= Solution
\b[Mixed endpoint spectrum.] For $y''+\lambda y=0$, integration against $y$ and <integration by parts>, using $y(0)=0,y'(\pi/2)=0$, gives $\lambda\int y^2=\int(y')^2>0$ for every nonzero real <eigenfunction>. Thus $\lambda=k^2>0$ and the first condition makes $y=A\sin(kx)$. The second requires $\cos(k\pi/2)=0$, so $k=2n+1$, $n\geq0$. Since
$$
\int_0^{\pi/2}\sin^2((2n+1)x)\,dx=\frac\pi4,
$$
the normalized <eigenfunctions> and <eigenvalues> are
$$
\boxed{u_n(x)=\frac2{\sqrt\pi}\sin((2n+1)x),\qquad\lambda_n=(2n+1)^2,\qquad n=0,1,\ldots.}
$$
The <orthogonality> also follows by applying <integration by parts> to two different <eigenfunctions> of the same <self-adjoint differential operator>.
\b[Forced expansion.] Let $k_n=2n+1$ and write $x=\sum f_nu_n$, $y=\sum c_nu_n$. The <Fourier coefficients> of the forcing are
$$
f_n=\frac2{\sqrt\pi}\int_0^{\pi/2}x\sin(k_nx)\,dx=\frac{2(-1)^n}{\sqrt\pi\,k_n^2},
$$
because the boundary cosine vanishes and $\sin(k_n\pi/2)=(-1)^n$. Applying $d^2/dx^2+\kappa$ to the <eigenfunction expansion> gives $(\kappa-k_n^2)c_n=f_n$. Since $\kappa$ is not an <eigenvalue>,
$$
\boxed{y(x)=\frac4\pi\sum_{n=0}^\infty\frac{(-1)^n\sin(k_nx)}{k_n^2(\kappa-k_n^2)}.}
$$
The coefficients are $O(k_n^{-4})$, so the series and its first two differentiated series converge absolutely and uniformly on the interval. Each term satisfies the two endpoint conditions, and $y''+\kappa y$ equals the mixed-endpoint <Fourier series> of $x$. The forcing coefficients are $O(k_n^{-2})$, giving uniform convergence of that series as well. Uniqueness follows because the homogeneous problem has a nonzero solution only at one of the excluded <eigenvalues>.
\b[Odd fourth-power sum.] Set $\kappa=0$. Direct integration of $y''=x$, with the same endpoint conditions, gives
$$
y(x)=\frac{x^3}6-\frac{\pi^2x}8,\qquad y(\pi/2)=-\frac{\pi^3}{24}.
$$
At $x=\pi/2$, the <eigenfunction expansion> instead gives $y(\pi/2)=-(4/\pi)\sum_{n\geq0}k_n^{-4}$. Equating the two values yields
$$
\boxed{\sum_{n=0}^\infty\frac1{(2n+1)^4}=1+\frac1{3^4}+\frac1{5^4}+\cdots=\frac{\pi^4}{96}.}
$$
Absolute and uniform convergence justify evaluation at the endpoint.
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