# Finding the G.S.(for equal roots) of the Euler-Cauchy diff. equation

1. Oct 11, 2007

### rock.freak667

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

$$x^2\frac{d^2y}{dx^2}+ax\frac{dy}{dx}+by=0$$

show that if there is one real double root of the aux. eq'n show that the G.S. is given by

$$y=c_1x^{n_1}+c_2x^{n_1}ln(x)$$

2. Relevant equations

Assume the trial solution $y=x^n$

3. The attempt at a solution
$$y=x^n \frac{dy}{dx}=nx^{n-1} \frac{d^2y}{dx^2}=n(n-1)x^{n-2} => n(n-1)x^n + anx^n +bx^n =n^2-(a-1)n+b=0 (Aux. Eq'n)$$

for equal roots $$(a-1)^2-4b=0$$

therefore $$n=\frac{1-a}{2}$$
hence one solution is $$y=x^\frac{1-a}{2}$$

so the other solution is $$y=vx^\frac{1-a}{2} \frac{dy}{dx}=\frac{dv}[dx}x^\frac{1-a}+ \frac{1-a}{2}Vx^\frac{-a-1}{2} \frac{d^2y}{dx^2}=\frac{d^2v}{dx^2}x^\frac{-1-a}{2} + (1-a)\frac{dv}{dx}x^\frac{-1-a}{2} -\frac{1-a^2}{4}vx^\frac{-a-3}{2}$$

sub. into the diff. eq'n....(this is where I think I am wrong)
(This is what I get)
$$\frac{d^2v}{dx^2}x^\frac{3-a}{2} + \frac{dv}{dx}x^\frac{3-a}{2} +\frac{vx^\frac{1-a}{2}((a-1)^2+4b)}{4}$$

Now I was hoping to get $(a-1)^2-4b$ in the last term there so that it would be zero but I apparently did something wrong..where did I go wrong?