= Solution
Let $L=-(1-x^2)D^2+xD$, so $Ly_n=n^2y_n$ and the forced equation is $(\gamma^2-L)Z=e^{\varepsilon x}$. The small-parameter <Taylor expansion> of the forcing is
$$
e^{\varepsilon x}=1+\varepsilon x+\frac{\varepsilon^2x^2}{2}+O(\varepsilon^3)
=y_0+\varepsilon y_1+\frac{\varepsilon^2}{4}(y_0+y_2)+O(\varepsilon^3).
$$
Each <Chebyshev polynomial> can be inverted separately under $\gamma^2-L$. None of the required denominators vanishes because $\gamma$ is a real noninteger. Thus the <regular solution of a forced Chebyshev equation> has
$$
\boxed{A=\frac{y_0}{\gamma^2},\qquad
B=\frac{y_1}{\gamma^2-1},\qquad
C=\frac{y_0}{4\gamma^2}+\frac{y_2}{4(\gamma^2-4)}}.
$$
Here we select the same regular endpoint domain as in part (a), with finite function values and finite endpoint <derivatives>. This qualification is needed: the printed forced equation alone does not state endpoint conditions. Without this selection one can add $c_1\cos(\gamma\arccos x)+c_2\sin(\gamma\arccos x)$, which remain finite but generally have singular endpoint <derivatives>. With $x=\cos\theta$, regularity gives $Z_\theta(0)=Z_\theta(\pi)=0$; the corresponding homogeneous equation has only the zero solution for noninteger $\gamma$. This makes the displayed expansion unique.
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