I am an independently studying student, and I am trying to solve the question in the attached link.
Essentially, I need help solving Equation (3) in the below attached picture for the ratio $v_r/v_s$. I do not understand how to separate the variables to create an equation that I can integrate each side. Also, I do not know how I would use my initial condition that $y(w) = 0$.
I have reproduced the necessary equations here:
Using the fact that $dx/dt$ is the component of $\vec{v}$ in the $x$-direction and $dy/dt$ is the component of $\vec{v}$ in the $y$-direction we see from Figure 2(b) that \begin{align} \newcommand{\abs}[1]{\left|#1\right|} \renewcommand{\vec}[1]{\mathbf{#1}} \begin{split} \vec{v} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) &= \vec{v}_s + \vec{v}_r = (-v_s \cos \theta, -v_s \sin \theta) + (0, v_r) \\ &= (-v_s \cos \theta, v_r - v_s \sin \theta), \end{split} \tag{1} \end{align} where the magnitude $\abs{\vec{v}_r} = v_r$ and $\abs{\vec{v}_s} = v_s$ are speeds. By using the fact that the corresponding components in (1) the above equation are equal and that $\cos \theta = x/\sqrt{x^2+y^2}$, $\sin \theta = y/ \sqrt{x^2 + y^2}$ we 've constructed the system of first-order differential equations \begin{align} \begin{split} \frac{dx}{dt} &= - v_s \frac{x}{\sqrt{x^2+y^2}} \\ \frac{dy}{dt} &= v_r - v_s \frac{y}{\sqrt{x^2+y^2}}. \end{split} \tag{2} \end{align} Because $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$, the solution curves of this system of differential equations satisfy the single first-order equation $$ \tag{3} \frac{dy}{dx} = \frac{y-(v_r/v_s)\sqrt{x^2+y^2}}{x}. $$
Edit: Thank you for the help with formatting!