With the <polytropic equation of state> and substitutions
$$
P=K\rho_c^{1+1/n}\theta^{n+1},
\qquad
\rho=\rho_c\theta^n,
\qquad
r=a\xi,
$$
we have
$$
\frac1\rho\frac{dP}{dr}
=(n+1)K\rho_c^{1/n}\frac{d\theta}{dr}.
$$
The pressure-support equation therefore becomes
$$
(n+1)K\rho_c^{1/n}\frac1{r^2}
\frac d{dr}\left(r^2\frac{d\theta}{dr}\right)
=-4\pi G\rho_c\theta^n.
$$
Choose
$$
\boxed{a^2=\frac{(n+1)K}{4\pi G}\rho_c^{(1-n)/n}.}
$$
Changing from $r$ to $\xi$ then yields the <Lane-Emden equation>
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}
\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n.}
$$
These are the <Lane-Emden variables for a stellar polytrope>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 14B / b / Solution

For the <nonrelativistic white dwarf>, combine the <hydrostatic pressure support equation> and <enclosed mass> equation:
$$
\frac{dP}{dr}=-\frac{Gm\rho}{r^2},\qquad
\frac{dm}{dr}=4\pi r^2\rho.
$$
Eliminating $m$ yields
$$
\boxed{\frac1{r^2}\frac d{dr}\left(\frac{r^2}{\rho}\frac{dP}{dr}\right)=-4\pi G\rho,\qquad P=K_{\rm NR}\rho^{5/3}.}
$$
Introduce the <Lane-Emden variables for a stellar polytrope>,
$$
\rho=\rho_c\theta^{3/2},\qquad r=a\xi,\qquad
 a^2=\frac{5K_{\rm NR}}{8\pi G}\rho_c^{-1/3}.
$$
The resulting <Lane-Emden equation> is
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}(\xi^2\theta')=-\theta^{3/2},\qquad
\theta(0)=1,\quad\theta'(0)=0.}
$$
The <white dwarf> surface is at the first zero $\xi_1$ of $\theta$, with $R=a\xi_1$.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 317 / 3 / ii / a / Solution

{c}

Writing
$$
\rho=\rho_c\theta^n,\qquad r=\alpha\xi,\qquad
\alpha^2=\frac{(n+1)K}{4\pi G}\rho_c^{(1-n)/n}
$$
reduces Newtonian hydrostatic balance to
$$
\frac1{\xi^2}\frac d{d\xi}\left(\xi^2\theta'\right)=-\theta^n,
\qquad \theta(0)=1,\quad\theta'(0)=0.
$$
The stellar surface is the first zero $\xi_1$ of $\theta$, and $R=\alpha\xi_1$.


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