I have been looking at the following differential equation
$$xy^{\prime \prime} + y^{\prime} + xy=0$$
Observe that since $x=0$ is a regular singular point we use the Frobenius method and substitute $y=\sum_{n=0}^{\infty}a_n x^{n+r}$ to get
$$r^2a_0x^{r-1} + (r+1)^2a_1x^r + \sum_{n=2}^{\infty}[(n+r)^2a_n+a_{n-2}]x^{n+r-1}=0$$
Since $a_0 \neq 0$ we get $r=0,0.$ giving $y_1 = a_0(1-\dfrac{x^2}{2^2}+\dfrac {x^4}{4^2\cdot2^2}-\dfrac{x^6}{6^2\cdot4^2\cdot2^2}......)$.
Now apparently to obtain the second solution I have to differentiate this solution w.r.t. $r$ and then evaluate this at $r=0$, but I really don't see where this comes from. Could anyone show me explicitly why exactly we do this? Or more generally -
If the roots of the indicial equation $r_1$ and $r_2$ are repeated, why is the second solution of the form
$$y_2= c\left(\frac {\partial y_1}{\partial r}\right)_{\large {r=r_1}}?$$
All the sources I've looked at just state this fact without really showing where this came from, so either I'm missing something really obvious or there is some other explanation.