Solution of the Cauchy-Euler equation

[Schmoozer]

Homework Statement

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

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

Hint: Use the following transformation:
x=e^t

Homework Equations

None

The Attempt at a Solution

I'm not really sure how to start this. Any help would be very much appreciated.

 

[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.

 

[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

 

[Schmoozer]

In any event, let's say I now have:

u(x)=[c₁+c₂ln|x|]x^m

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=x^mln(x)

 

[HallsofIvy]

Do you know what "variation of parameters" IS?

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

 

[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?