How to integrate this looking simple ODE?

Question

I meet an ODE about $V(\theta)$

$$\frac{\mathrm d^2V}{\mathrm d\theta^2}+\frac{1}{2V}=0.$$ But I can not figure out how to integrate it to yield

$$\left(\frac{\mathrm dV}{\mathrm d\theta}\right)^2+\log V=C_1$$ or $$\theta=\int^V\frac{\mathrm d V}{\sqrt{C_1-\log V}}+C_2,$$ where $C_1$, $C_2$ are constant.

If I check with Mathematica using

DSolve[D[V[\[Theta]], {\[Theta], 2}] + 1/(2 V[\[Theta]]) == 0, 
      V[\[Theta]], \[Theta]]

I obtain

(*{{V[\[Theta]] -> E^(
C[1] - InverseErf[-(Sqrt[E^(-2 C[1]) (\[Theta] + C[2])^2]/
  Sqrt[\[Pi]])]^2)}, {V[\[Theta]] -> E^(
C[1] - InverseErf[Sqrt[E^(-2 C[1]) (\[Theta] + C[2])^2]/
 Sqrt[\[Pi]]]^2)}}*)

Thanks!

Answer 1

Multiply everything by $\frac{dV}{d\theta}$ in your first expression. Then you get

$$ \frac{dV}{d\theta}\frac{d^2 V}{d\theta^2} + \frac{1}{2}\frac{1}{V}\frac{dV}{d\theta} = 0.$$

The left hand side can be recognized as

$$\frac{1}{2}\frac{d}{d\theta}\left(\frac{dV}{d\theta}\right)^2 + \frac{1}{2}\frac{d}{d\theta}\log V.$$