= Solution
Set $\rho=\rho_c e^{-\Psi(\xi)}$ and $r=a\xi$. Then $d\ln\rho/dr=-a^{-1}\Psi'$ and the equation from part (a) becomes
$$
-\frac1{a^2\xi^2}\frac d{d\xi}(\xi^2\Psi')
=-\frac{4\pi G\rho_c}{K}e^{-\Psi}.
$$
With $a^2=K/(4\pi G\rho_c)$, the coefficients cancel, giving the <isothermal Lane-Emden equation>
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}(\xi^2\Psi')=e^{-\Psi}.}
$$
The regular central <boundary conditions> are
$$
\boxed{\Psi(0)=0,\qquad\Psi'(0)=0.}
$$
The first makes the central <mass density> equal to $\rho_c$. The second follows from smooth <spherical symmetry> and the vanishing central <gravitational acceleration>; a preferred radial slope at the origin would be singular. Substitution of a regular <power series> gives
$$
\Psi(\xi)=\frac{\xi^2}{6}-\frac{\xi^4}{120}+O(\xi^6),
$$
which supplies nonsingular starting data for numerical integration. The central conditions select the regular solution, not the <singular isothermal sphere>.
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