I have been doing some exercises about solving differential equations, but I am not be able to solve this one:
Find the implicit solutions of the following DE $$\dfrac{xx'}{\sqrt{x^2+r(x')^2}}=c,$$ with $r$ a positive constant and $c$ a real number
I tried to find a suitable change of variable, but it has not worked.
Thanks in advance: Any help will be greatly appreciated.
Finding $x(t)$ by solving $$xx'=c\sqrt{x^2+rx'^2} \implies x^2x'^2=c^2x^2+rc^2x'^2 \implies x'^2(x^2-rc^2)=c^2x^2$$ $$\implies \frac{dx}{dt}=\pm \frac{cx}{\sqrt{x^2-rc^2}} \implies \int \frac{\sqrt{x^2-a^2}}{x} dx= \pm \int c ~dt $$ Let $x=a~ sec~ u \implies dx=a ~sec~ u \tan u$, then $$a \int \tan^2 u ~ du =\pm \int du \implies a \int (sec^2 u -1) du =\pm ct+K$$ $$\implies a(\tan u-u)=\pm ct+K \implies \sqrt{x^2-a^2}-a ~sec^{-1}(x/a)=\pm ct+K,~ a=c\sqrt{r}$$
The sign $\pm$ will be fixed by the intial condition,