Integrate and derivative

Question

i'm not able to explain the following step:

$\frac{1}{k+v(x)}=\frac{d^2 v}{dx^2}$ by integrating this equation:

$(C-\frac{1}{k+v(x)})^{\frac{1}{2}}=\frac{dv}{dx}$

Please, if somebody can help i'll be very grateful!

Answer 1

Multiplying both sides by $v'dx=dv$, you have $$\frac{dv}{k+v}=v'dv'.$$ Integrating, $$\ln(k+v)+C=\frac12v'^2,$$and $$(2\ln(k+v)+C)^{1/2}=\frac{dv}{dx}.$$ Not the expected answer.

Likely fix: $$\frac{dv}{(k+v)^\color{red}2}=\color{red}2v'dv',$$ $$C-\frac1{k+v}=v'^2,$$ $$(C-\frac1{k+v})^{1/2}=\frac{dv}{dx}.$$