= Solution
{c}
Continue with the active sign convention for the <linearized coordinate gauge transformation>. Choose local <functions> $A,B,C,F$ of $u$ by <integration> so that
$$
A''=\frac c2,\qquad B''=\frac d2,\qquad C''=\frac e2,\qquad F'=-\frac f2.
$$
For example, for any fixed $u_0$ one may take
$$
A(u)=\frac12\int_{u_0}^u(u-s)c(s)\,ds,\quad
B(u)=\frac12\int_{u_0}^u(u-s)d(s)\,ds,\quad
C(u)=\frac12\int_{u_0}^u(u-s)e(s)\,ds,\quad
F(u)=-\frac12\int_{u_0}^u f(s)\,ds.
$$
A generator, specified as a <covector> with components in the <null coordinates> $(u,v,x,y)$, is
$$
\boxed{\chi=A\,dv+B\,dx+C\,dy+
\bigl(-A'v-B'x-C'y+F\bigr)\,du.}
$$
Indeed, with $\delta h_{AB}=\partial_A\chi_B+\partial_B\chi_A$, the mixed components cancel:
$$
\delta h_{uv}=A'-A'=0,\qquad
\delta h_{ux}=B'-B'=0,\qquad
\delta h_{uy}=C'-C'=0.
$$
All components without a $u$ index vanish as well, and
$$
\delta h_{uu}=-2A''v-2B''x-2C''y+2F'
=-cv-dx-ey-f.
$$
Since $n^\flat=du$, only the $uu$ component needs changing. Thus
$$
\boxed{\mathring h=\psi\,du^2,\qquad
\psi=a(u)(x^2-y^2)+b(u)xy.}
$$
\b[Yes, $\mathring h$ satisfies both equations (3).] Its <trace> is zero, $\partial_v\psi=0$ gives zero <divergence>, and the <harmonic quadratic plane-wave profile> gives
$$
\partial^A\overline{\mathring h}_{AB}=2\partial_v\psi\,\delta_B{}^u=0,
\qquad
\Box\psi=4\partial_u\partial_v\psi+\partial_x^2\psi+\partial_y^2\psi=2a-2a=0.
$$
The two surviving <functions> encode the two <gravitational wave polarizations>; constant and linear transverse terms have been removed by the explicit <linearized coordinate gauge transformation>.
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