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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 $2\pi RSdR$ 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 givesIf 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, isHere $\nu$ 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 $f{Q}^{+}$ givesCombining this with mass conservation yields the wind-powered mass loss from a steady accretion disk relation

Define the outward viscous torque in an accretion disk asThe net inward angular momentum flux is $\dot{M}h-\mathcal{G}$. Because each unit mass lost to the wind carries the local specific angular momentum $h$, annular conservation of angular momentum readsIn the wind region, $\mathcal{G}=Rh{\dot{M}}^{\prime }/(3f)$. For constant $f$, differentiation and cancellation of $h$ givesThus the requested Cauchy-Euler differential equation is

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 $R_{\ast }\le R\le R_0$. A zero inner viscous torque in an accretion disk implies ${\dot{M}}^{\prime }(R_{\ast })=0$, so $B=A{R}_{\ast }^{3/2}/2$. There is no point mass sink at $R_0$, so the accretion rate is continuous there and equals ${\dot{M}}_0$. The complete result isIts 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 ${\dot{M}}^{\prime }(R_0)=0$ on its inner limit. The viscous torque in an accretion disk remains continuous and extends outside via ${\mathcal{G}}^{\prime }={\dot{M}}_0{h}^{\prime }$.

At the inner boundary the exact accreted fraction isThe 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 $R_{\ast }/R_0$ is small.