 Solution

Put $s_n(x)=\sin (q_{n}x)$ and $s_m(x)=\sin (q_{m}x)$. Their differential equations are $-s_n^{\prime \prime }=q_n^2{s}_n$ and $-s_m^{\prime \prime }=q_m^2{s}_m$. Subtract the two cross-multiplied equations and use integration by parts:The lower boundary term is zero because both functions vanish there. At the upper endpoint, the Robin boundary condition gives $s_j^{\prime }(1)=s_j(1)/\alpha$, soSince distinct positive roots have $q_n^2\ne q_m^2$, we concludeThis orthogonality holds for every $\alpha >0$, even when the sine family requires the additional nonpositive mode for completeness as explained in part (c).