1-a,\sqrt {1/2\, \left( a-1 \right) ^{2}-2\,b} \right)$$

???

I'm not sure if that latex stuff worked:

L2+(a-1)L+b=0

and

L=(1-a)(+/-)(((a-1)^2-4b)/2)1/2

Last edited: Oct 19, 2008
4. Oct 19, 2008

### Schmoozer

In any event, lets say I now have:

u(x)=[c1+c2ln|x|]xm

or

u(x)=[c1+c2ln|x|]x-(a-1)/2

How do I use the variations of parameters to show the second independent solution:

y2=xmln(x)

5. Oct 20, 2008

### HallsofIvy

Do you know what "variation of parameters" IS?

Since you already know that xm is a solution, try a solution of the form y= u(x)xm. Then y'= u' xm+ m u xm-1 and y"= u" xm+ 2m u' xm-1+ m(m-1)u xm. Put those into the equation and use the fact that xm itself satifies the equation.

6. Oct 20, 2008

### Schmoozer

No not really. I have notes on it buts I can't make heads or tails of it.

Where is the 'u' coming from?

Last edited: Oct 20, 2008