= Solution
Interpret $\sigma_{\mathrm{DM}}$ as a constant one-dimensional, isotropic <velocity dispersion>. The effective <pressure> is $p=\rho\sigma_{\mathrm{DM}}^2$, corresponding to the stated gas-temperature analogy. For $\rho=A/r^2$, the spherical enclosed mass is
$$
M_{\mathrm{sph}}(r)=4\pi\int_0^r s^2\rho(s)\,ds=4\pi Ar.
$$
The two sides of the <hydrostatic equilibrium> equation are then
$$
\frac1\rho\frac{dp}{dr}=-\frac{2\sigma_{\mathrm{DM}}^2}{r},\qquad
-\frac{GM_{\mathrm{sph}}(r)}{r^2}=-\frac{4\pi GA}{r}.
$$
Equating them fixes the <singular isothermal sphere>:
$$
\boxed{A=\frac{\sigma_{\mathrm{DM}}^2}{2\pi G},\qquad
\rho(r)=\frac{\sigma_{\mathrm{DM}}^2}{2\pi Gr^2}.}
$$
For collisionless <dark matter> this is an isotropic Jeans-equation construction, not a claim that the particles thermalize by gas collisions. The central singularity and infinite total mass show that the untruncated profile is an idealization.
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