Background:

I am reading the paper: Best constant in Sobolev Inequality by Talenti (see here) and I am trying to understand the following step.

On Pg. 365, the author is arguing that the solutions to the following equation

$$\left(r^{m-1}\left|u^{\prime}\right|^{\dot{p}-1} \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0 \quad(C=\text { a positive constant })$$

take the form $(a+br^p)^{1-m/p}.$ Here $m$ is the dimension of the space, $q=\frac{2m}{m-2}$ and $1<p<m.$ The above ode corresponds to the radial solutions for the Euler Lagrange equation associated to the Sobolev inequality.

He begins by considering the case $p=2$. Then we have,

$$\left(r^{m-1}\left|u^{\prime}\right| \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0.\quad \quad (28)$$ By considering $u(r)=r^{1-m/2}v(r)$ we obtain the following ode for $v$, $$r\left(r v^{\prime}\right)^{\prime}=(1-m / 2)^{2} v-C v^{q-1}.$$ By multiplying the above ode by $v$ and integrating this is equivalent to solving, $$\left(r v^{\prime}\right)^{2}=(1-m / 2)^{2} v^{2}-(2 C / q) v^{q}+\text { constant }.\quad \quad (*)$$ From this the author concludes that all the solutions of $(28)$ that are positive decreasing and satisfy the following decay conditions $$u(r)=o(r^{1-m/2}),u'(r)=o(r^{-m/2})$$ must be of the form $u(r)=(a+br^2)^{1-m/2}$ for constants $a$ and $b$ related to $C$.

Question: I am not sure how looking at the ode $(*)$ allows the author to deduce that the solutions must take the form $u(r)=(a+br^2)^{1-m/2}$?