**Background:**

I am reading the paper: *Best constant in Sobolev inequality* by Talenti (see [here][1]) and I am trying to understand the following step. 

On p. 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,

$$\left(r^{m-1}\left|u^{\prime}\right| \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0.\label{1}\tag{28}$$ 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,
$$\left(r v^{\prime}\right)^{2}=(1-m / 2)^{2} v^{2}-(2 C / q) v^{q}+\text { constant }.\label{2}\tag{$\ast$}$$
From this the author concludes that all the solutions of \eqref{1} 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 \eqref{2} allows the author to deduce that the solutions must take the form $u(r)=(a+br^2)^{1-m/2}$?



  [1]: https://math.jhu.edu/~js/Math646/talenti.sobolev.pdf