The solution to this differential equation
$$r^2(r^2-{r_+}^2)(\frac{d\phi}{dr})^2+e^{2\phi(r)}=2r^2-{r_+}^2$$
is given by $\phi(r)=\ln{r}$ and $(r>r_+)$. I want to solve this equation explicitly as I have to solve this equation again with a different inhomogeneous term i.e.,$(2r^2-{r_+}^2)$ replaced by $2r^2$.
In order to get rid of the exponential I substituted $\phi(r)=\ln{y(r)}$ in the equation
$$\frac{r^2}{y^2}(r^2-{r_+}^2)(\frac{dy}{dr})^2+y^2=2r^2-{r_+}^2$$ I am stuck here as I am unable to figure out the next substitution.Any comments on the solution after changing the inhomogeneous term will also be helpful.
Rearrange the first equation to obtain
$$(r^2-r_+^2)\left[\left(\frac{d\phi}{d\log r}\right)^2-1\right] = r^2-e^{2\phi}$$
This motivates checking for solutions at the phase transitions (or pseudo equilibria) of the equation
$$\frac{d\phi}{d\log r} = \pm 1$$
of which only $+1$ returns a valid solution. To continue with a phase analysis, consider the sign of $r^2-e^{2\phi}$ when $r>r_+$ and when $r<r_+$. What does this imply about the growth of the function (and hence the (non?)existence of certain solutions) ?
An application of Picard-Lindelof can prove the uniqueness of a solution for this equation by taking cases.