= Solution
By part (c), write $\lambda=\mu^2$ with $\mu>0$. The equation is $y''+\mu^2y=0$, and the left <Dirichlet boundary condition> eliminates the cosine term. Thus $y=A\sin(\mu x)$, with $A\ne0$. The right <Robin boundary condition> becomes
$$
\mu\cos(\mu\ell)+k\sin(\mu\ell)=0.
$$
Put $z=\mu\ell$ and $\kappa=k\ell>0$. A root cannot have $\cos z=0$, so this is equivalent to $\tan z=-z/\kappa$.
On each interval $((n-1/2)\pi,n\pi)$, $n\geq1$, the function $\tan z+z/\kappa$ rises from $-\infty$ to a positive value and has derivative $\sec^2z+1/\kappa>0$. The <intermediate value theorem> and strict <monotonicity> therefore give exactly one root $z_n$ there. On the intervening positive-tangent intervals both terms are positive, so there are no further positive roots. Consequently
$$
\boxed{\lambda_n=\frac{z_n^2}{\ell^2},\qquad y_n(x)=A_n\sin\frac{z_nx}{\ell},\qquad(n-\tfrac12)\pi<z_n<n\pi,\quad n=1,2,\ldots.}
$$
These intervals establish $\lambda_1<\lambda_2<\cdots$ and an infinite sequence of <eigenvalues>. The apparent algebraic root $z=0$ of the sine equation is excluded: it gives a zero sine function, and the direct zero-eigenvalue solution $y=Ax$ would require $A(1+k\ell)=0$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2-2016-ib-robin-roots.png]
{title=One intersection of $\tan z$ with $-z/\kappa$ on each negative-tangent branch; the illustration uses $\kappa=1$}
For <vector normalization> to unit <norm>, the <Dirichlet-Robin eigenvalues on a finite interval> give
$$
A_n=\left[\frac\ell2+\frac{k}{2(k^2+\lambda_n)}\right]^{-1/2}.
$$
The eigenfunctions can otherwise be scaled by any nonzero real constant.
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