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Oct 26, 2021 at 15:20 comment added Kashmiri Let us continue this discussion in chat.
Oct 26, 2021 at 14:49 comment added Eli @kashmiri if Eq. (2) is correct $\frac{d^2u(-\theta)}{d\theta^2}+u(-\theta)=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u(-\theta)}\Big)~$ then substitute $~u(-\theta)=v~,du=dv~$ and you obtain the differential equation (1)
Oct 26, 2021 at 14:14 comment added Eli We can try but we can’t write equations?
Oct 26, 2021 at 14:07 comment added Kashmiri I had a related doubt can we have a chat?
Oct 26, 2021 at 14:07 comment added Kashmiri Your answer helped me. +10 :)
Oct 26, 2021 at 13:05 history edited Eli CC BY-SA 4.0
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Oct 26, 2021 at 13:03 comment added Eli Do you mean that i should put the argument , i will do it thanks
Oct 26, 2021 at 12:16 comment added Kashmiri I.e your second equation is :$$\frac{d^2u(-\theta) }{d\theta^2}+u(-\theta) =-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u (-\theta) }\Big)$$
Oct 26, 2021 at 12:13 comment added Kashmiri Hi Eli, in equation 2 is $u=u(-\theta)$ ?
Sep 17, 2021 at 18:46 vote accept Manu
Sep 17, 2021 at 18:44 comment added Eli \begin{align*} &\text{lets say that u is a function of $~x^2~$ thus}\\ &y=u(x^2)\\ & \text{what is the derivative $~\frac{dy}{dx}~$ ?}\\ & \frac{dy}{dx}=\frac{du}{dx}\,2x \quad, \text{not}\,~ \frac{du}{dx^2} \end{align*}
Sep 17, 2021 at 17:17 comment added Manu Why $\frac{du(-\theta)}{d(-\theta)}=\frac{du(\theta)}{d\theta}$? May you please explain this.
Sep 17, 2021 at 17:13 comment added Manu Thanks for the reply. In your edit shouldn't, $\frac{dy}{d\theta}=\frac{du(f(\theta))}{df(\theta)}\frac{df}{d\theta}=\frac{du(-\theta)}{d(-\theta)}\frac{d(-\theta)}{d\theta}=-\frac{du(-\theta)}{d(-\theta)}$
Sep 17, 2021 at 16:48 history edited Eli CC BY-SA 4.0
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Sep 17, 2021 at 16:29 history answered Eli CC BY-SA 4.0