why is $\dfrac{dr}{r~d\theta} = \cot \psi$ ?
Extracted from Ordinary Differential Equations, Garrett Birkhoff, in the chapter of Linear Fractional Equations (First order Differential Equations).
When solving an homogeneous linear fractional equation $$\frac{dy}{dx}=\frac{cx+dy}{ax+by}$$
We came up with:
$$\frac{dv}{F(v)-v}=\frac{dx}{x}=d(\ln x)$$ where $v={y\over x}$
Finding the solution of $x$ is straightforward. However, I do not understand when we "introduce polar coordinates".
The book says: $x=r \sin\theta$ and $y=r \cos\theta$. If $\psi=\gamma - \theta$ is the angle between the tangent direction $\gamma$ and the radial direction $\theta$, then:
$$\frac{dr}{r~d\theta} = \cot \psi$$
Also, why do we do that? is it equal to $d (\ln(x))$? May you explain it?
$\frac{dr}{r~d\theta} = \cot \psi$ is a general property which can be analytically proved. It can be intuitively understood on the next figure :