Question 1 of 7: Sturm–Liouville Eigenvalue Problem
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2018 — 04-BS-5 Advanced Mathematics, 3 hours, closed book (approved Casio/Sharp calculator and one double-sided aid sheet permitted). Any five of the seven questions constitute a complete paper and all questions are of equal value; all seven are answered below as a full study resource.
Check The extraction prints the boundary condition as "$y(1)=y(e^2)$," which is not a valid pair of homogeneous end conditions for a regular Sturm–Liouville problem (an S-L problem needs one condition at each end, not an equality between the two end values). The coefficients $x^{-3}$, $x^{-5}$, $(4+\lambda)$ belong to the standard textbook family $(x^{-(2k-1)}y')'+(k^2+\lambda)x^{-(2k+1)}y=0$ with $k=2$, whose paired sibling problems on this exam (and in Kreyszig) are always posed with the Dirichlet pair $y(1)=0,\ y(e^2)=0$. That reading is adopted below; it is the only one consistent with the printed coefficients and reproduces a clean discrete eigenvalue set.
Given. The self-adjoint (Sturm–Liouville) equation $(x^{-3}y')'+(4+\lambda)x^{-5}y=0$ on $1\le x\le e^2$, with $p(x)=x^{-3}$, weight $w(x)=x^{-5}$, and (reconstructed) homogeneous boundary conditions $y(1)=0,\ y(e^2)=0$.
Find. All eigenvalues $\lambda_n$ and the corresponding eigenfunctions $y_n(x)$.
Approach. Multiply through by $x^5$ to collapse the SL form into a Cauchy–Euler equation, solve its characteristic equation, discard the cases that cannot satisfy two homogeneous end conditions, then apply the boundary conditions to the surviving oscillatory family.
Reduce to Cauchy–Euler form. Expanding the derivative, $(x^{-3}y')'=x^{-3}y''-3x^{-4}y'$, so the equation reads
$$x^{-3}y''-3x^{-4}y'+(4+\lambda)x^{-5}y=0$$
Multiplying every term by $x^{5}$ clears the negative powers:
$$\boxed{x^{2}y''-3xy'+(4+\lambda)y=0}$$
a linear Cauchy–Euler (equidimensional) equation.
Solve the characteristic equation. Trying $y=x^{m}$ gives $m(m-1)x^m-3mx^m+(4+\lambda)x^m=0$, so
$$m^{2}-4m+(4+\lambda)=0\ \Rightarrow\ m=2\pm\sqrt{4-(4+\lambda)}=2\pm\sqrt{-\lambda}$$
For $\lambda\le0$, $m$ is real and $y=C_1x^{m_1}+C_2x^{m_2}$; enforcing $y(1)=0$ and $y(e^2)=0$ on such a solution forces $C_1=C_2=0$ (only the trivial solution), so no non-positive $\lambda$ is an eigenvalue — the eigenvalues must be positive.
Positive $\lambda$: oscillatory solution. Write $\lambda=\mu^2$ ($\mu\gt0$), so $m=2\pm i\mu$ and
$$y(x)=x^{2}\big[C_1\cos(\mu\ln x)+C_2\sin(\mu\ln x)\big]$$
Apply $y(1)=0$. Since $\ln 1=0$, $y(1)=1^2[C_1\cos0+C_2\sin0]=C_1=0$.
Apply $y(e^2)=0$. With $C_1=0$ and $\ln(e^2)=2$,
$$y(e^2)=e^{4}\,C_2\sin(2\mu)=0$$
For a non-trivial eigenfunction $C_2\ne0$, so $\sin(2\mu)=0\Rightarrow 2\mu=n\pi,\ n=1,2,3,\dots$
Assemble the eigenpairs. $\mu_n=n\pi/2$, hence
$$\boxed{\lambda_n=\dfrac{n^2\pi^2}{4},\qquad y_n(x)=x^{2}\sin\!\Big(\dfrac{n\pi}{2}\ln x\Big),\qquad n=1,2,3,\dots}$$
The first three numerical eigenvalues are $\lambda_1=\pi^2/4\approx2.467401$, $\lambda_2=\pi^2\approx9.869604$, $\lambda_3=9\pi^2/4\approx22.206610$.
