= Solution
Seek a <Frobenius series> $y=x^r\sum_{n\geq0}a_nx^n$, with $a_0\ne0$, 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 gives
$$
a_{n+2}=-\frac{n+r+\lambda}{(n+r+2)(2n+2r+1)}a_n,\qquad n\geq0.
$$
The 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 are
$$
f_0(x)=1-\frac\lambda2x^2+O(x^4),\qquad
f_{3/2}(x)=x^{3/2}\left[1-\frac{2\lambda+3}{28}x^2+O(x^4)\right].
$$
Thus the general local solution is $C_0f_0+C_1f_{3/2}$. The endpoint condition at zero forces $C_0=0$. For a nontrivial solution, $C_1\ne0$, so $y(x)\sim C_1x^{3/2}$ and \b[$\alpha=3/2$], 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.<NIST Digital Library of Mathematical Functions>[§2.7(i), equations 2.7.3–2.7.4]{ref}
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