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Jun 28, 2022 at 11:18 answer added Mozibur Ullah timeline score: -2
Jun 27, 2022 at 12:04 answer added Peter timeline score: 0
Jun 27, 2022 at 12:00 history tweeted twitter.com/StackPhysics/status/1541390921010405376
Jun 27, 2022 at 11:46 answer added Umaxo timeline score: 0
Jun 27, 2022 at 11:25 comment added Agnius Vasiliauskas Integrable parameter $\lambda$ doesn't have to be constant but rather it can be function of $x$,-> $\lambda = \lambda (x)$, or even function of multiple parameters such as $\lambda = \lambda (x_1,x_2,...,t)$. Then it can be split into partial derivatives, etc. As always,- at first we define initial conditions, then we think of how to integrate that expression.
Jun 27, 2022 at 10:47 answer added jensen paull timeline score: 0
Jun 27, 2022 at 10:44 answer added M.S. timeline score: 2
Jun 27, 2022 at 9:20 vote accept Gino Gamboni
Jun 27, 2022 at 9:19 answer added ɪdɪət strəʊlə timeline score: 5
Jun 27, 2022 at 9:15 comment added Ankit @Gino Gamboni if I get your questions correctly , I think each infinitesimal pieces of length "dx" in case of uniform distribution of mass has same infinitesimal masses... But if the mass density varies from point to point something like $ \lambda =a+bx$ then the infinitesimal pieces will have different masses..
Jun 27, 2022 at 9:07 answer added Enormity timeline score: 1
Jun 27, 2022 at 8:46 comment added schris38 Hi and welcome to Stackexchange. What is it exactly that you want to ask?
Jun 27, 2022 at 8:20 history edited Qmechanic CC BY-SA 4.0
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S Jun 27, 2022 at 8:04 review First questions
Jun 27, 2022 at 8:31
S Jun 27, 2022 at 8:04 history asked Gino Gamboni CC BY-SA 4.0