Recent content by affirmative

Thank you very much, that actually kind of makes sense now!

I have no idea unfortunately. My question is how did the examiner arrive at the answer above (it is the verbatim answer for the past paper I'm currently working on).

My question is with part c, more specifically the calculating of the amplitude part. http://img691.imageshack.us/img691/1750/shmquestion.jpg The answer to the question is: [PLAIN][PLAIN]http://img80.imageshack.us/img80/8921/shmanswer.jpg I do not understand how to arrive at this...

Yes, it actually does. Thank you Mark.

I can only help you find the inverse in terms of y=something man. Anything other than that and I'm sorry -- have to wait for someone else to help you... :(

What you want to be doing, I think, is to substitute f(x)=y. Then, you need to switch x and y in the equation and rearrange it to get y=something. Did I explain myself clearly?

My question is mostly down to staggering ignorance of basic notation. I have been unable to find a straightforward answer anywhere - so I assume it must be pretty basic. I have a function: f(x,y)=x^3+y^3-3x-12y+20 Given that I know a value of x, ie x=-1, how do I extract a value of y using...

Actually yes... Surprisingly. Just looking at it now I can understand what you meant when you said a fair amount of work! Thank you.

I'm at a total loss for what to do next. I think and think but nothing occurs to me. I don't just have two variables, no matter what I do I will always get A and B in terms of two other things; I can never solve for them since there are two letters a and u and not numbers.

Maybe I just don't get differential equations - I don't know. All I manage to get for my constants A and B is the one in terms of the other and a or u. x(0) yields: A+B=a x'(0) yields: (1/2)[k(-A-B) + \sqrt[2]{k^2-4n^2}(A-B)] = u As a result of this I can only get the constants in terms...

The general solution I obtain: y = Ae^{(1/2)(-b + \sqrt[2]{k^2-4n^2})x} + Be^{(1/2)(-b - \sqrt[2]{k^2-4n^2})x} I was able to get this far when doing it by myself, it is here that I hit the wall. I fail to see what anything I have done so far has to do with proving the Integral at the...

x'' + kx' + (n^2)x = 0, let x = ce^{Yx} So: x' = Yce^{Yx} And: x'' = (Y^2)ce^{Yx} So eventually: Y^2 + kY + n^2 = 0 I believe the next step would be to factorise this, and then depending on whether there are two roots or a repeated root it would be of the form y = Ae^{(Y1)x} + Be^{(Y2)x} or y...

I would very much appreciate if anyone can help me with this problem; I have approached it from many different angles to no avail. The position x(t) of a particle moving along the x-axis is governed by the differential equation: x'' + kx' + (n^2)x = 0 , and initially x(0) = a, x'(0) = u...