The problem
Suppose that a block sliding with a positive velocity $v$ along the $x$-axis experiences a drag force, $$F(v)=-F_0e^{Kv},$$ where $F_0$ and $K$ are positive constants. At time $t=0$, the block is at $x=0$, and has initial velocity $v=v_0$. Find the time at which the block's velocity momentarily goes to $0$, and also find the block's displacement at that time.
Attempt at a solution
We have, $$F=ma=m\frac{dv}{dt}=-F_0e^{Kv}.$$ This can be rearranged to, $$e^{-Kv}dv=-\frac{F_0}{m}dt.$$ Then, we can integrate both sides subject to specified initial conditions. We have, $$\int_{v_0}^{v}e^{-Kv'}dv'=\int_{0}^{t}-\frac{F_0}{m}dt'\rightarrow -\frac{1}{K}(e^{-Kv})\bigg|_{v_0}^{v}=-\frac{F_0}{m}t\rightarrow \frac{1}{K}(e^{-Kv}-e^{-Kv_0})=\frac{F_0}{m}t.$$ We can rearrange to solve for velocity by, $$e^{-Kv}=\frac{F_0K}{m}t+e^{-Kv_0}\rightarrow -Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=-\frac{1}{K}\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}.$$ Then, to find the instant at which $v=0$, we solve for the zeroes of this express. Because $K$ is a constant, it remains that, $$\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}=0,$$ such that, $$\frac{F_0K}{m}t+e^{-Kv_0}=1.$$ Solving for $t$ we have, $$t=\frac{m}{F_0K}(1-e^{-Kv_0}),$$ when $v=0$. It remains to find position as a function of time, such that we can find the position of the particle when velocity is $0$. We note that, $$F=m\frac{dx}{dt}\frac{dv}{dx}=mv\frac{dv}{dx}.$$ Then we have that, $$mv\frac{dv}{dx}=-F_0e^{Kv}\rightarrow ve^{-Kv}dv=-\frac{F_0}{m}dx.$$ Again, integrating both sides we have, $$\int_{v_0}^{v}v'e^{-Kv'}dv'=\int_{0}^{x}-\frac{F_0}{m}dx'.$$ Integrating by parts we have, $$-\frac{1}{K}v'e^{-Kv'}\bigg|_{v_0}^{v}+\frac{1}{K}\int_{v_0}^{v}e^{-Kv'}dv'=-\frac{F_0}{m}x.$$ Then, $$-\frac{1}{K}(ve^{-Kv}-v_0e^{-Kv_0})-\frac{1}{K^2}(e^{-Kv'}\bigg|_{v_0}^{v})=-\frac{F_0}{m}x,$$ and, $$-\frac{1}{K}(ve^{-Kv}-v_0e^{-Kv_0})-\frac{1}{K^2}(e^{-Kv}-e^{-Kv_0})=-\frac{F_0}{m}x.$$ It remains to find $x$ when $v=0$. We have (when $v=0$), $$\frac{v_0}{K}e^{-Kv_0}+\frac{1}{K^2}e^{-Kv_0}-\frac{1}{K^2}=-\frac{F_0}{m}x.$$ Rearranging gives, $$x=-\frac{mv_0}{F_0K}e^{-Kv_0}-\frac{m}{F_0K^2}e^{-Kv_0}+\frac{m}{F_0K^2},$$ when $v=0$.
Just wondering if I've made a silly calculation error (or logic error), as the solution doesn't look as nice as I'd expected.