= Solution
The supplied PDF labels this equation \i[clarified from original for tutorial purposes]. Use the model of an <eccentric wheel with missing-mass kinetic energy neglected>: neglect the missing disc's contribution to <kinetic energy>, while retaining its subtraction from gravitational <potential energy>. Do not silently replace the gravitational <mass> $W$ by the full-disc <mass> $D$ as well.
Choose $\theta$ as the clockwise <angle> of $\overrightarrow{CH}$ from the upward vertical, and set $s=R\theta$. The <rolling without slipping> constraint makes changes in $s$ equal changes in the distance C travels down the slope. Choose its additive origin so $s=0$ when the hole is vertically above C. Then, up to an irrelevant height constant,
$$
y_C=-s\sin\alpha+\mathrm{constant},\qquad y_H=y_C+r_o\cos(s/R).
$$
The approximate <kinetic energy> is that of the complete uniform disc rolling at $\dot s=R\dot\theta$:
$$
T\simeq\frac12D\dot s^2+\frac12\left(\frac12DR^2\right)\frac{\dot s^2}{R^2}=\frac34D\dot s^2.
$$
Subtracting the missing disc's gravitational <potential energy> gives
$$
U=Dgy_C-hgy_H=-Wgs\sin\alpha-hgr_o\cos(s/R)+\mathrm{constant}.
$$
Apply the <Euler-Lagrange equation> to $\mathcal L=T-U$:
$$
\frac32D\ddot s+\frac{hgr_o}{R}\sin(s/R)-Wg\sin\alpha=0.
$$
Therefore
$$
\boxed{K_1=\frac{2hgr_o}{3DR}=\frac{2gr_o^3}{3R^3},\qquad K_2=-\frac{2Wg\sin\alpha}{3D}=-\frac{2g}{3}\left(1-\frac{r_o^2}{R^2}\right)\sin\alpha,\qquad\gamma=\frac1R.}
$$
Here $K_1,K_2$ have dimensions of <acceleration> and $\gamma$ has dimensions of inverse <length>. The initial conditions are $s(0)=\dot s(0)=0$ under the vertical-above interpretation. If an initial orientation is instead specified relative to the slope normal, the same coordinate convention produces a different initial $s$, not different constants. With $r_o\to0$ the equation becomes $\ddot s=(2/3)g\sin\alpha$, the familiar solid-disc result. Retaining the hole's exact <kinetic energy> would introduce coordinate-dependent inertia and extra $\dot s^2$ terms; the printed neglect is what permits the requested constant-coefficient form.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014-perforated-wheel-geometry.png]
{title=The axle, geometric centre and <centre of mass> of the perforated wheel}
{description=Original geometric constructions. Left: fixed-axis inertia subtracts a missing disc about H. Right: the rolling coordinate is the hole angle from the upward vertical; the centre travels downhill by R times the angle, while the missing mass contributes a cosine to potential energy.}
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