I have a second order linear ode which I am solving about a regular singular point $x=0$. The ode is $$ xy"+2y'+xy=0$$ I found the equation as $$ (r^2+r)a_0x^{r-1}+ (r^2+3r+2)a_1x^r +$$ the recurrence relation $$a_m=\dfrac{-a_{m-2}}{(m+r)(m+r+1)}$$
Now my question are
- I am getting two quadratic equations, the equation corresponding to lowest power is the indicial equation giving me roots $0,-1$. But I also get another quadratic equation which has roots $2,-1$ from the coefficient of $x^r$, is there anything significant if I get two quadratic equations having a common root.? Does it have any implication. The reason I am asking this is because the roots of the indicial equations differe by an integer. But putting $r=-1$ is giving two LI solutions. Also, when I put $r=0$, I am getting another set of two Linearly independent solutions but containing a part of the previous solution. I mean either in this type of case you get only one LI solutions, but in this it seems as if I am getting more than what is needee. How do I explain this phenomenon of what is making this two sets of LI solutions possible?
The equation can be written $(xy)''+(xy)=0$ with obvious solutions.
With $r=-1$ you get both basis solutions, one with coefficient $a_0$ and one with $a_1$, with $r=0$ you get again the second basis solution from the $r=-1$ case. This is an exceptional case, usually you will only get one basis solution, with the second obtained via reduction-of-order and containing a logarithmic term.