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?
Best Answer
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 $m=1$?