For a <Keplerian orbit>, write $h=Kr^{1/2}$, so $h'=h/(2r)$. Substituting the stated torque laws into
$$
\partial_r\mathcal G-r\mathcal T=\dot Mh'
$$
gives
$$
y'+\left(\frac1{2r}-\frac{\beta}{3\nu r^{1/2}}\right)y
=\frac{\dot M}{6\pi r}.
$$
With
$$
x=\frac r{r_{\rm in}},
\qquad
\lambda=-\frac{2\beta\sqrt{r_{\rm in}}}{3\nu},
$$
the <integrating factor> is $x^{1/2}e^{\lambda\sqrt x}$. The boundary condition $y(1)=0$ then gives
$$
y(x)=\frac{\dot M}{3\pi\lambda}x^{-1/2}
\left[1-e^{\lambda(1-\sqrt x)}\right].
$$
Thus
$$
\boxed{f(x)=1-\sqrt x}
$$
and
$$
\boxed{\nu\Sigma
=\frac{\dot M}{3\pi\lambda}x^{-1/2}
[1-\exp(\lambda f(x))]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-321/1/c/ii/solution