= Solution
Use the outer expansions
$$
y=y_0+\epsilon y_1+\cdots,
\qquad
\omega^2=\omega_0^2+\epsilon\omega_1^2+\cdots.
$$
The reduced leading equation is
$$
y_0''+\omega_0^2y_0=0.
$$
For the symmetric fundamental mode, $y_0=a\cos(\omega_0x)$. The reduced second-order problem can impose the displacement conditions $y_0(\mathord\pm1)=0$ but not both clamped-slope conditions. The smallest positive root is
$$
\boxed{
\omega_0=\frac\pi2,
\qquad
y_0=a\cos\frac{\pi x}{2}}.
$$
Near the right endpoint introduce the stretched coordinate
$$
X_+=\frac{1-x}{\epsilon},
$$
and near the left endpoint use $X_-=(1+x)/\epsilon$. The leading outer solution is $O(\epsilon)$ in either inner region, so put $y=\epsilon Y_0+\cdots$. The dominant inner equation is
$$
Y_0''''-Y_0''=0.
$$
At either endpoint, clamping gives $Y_0(0)=Y_0'(0)=0$, bounded matching excludes the growing exponential, and the outer behavior requires
$$
Y_0\sim\frac{a\pi}{2}X_\pm
\qquad(X_\pm\to\infty).
$$
Thus the two leading inner solutions have the same form:
$$
\boxed{
y_{\rm in}^{\pm}
=\epsilon\frac{a\pi}{2}
\left(X_\pm-1+e^{-X_\pm}\right)}.
$$
They show explicitly that each clamped endpoint has a <boundary layer> of width $O(\epsilon)$.
At order $\epsilon$, the outer equation is
$$
y_1''+\frac{\pi^2}{4}y_1
=-\omega_1^2a\cos\frac{\pi x}{2}.
$$
The symmetric solution is
$$
\boxed{
y_1=b\cos\frac{\pi x}{2}
-\frac{\omega_1^2a}{\pi}
x\sin\frac{\pi x}{2}},
$$
which is the stated resonant particular solution plus a homogeneous normalization term.
As $x=1-\epsilon X_+$ approaches the right endpoint from the outer region,
$$
y_0+\epsilon y_1
\sim\epsilon
\left[
\frac{a\pi}{2}X_+
-\frac{\omega_1^2a}{\pi}
\right].
$$
The large-$X_+$ inner expansion is
$$
y_{\rm in}^+
\sim\epsilon
\left[
\frac{a\pi}{2}X_+
-\frac{a\pi}{2}
\right].
$$
Matching the constant terms gives
$$
\omega_1^2=\frac{\pi^2}{2}.
$$
The left endpoint gives the same condition by symmetry. Therefore
$$
\boxed{
\omega^2
\sim\frac{\pi^2}{4}
+\frac{\pi^2}{2}\epsilon
+O(\epsilon^2)}.
$$
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