Solutions to an ODE using Frobenius when roots are repeated

Question

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.

Answer 1

If you modify the equation to $$ xy''+(1-\varepsilon)y'+xy=0,~~~ε\approx 0 $$ then the coefficient equation changes to $$ x^{n-1+r}:~~ (n+r)(n+r-ε)a_n+a_{n-2}=0 $$ The indicial equation at $n=0$ now has two different solutions $r=0$ and $r=ε$ that have no integer distance, so both give a Frobenius power series solution. $a_k=0$ for odd $k$ results in both cases. The coefficient recursions for the even-index coefficients, $n\ge 2$ even, are $$ r=0:~~a_n=-\frac{1}{n(n-ε)}a_{n-2} \\ r=ε:~~a_n=-\frac{1}{(n+ε)n}a_{n-2} $$ Thus we get the first solution $y_1(x)=y(-ε,x)$ and the second $y_2(x)=x^εy(ε,x)$, where $y(ε,x)$ was set as the power series with the coefficients of the second solution with $a_0=1$. In the limit $ε\to 0$ both give the same solution.

Now as also linear combinations of both are solutions, we get $$ y_3(x)=\frac{y_2(x)-y_1(x)}{ε}=\frac{x^ε-1}{ε}y(ε,x)+2\frac{y(ε,x)-y(-ε,x)}{2ε} $$ as a third solution. In the limit the difference quotients become differential quotients, the limit is $$ y_3(x)=\ln(x)y(0,x)+2\left.\frac{\partial y(ε,x)}{\partial ε}\right|_{ε=0}. $$ This shows how a derivative can be introduced into the solution process. Now try to map this to the somehow parametrized coefficient sequence of the unmodified solution. I think it will work, but I see not how you could get a direct justification from that. In the direct interpretation of what you wrote, you would consider $$ y(r,x)=x^r\sum_{n=0}^\infty a_n(r)x^k $$ where $$ a_n(r)=\frac1{(n+r)^2}a_{n-2} $$ so that $y_1(x)=y(0,x)$ and the $r$-derivative of $y(r,x)$ at $r=0$ results in the same formula for $y_3$ at $r=ε=0$.