I have the following problem: Find a power series solution of the Cauchy-Euler Equation around $x=0$: $$(2x+5)y'' + y'= 0, \quad y(0)=2, \quad y'(0)=3.$$
When I tried solving it, the recurrence I got for the coefficients of $y$, assuming that $y=a_0+a_1 x^1+a_2 x^2+\ldots$, is $a_{n+1}=-((2n-1)/5n+5)a_n$ for all $n \ge 2$. Given $y(0) = 2$, then $a_0 = 2$, and accordingly from $y'(0)=3 \implies a_1=3$.
However, the recurrence here does not work for $a_1$ and $a_0$. $a_1=-((2 \cdot 0-1)/5 \cdot 0+5) \cdot 2=2/5$ instead of the given $a_1=3$. Why is this?
While combining the power series to get the recurrence, I also independently got $10 a_2=-a_1$, hence $a_2=-3/10$. This matches with the recurrence formula if we calculate $a_2$ using $a_1$ and the formula. This is inspite of the fact that the recurrence formula I got from a power series where $n \ge 2$, so it should not work for $n=1$, yet it does.
So why does $a_0$ not fit in the formula and how does it fit in the final answer, i.e., in the power series representation of the solution?
Should I have substituted $y'$ with something like $p$ and solved the equation via a power series as an FOD equation instead and then integrated the power series solution?