solution to the differential equation $ma=mg-kv^2$ (free fall)

Question

The differential equation $ma=mg-kv^2$ is to describe the motion of a particle that is falling from a tall building, and the air resistance is proportional to the square of the velocity of the particle.

I solve the DE by integration using the initial condition $x=0$ when $t=0$ and the assumption that $g \ge kv^2$, and obtained $v=\sqrt{\dfrac{g}{k}(1-e^{-2kx})}$, where $x$ is the distance the particle travels downwards in metres.

Did I obtain the correct particular solution?

Answer 1

$$ma=mg-kv^2$$ $$mv\frac{dv}{dx}=mg-kv^2$$ $$\int mv\frac{dv}{mg-kv^2}=\int dx$$ $$-\dfrac{m\ln\left(\left|kv^2-gm\right|\right)}{2k}+C=x$$

at $x=0$ $v=0$ $$C=\dfrac{m\ln\left(\left|gm\right|\right)}{2k}$$

$$-\dfrac{m\ln\left(\left|kv^2-gm\right|\right)}{2k}+\dfrac{m\ln\left(\left|gm\right|\right)}{2k}=x$$

We have $mg\ge kv^2$

thus $$v^2=\frac{mg}{k}(1-e^{-2kx/m})$$

Answer 2

I solved the differential equation and came up with $$v=\sqrt{\dfrac{mg}{k}(1-e^{-2kx/m})}$$ Which is slightly different from your answer.

Are you assuming $m=1$?