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I need to find out a solution to the equation:

z r[z] r''[z] - (d - 1) r[z] (r'[z])^3 - (d-2) z (r'[z])^2 - 
  (d - 1) r[z] r'[z] - (d - 2) z == 0

subject to the boundary conditions: r[0] == R and r'[z0] = ∞.

It's known that a solution exists and it is: r^2 + z^2 == R^2, I tried:

DSolve[
  {z r[z] r''[z] - (d - 1) r[z] (r'[z])^3 - (d - 2) z (r'[z])^2 - 
    (d - 1) r[z] r'[z] - (d - 2) z == 0, 
   r[0] == R, r'[z0] == ∞}, 
  r[z], z]

The equation and the solution can be found in arXiv: hep-th/0605073 page 43, eqn. 7.8

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  • $\begingroup$ You already have the answer:r^2 + z^2 = R^2, so what's the problem.? $\endgroup$ – Mariusz Iwaniuk May 12 at 9:37
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    $\begingroup$ what is zr[z] ? Is this supposed to be z*r[z] and what is r'[SubStar[z]] == is supposed to be? if SubStar[z] is supposed to be some constant, why not write r'[z0] or such so it is clear? $\endgroup$ – Nasser May 12 at 9:38
  • $\begingroup$ @MariuszIwaniuk I have the solution but don't know how to solve the equation. $\endgroup$ – Sabyasachi Maulik May 12 at 11:02
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    $\begingroup$ You believe the solution is independent of d? Why? $\endgroup$ – m_goldberg May 12 at 12:45
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    $\begingroup$ @Michael E2 Yes z0 is equal to R. $\endgroup$ – Sabyasachi Maulik May 12 at 14:17
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d = 2; sol = 
 DSolve[{z r[z] r''[z] - (d - 1) r[
       z] (r'[z])^3 - (d - 2) z (r'[z])^2 - (d - 1) r[z] r'[
       z] - (d - 2) z == 0}, r[z], z]

(*Out[]= {{r[z] -> 
   0}, {r[z] -> -I E^-C[1] Sqrt[-1 + E^(2 C[1]) z^2] + C[2]}, {r[z] ->
    I E^-C[1] Sqrt[-1 + E^(2 C[1]) z^2] + C[2]}}*)
 s = r[z] /. sol[[3]] /. C[2] -> 0

(*Out[]= I E^-C[1] Sqrt[-1 + E^(2 C[1]) z^2]*)
Solve[s^2 + z^2 == R^2, C[1]]

(*Out[]= {{C[1] -> 
   ConditionalExpression[1/2 (2 I \[Pi] C[2] + Log[1/R^2]), 
    C[2] \[Element] Integers]}}*)

Consequently $r(z)=\pm \sqrt {R^2-z^2}$

Note that in the article Aspects of Holographic Entanglement Entropy the authors cited another equation on p.34 $$ rzz′′ + (d−1)z(z′)^3 + (d−1)zz′ + dr(z′)^2 + dr = 0. $$ For this equation, the solution $z^2+r^2=R^2$ exists for any $d$.

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Let $$\rho^2 = z^2+r^2\quad\text{and}\quad \left({ds\over dz}\right)^2 = 1 + \left({dr \over dz}\right)^2\,.$$ Then the ODE is equivalent to $$z\,{d^2(\rho^2) \over dz^2} = (d-1)\left({ds\over dz}\right)^2 {d(\rho^2) \over dz} \,.$$ By inspection $\rho^2 = \text{constant}$ is a solution.

Check (there's a factor of two difference between the equations):

ode = z r[z] r''[z] -
 (d - 1) r[z] (r'[z])^3 - (d - 2) z (r'[z])^2 - (d - 1) r[z] r'[z] -
 (d - 2) z;
With[{ds = Sqrt[1 + r'[z]^2], ρ = Sqrt[z^2 + r[z]^2]},
 2 ode - (z D[ρ^2, {z, 2}] - (d - 1) ds^2 D[ρ^2, z]) // 
  Simplify
 ]
(*  0  *)
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