= Solution
Set $z=x^3/3$ and write $y(x)=Y(z)$. Then
$$
y'=x^2Y_z,\qquad y''=2xY_z+x^4Y_{zz},
$$
so the two first-derivative contributions cancel exactly, leaving
$$
-Y_{zz}-aY=\lambda Y,\qquad Y(0)=Y(1/3)=0.
$$
The Dirichlet spectrum of $-d^2/dz^2$ on this interval is $(3n\pi)^2$, $n\geq1$. If $\lambda+a\leq0$, the linear or hyperbolic solutions cannot satisfy both zero endpoints nontrivially, so no additional modes have been omitted. Thus
$$
\boxed{\lambda_n=9n^2\pi^2-a,\qquad y_n(x)=\sqrt6\sin(n\pi x^3),\qquad n=1,2,\ldots.}
$$
To verify normalization, use $dz=x^2dx$:
$$
\int_0^1x^2\sin(n\pi x^3)\sin(m\pi x^3)dx=\int_0^{1/3}\sin(3n\pi z)\sin(3m\pi z)dz=\frac{\delta_{nm}}6.
$$
This proves weighted orthonormality and also transfers completeness from the ordinary <Fourier sine series>. The construction is the <cubic-coordinate reduction of a singular Sturm-Liouville operator>. The positive constant $a$ shifts the entire spectrum down; a zero <eigenvalue> occurs if $a=9n^2\pi^2$, so the no-zero-eigenvalue proviso in part (a)(iii) must be imposed separately if that inverse formula is used here.
Back to article page
