= Solution
{c}
Use the unweighted <inner product> $\langle w,y\rangle=\int_0^a\overline w y\,dx$. Two applications of <integration by parts> give
$$
\int_0^a\overline w(y''+2y'+2y)\,dx
=\int_0^a(\overline w''-2\overline w'+2\overline w)y\,dx
+[\overline w y'-\overline w'y+2\overline w y]_0^a.
$$
The given <boundary conditions> set $y=0$ at both ends, but leave $y'(0)$ and $y'(a)$ independently variable. The remaining boundary term vanishes for every admissible $y$ precisely when $w(0)=w(a)=0$. Consequently the <formal adjoint> and the boundary conditions specifying the <adjoint operator> are
$$
\boxed{\mathcal L^*w=w''-2w'+2w,\qquad w(0)=w(a)=0}.
$$
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