Trajectory of a swimmer trying to reach the opposite bank I came across the following question 2 days ago:

$Q$. A swimmer is swimming in a river with a speed $u$ with respect to the speed of water which is $v$. It is assumed that the river has straight parallel banks and the speed of water is the same at all points and has a direction parallel to the banks. Assume that the swimmer starts swimming at a point $A$. He always faces a point $B$ on the opposite bank. $B$ is the point that is directly opposite to $A$. Find the trajectory of the swimmer. The width of the river is $w$.

I assumed that point $A$ is the origin i.e., $A\equiv (0,0)$ and $B\equiv (0,w)$. Let's say the swimmer is at point $P\equiv (x,y)$ at some point of time. At that point, he makes an angle $\theta$ with the horizontal with respect to point $B$. Therefore $\cos\theta=\frac x{\sqrt{x^2+(w-y)^2}}$ and $\sin\theta=\frac {w-y}{\sqrt{x^2+(w-y)^2}}$. Now I got the following two equations:-$$\frac {dx}{dt}=v-u\cos\theta=v-\frac {ux}{\sqrt{x^2+(w-y)^2}}\quad and \quad\frac {dy}{dt}=u\sin\theta=\frac {u(w-y)}{\sqrt{x^2+(w-y)^2}}$$Dividing them I got this differential equation:-$$\frac {dy}{dx}=\frac {u(w-y)}{v\sqrt{x^2+(w-y)^2}-ux}$$How do I solve this differential equation? Is there any solution that doesn't use calculus in such type of questions? You may also use the fact that $$\frac d{dt}(\sqrt{x^2+(w-y)^2})=v\cos\theta-u=\frac {vx}{\sqrt{x^2+(w-y)^2}}-u$$
 A: Let me suppose the swimmer begins at a point $I = (\omega, 0)$ and that your target is the origin $ F = (0,0)$ - I think some of the equations are simpler if the initial position, rather than the target position, is at the origin. In my co-ordinates, the river bank is in the $y$-direction.
The river flows at speed $v$ and the swimmer's speed is $u$. We have that
$$
\begin{align}
\dot x =& -u\cos\theta\\
\dot y =& -u\sin\theta + v
\end{align}
$$ 
We ultimately want $y(x)$ so we can plot the swimmers trajectory. We can write that 
$$
\frac{dy}{dx} = \frac{\dot y}{\dot x} = \tan\theta - \frac vu \frac1{\cos\theta} = \frac yx - \frac vu \sqrt{1 + \left(\frac yx\right)^2}
$$
Now let $p = \frac yx$ such that $\frac{dy}{dx} = p + x\frac{dp}{dx}$. We find
$$
\begin{align}
p + x\frac{dp}{dx} =& p - \frac vu \sqrt{1 + p^2}\\
- \frac vu \frac{dx}{x} =& \frac{dp}{\sqrt{1 + p^2}}
\end{align}
$$
You can integrate the RHS with the subsitution $p=\tan \theta$.
$$\begin{align}\int {\frac{dp}{\sqrt{1 + p^2}}}&=\int {\sec^2 \theta d\theta \over \sec \theta}\\&=\int{\sec\theta d\theta}\\&=\ln(\sec\theta+\tan\theta)+C\\&=\ln(\sqrt{1+p^2}+p)+C\end{align}$$ 
Hence
$$
- \frac vu \ln x = \ln(\sqrt{1 + p^2} + p) + C
$$
which you can solve to find that
$$
p = \frac12 \left[ e^C x^{-v/u} - e^{-C} x^{v/u}\right]
$$
Finally you can find that
$$
y = \frac12 \left[ e^C x^{1-v/u} - e^{-C} x^{1+v/u}\right]
$$
Now put it your initial condition that $y(\omega) = 0$, you can solve for $C$ and find that,
$$
y = \frac\omega2 \left[ \left(\frac{x}{\omega}\right)^{1-v/u} - \left(\frac{x}{\omega}\right)^{1+v/u}\right]
$$
If $v/u < 1$, the swimmer hits the target from a direction parallel to the riverbank, hitting the bank exactly at the target, swimming directly against the current when he hits the target. The bigger $v/u < 1$, the further he travels downstream before making it to his target (though this is bounded at $\omega/2$ - the longest paths approximately hit the river bank at $\omega/2$ then move towards the target). If $v/u > 1$, then $y(0)$ divergences - the current takes you infinitely down the river, asymptotically reaching the shore-line. If $v/u=1$, I think you come to a rest on the riverbank at $\omega/2$ downstream of your target.
I doubt there are any solutions without calculus, other than general reasoning about the swimmer's trajectory in the 3 cases above.
