= Solution
Take $\dot M>0$ to mean inward <accretion rate>, and let $S$ count loss through both faces per unit projected area of the <accretion disk>. An annulus of width $dR$ loses $2\pi R S\,dR$ to the wind. In steady state, its incoming mass flux at its outer edge equals the inward flux at its inner edge plus this loss. Hence <mass conservation> gives
$$
\boxed{\frac{d\dot M}{dR}=2\pi RS.}
$$
If a loss rate is instead defined per individual face, it must be doubled before substituting for $S$ here.
Use <Keplerian rotation> $\Omega^2=GM/R^3$. The vertically integrated <viscous dissipation>, summed over both disk faces, is
$$
Q^+=\nu\Sigma\left(R\frac{d\Omega}{dR}\right)^2=\frac94\nu\Sigma\Omega^2.
$$
Here $\nu$ is the <kinematic viscosity>, and $\Sigma$ is the <surface density of a disk>; for height-dependent viscosity the product means $\int\rho\nu\,dz$. Material already in a circular orbit has <specific orbital energy> $v_K^2/2-GM/R=-GM/(2R)$. Giving it zero total binding energy therefore costs an additional $GM/(2R)$ per unit mass. The total launch speed is the <escape velocity> $\sqrt{2GM/R}$, but one must not charge the wind for the orbital <kinetic energy> it already possesses. Equating wind power to $fQ^+$ gives
$$
f\frac94\nu\Sigma\frac{GM}{R^3}=S\frac{GM}{2R}.
$$
Combining this with <mass conservation> yields the <wind-powered mass loss from a steady accretion disk> relation
$$
\boxed{\nu\Sigma=\frac{2SR^2}{9f}=\frac{R}{9\pi f}\frac{d\dot M}{dR}.}
$$
Define the outward <viscous torque in an accretion disk> as
$$
\mathcal G=-2\pi R^3\nu\Sigma\frac{d\Omega}{dR}=3\pi\nu\Sigma h,\qquad h=R^2\Omega=\sqrt{GMR}.
$$
The net inward <angular momentum> flux is $\dot Mh-\mathcal G$. Because each unit mass lost to the wind carries the local <specific angular momentum> $h$, annular <conservation of angular momentum> reads
$$
\frac{d}{dR}(\dot Mh-\mathcal G)=h\frac{d\dot M}{dR},\qquad
\mathcal G'=\dot M h'=\frac{\dot M h}{2R}.
$$
In the wind region, $\mathcal G=Rh\dot M'/(3f)$. For constant $f$, differentiation and cancellation of $h$ gives
$$
R\dot M''+\frac32\dot M'=\frac{3f}{2R}\dot M.
$$
Thus the requested <Cauchy-Euler differential equation> is
$$
\boxed{R^{1/2}\frac{d}{dR}\left(R^{3/2}\frac{d\dot M}{dR}\right)=\frac32f\dot M.}
$$
For $f=1/3$, a power-law trial $\dot M\propto R^p$ gives $p^2+p/2-1/2=0$, with roots $p=1/2,-1$. Therefore $\dot M=A\sqrt R+B/R$ in $R_*\le R\le R_0$. A zero inner <viscous torque in an accretion disk> implies $\dot M'(R_*)=0$, so $B=AR_*^{3/2}/2$. There is no point mass sink at $R_0$, so the <accretion rate> is continuous there and equals $\dot M_0$. The complete result is
$$
\boxed{\dot M(R)=\begin{cases}
\displaystyle\dot M_0\frac{\sqrt R+R_*^{3/2}/(2R)}{\sqrt{R_0}+R_*^{3/2}/(2R_0)},&R_*\le R\le R_0,\\
\dot M_0,&R\ge R_0.
\end{cases}}
$$
Its derivative is nonnegative throughout the wind region, as required for a nonnegative loss rate $S$. The derivative can jump to zero outside $R_0$: the energy relation applies only inside the wind zone and does not impose $\dot M'(R_0)=0$ on its inner limit. The <viscous torque in an accretion disk> remains continuous and extends outside via $\mathcal G'=\dot M_0h'$.
At the inner boundary the exact accreted fraction is
$$
\boxed{\frac{\dot M(R_*)}{\dot M_0}=\frac{\tfrac32\sqrt{R_*/R_0}}{1+\tfrac12(R_*/R_0)^{3/2}}
\simeq\frac32\left(\frac{R_*}{R_0}\right)^{1/2}\quad(R_*\ll R_0).}
$$
The last expression is a leading-order asymptotic equality, not the exact finite-radius fraction. Most of the supplied mass is expelled before reaching the star when $R_*/R_0$ is small.
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