How to solve this autonomous DE with RK 4? I have this equation :
$$\frac{d\alpha}{dz} = - \frac{dr}{dz} * \frac{\tan(\alpha)}r $$
I searched for some similar examples but non of these equations was like this one.
I'm confused about this one. As far as I know, I used RK4 for equations like this : $$y'(t) = F(t,y(t))$$
Thank you for helping me !

Here's the context for the equation.

I just deleted the lambda part to make the equation easier. But I still don't figure out how to solve it!
 A: One has step by step :
$$
\begin{array}{rcl}
\displaystyle\frac{\mathrm{d}\alpha}{\mathrm{d}z} = -\frac{\tan\alpha}{r}\frac{\mathrm{d}r}{\mathrm{d}z}
   &\Leftrightarrow& \displaystyle
   \frac{\mathrm{d}\alpha}{\mathrm{d}z} = -\tan\alpha\,\frac{\mathrm{d}}{\mathrm{d}z}\ln r \\
   &\Leftrightarrow& \displaystyle
   \frac{\mathrm{d}\alpha}{\tan\alpha} = -\mathrm{d}(\ln r) \\
   &\Leftrightarrow& \displaystyle
   \int_{\alpha_0}^\alpha\frac{\cos\alpha}{\sin\alpha}\mathrm{d}\alpha = -\int_{r_0}^r\mathrm{d}(\ln r) \\
   &\Leftrightarrow& \displaystyle
   \frac{\mathrm{d}\alpha}{\tan\alpha} = -\mathrm{d}(\ln r) \\
   &\Leftrightarrow& \displaystyle
   \ln\left(\frac{\sin\alpha}{\sin\alpha_0}\right) = -\ln\left(\frac{r}{r_0}\right) \\
   &\Leftrightarrow& \displaystyle
   \sin\alpha = \frac{r_0}{r}\sin\alpha_0 \\
   &\Leftrightarrow& \displaystyle
   \alpha = \arcsin\left(\frac{r_0}{r}\sin\alpha_0\right)
\end{array}
$$
with $r_0 = r(z_0)$ and $\alpha_0 = \alpha(z_0)$.
A: Without knowing something more than what you've given us, you can't "solve" this equation for $\alpha$ and $r$ as functions of $z$.  What you can do is write it as
$$ \dfrac{d\alpha}{dr} = -\frac{\tan \alpha}{r} $$
and solve for $\alpha$ as a function of $r$ (whether in closed form, or using RK4, as you wish).
A: Interpreting $(r,z)$ as meridian coordinates among the as yet undefined cylindrical coordinates in three space
$(r, z, \theta)$ and $\alpha$ is some angle the
arc makes to meridian on the surface..
we have after removing $z$.. an ODE with straight separation of variables:
$$ - {\cot\alpha}\cdot d\alpha =\frac{ dr} {r} $$
Integrating
$$-\log \sin \alpha = \log~ r -\log ~ r_{min} $$
with an arbitrary constant $r_{min}$ brought in
$$ r\cdot \sin \alpha = ~ r_{min} $$
which is the geodesics describing Clairaut's Law for all arcs on a surface of revolution... tangential to a constant minimum circle $ r= r_{min}$.  Note that it did not require RK4 for a solution.
In order to further solve for variables $(r, z, \theta)$  it needs a longer answer. We need more independent equations e.g.,
$$ \frac {dr}{dz}= \tan \phi = 1 $$
for a cone of semi-vertical angle $ 45^{\circ}$
and... coupled differential equations w.r.t. an independent variable $ \theta $ or arc $s.$
As an example with independent variable arc $s$ we can solve either analytically in simple cases or numerically using RK4 a set of simultaneous / coupled set of differential equations (primed on $~s):$
$$  \alpha'= f(s), \phi'=g(s), r'= \sin \phi \cos \alpha, z' =\cos \alpha \cos \phi, r\theta' = \sin \alpha. $$
A: Thank you all for helping me.
Here's the context for the equation.

I just deleted the lambda part to make the equation easier. But I still don't figure out how to solve it !
