Recent content by Ebby

Thank you. I now understand. I think if someone had said, "Think that you are choosing to make ## \frac {du} {dx} = \frac u x ## so that the ## v ## term disappears, rather than thinking ## \frac {du} {dx} - \frac u x = 0 ##," it would have been clearer to me. Anyway, thank you all. This has...

OK, I see the typo now. Phew. But let me ask the really dumb question: Given that: $$u \frac {dv} {dx} + v(\frac {du} {dx} - \frac u x) = 1$$ Why can we say that, algebraically: $$\frac {du} {dx} - \frac u x = 0$$ ? I know, I'm really stupid...

I'm a bit confused because apparently I've made a typo but I simply used an example from this site about DEs (it's Example 1): https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html Where exactly did I go wrong?

If we have a first order d.e. like: $$\frac {dy} {dx} - \frac y x = 1$$ I would use two subs, namely: ##y = uv## and ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx}## So I get: ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx} - \frac {uv} {x} = 1## I then factor like this...

This was interesting to read. Thank you all for helping.

If I reason this as follows, I run into problems. Please help me understand what is wrong with reasoning like this. a) I start with the left hand side of the equation and let that x be -2. b) I square it. This gives me 4. So I now have the square root of 4. c) The square root of 4 is +/- 2. The...

Thank you. This is very helpful, especially @kuruman's explanation.

What is the work done by gravity in moving the particle from a distance of ##\infty## to a distance ##R## from the centre of the Earth (where ##R## > the radius of the Earth)? The answer is obvious, since the displacement and the force of gravity are in the same direction. Therefore, gravity...

Thank you all. I think the question, as set, could have been a little clearer.

I approach this by considering the four springs in parallel each with spring constant ##k## as one spring with four times the spring constant ##k' = 4k##. The car is dropped and at the moment its tyres touch the ground I assume that the spring is in its resting position. As the car continues to...

Thank you :)

The question is easy. I merely have a query: When the bow string is released, the potential energy stored in it ##U = \frac {kx^2} {2}## is all transformed to kinetic energy ##K = \frac {mv^2} {2}##, so we have:$$v = \sqrt {\frac {k} {m}}x$$ I now need to eliminate ##k##, so I can use ##k =...

In message 21 above, is the integral correct? Thanks.

Ah that was careless. I did not mean to write that. Yes I understand the unit vectors now. Decomposing, this means that ##\vec F = R \hat \theta + 0 \hat r##. And now to form the work integral. I imagine it's the integral over ##d\theta## of ##\vec F## dotted with something? Perhaps...

I drew it out using @PeroK's nomenclature. I still have questions: