The pilot's apparent (to himself) velocity is the relative velocity, $\displaystyle \mathbf{v'}$ , given by
$\displaystyle \mathbf{v'} = \mathbf{v} - \mathbf{w}$
where $\displaystyle \mathbf{v}$ is the actual velocity and
$\displaystyle \mathbf{w} = w \hat{\mathbf{j}}$
is the wind velocity. Now, since he is always flying towards $\displaystyle O$ we have that
$\displaystyle \mathbf{v'} = -v \hat{\mathbf{e}}_r = -\frac{v}{\sqrt{x^2 +y^2}} (x \hat{\mathbf{i}} + y\hat{\mathbf{j}})$
so
$\displaystyle \mathbf{v} = \mathbf{v'} + \mathbf{w}$
$\displaystyle = -v \hat{\mathbf{e}}_r +w \hat{\mathbf{j}}$
$\displaystyle = -\frac{v x}{\sqrt{x^2 +y^2}} \hat{\mathbf{i}} + \left(w -\frac{vy}{\sqrt{x^2 +y^2}} \right)\hat{\mathbf{j}}$.
Now, note that $\displaystyle v_x = \dot{x}$ and $\displaystyle v_y = \dot{y}$ so
$\displaystyle \frac{dy}{dx} = \frac{v_y}{v_x}$
$\displaystyle \Leftrightarrow \frac{dy}{dx} = \frac{y}{x} - \frac{w}{v} \sqrt{1+ \left(\frac{y}{x}\right)^2}$.
Now let $\displaystyle y = z x$ so
$\displaystyle \frac{dy}{dx} = x \frac{dz}{dx} + z$ so our equation becomes
$\displaystyle x \frac{dz}{dx} + z = z - \frac{w}{v} \sqrt{1 + z^2}$
$\displaystyle \Leftrightarrow x \frac{dz}{dx} = - \frac{w}{v} \sqrt{1 + z^2}$
Integrating and using the fact that at $\displaystyle x=D$, $\displaystyle y = 0$ we have
$\displaystyle \int_{0}^{z} \frac{1}{\sqrt{1+{z'}^2}} \, \mathrm{d}z' = -\frac{w}{v} \int_{D}^{x} \frac{1}{x'}\, \mathrm{d}x'$
giving us
$\displaystyle \sinh^{-1} z = \frac{w}{v} \ln \left(\frac{D}{x} \right)$,
$\displaystyle \Leftrightarrow z = \sinh \left[ \frac{w}{v} \ln \left(\frac{D}{x} \right) \right]$,
$\displaystyle \boxed{ \Leftrightarrow y = x \sinh \left[ \frac{w}{v} \ln \left(\frac{D}{x} \right) \right]}$.