Past exam of the differential equations 2 course of the University of Oxford / 2021 / 2 / b / Solution

For the first Frobenius solution, take the larger indicial root $r=a$ and normalize the first coefficient to one:Substitution and comparison of the coefficient of $x^{n+a}$ giveEvery denominator is nonzero because $a\ge 0$.

For $\beta =1$, the recurrence relation is $c_n=-b{c}_{n-1}/[n(n+2a)]$, soFor $\beta =2$, $c_1=0$ and all odd coefficients vanish; the even coefficients satisfy $c_{2k}=-b{c}_{2k-2}/[2k(2k+2a)]$. HenceFor any positive integer $\beta$, induction in the recurrence relation shows that only indices divisible by $\beta$ survive:The empty product for $k=0$ equals one. Successive terms in the power series factor have ratio of magnitudesThus the ratio test gives convergence for every finite $x$; a nonintegral $a$ still requires a chosen side of zero or a complex branch for the factor $x^a$.