 Solution

Seek a Frobenius series $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$, on the positive side of zero. The lowest coefficient gives the indicial equation from part (a); the next gives $(r+1)(2r-1)a_1=0$. For either permitted indicial exponent, $a_1=0$. Matching the subsequent powers givesThe two indicial exponents differ by a noninteger, so there are two independent Frobenius solutions without a logarithmic companion. Normalizing each leading coefficient to one, their forms areThus the general local solution is $C_0{f}_0+C_1{f}_{3/2}$. The endpoint condition at zero forces $C_0=0$. For a nontrivial solution, $C_1\ne 0$, so $y(x)\sim C_1{x}^{3/2}$ and $\alpha =3/2$, with the arbitrary nonzero amplitude understood.

The condition at one selects values of $\lambda$ for which this branch also vanishes there; it does not change the selected local exponent. The zero solution, available for every $\lambda$, has no nonzero leading coefficient and is excluded when asking for a leading exponent. This is a direct application of the Frobenius method and the noninteger-separated-root case.^[ref]