- #1
Radek Vavra
- 4
- 0
So, as a result of a thought experiment, I've got a differential equation, which I can't solve:
[itex]
R r'' \sin \frac{r}{R} - 2 (r')^2 \cos \frac{r}{R} - R^2 \cos \frac{r}{R} \sin^2 \frac{r}{R} = 0
[/itex], [itex]R > 0[/itex]
To make the matters worse, the function [itex]r(\varphi)[/itex] will probably depend on multiple parameters, because when I put [itex]r << R[/itex], I could approximate the equation:
[itex]
r r'' - 2 (r')^2 - r^2 = 0
[/itex]
which gave solution (mostly by lucky guess):
[itex]
r = \frac{a}{\sin \varphi + b \cos \varphi}
[/itex], [itex]a\in\mathbb R[/itex], [itex]b\in\mathbb R[/itex]
Since I'm used only to the simplest types of differential equations, could you please help me and describe every step :shy:
[itex]
R r'' \sin \frac{r}{R} - 2 (r')^2 \cos \frac{r}{R} - R^2 \cos \frac{r}{R} \sin^2 \frac{r}{R} = 0
[/itex], [itex]R > 0[/itex]
To make the matters worse, the function [itex]r(\varphi)[/itex] will probably depend on multiple parameters, because when I put [itex]r << R[/itex], I could approximate the equation:
[itex]
r r'' - 2 (r')^2 - r^2 = 0
[/itex]
which gave solution (mostly by lucky guess):
[itex]
r = \frac{a}{\sin \varphi + b \cos \varphi}
[/itex], [itex]a\in\mathbb R[/itex], [itex]b\in\mathbb R[/itex]
Since I'm used only to the simplest types of differential equations, could you please help me and describe every step :shy: