Recent content by Gatsby88

Thank you. Ah. Yes, we've done integrating factors earlier in the course. The integrating factor will be e^{\int {c}} = e^{c\theta} multiplying by this gives e^{c\theta} \frac{dZ}{d\theta} + C e^{c\theta} Z = e^{c\theta}(a cos \theta + b sin \theta) And then...

Your intuition with the marbles is correct. We have an initial situation which undergoes a process and then a final situation. So, Energy lost = Initial - Final. Although, Ill give you a standard disclaimer that 'lost' isn't the correct term to be using here. I think your problem comes...

Homework Statement I have a differential equation of the form \frac{dZ}{d\theta} + cZ = a cos \theta + b sin \theta Where Z = \frac{1}{2}\dot{\theta}^{2} I need to find the general solution of this equation. a, b and c are all constants. Homework Equations The questions suggests using...

Rather than simplifying it and then accounting for various things, it might be clearer what you need to do if you label the ends A1 A2 B1 B2 C1 C2 And consider how many different ways there are to arrange this such that A1 is next to A2, B1 is next to B2 and C2 is next C3 And remember that...

There are different types of comparison rate. I assume you mean APR (which I believe is just the % rate per year over the course of the loan term) If we knew that the APR was say 15% on an initial amount of £3000 over 10 years then Id do 3000 * (1.15)^10 = 12,136 Now, you have some of these...

Awesome. Will do, Thanks. Its a set question, but one that doesn't impact upon my grade at all.

FWIW, looking here gives me a lot of help and I think that ∫x2e-2ax2 dx = 1/2 * √[pi/(2a)^3] Its 1/2 and not 1/4 because the standard integral on wiki goes from 0 to inf and we need the area from -inf to inf. Since I am dealing with an even function I can just double the area. Grateful...

How do I go about manipulating gaussian (standard) integrals? From this question I can say that ∫e-2ax2 dx = √(pi/2a) I have a follow on question that, after doing some manipulation, means I need to find ∫x2e-2ax2 dx

You should really start from some assumptions and work through and it seems as though youve assumed it to be true and worked backwards. Working backwards is good for proof by contradiction, but for a straightforward proof like this just go through the algebra. For example, your starting points...

Absolutely. Id forgotten that it was √pi and not merely pi Carrying this through gives √pi/√pi which is 1 as required! Thank you for the help, I feel I understand where I was going wrong much better now.

ψ(x,0) = C1/4 * eax2-ikx a and k are +ve real constants So C1/4 * e-ax2-ikx = C1/2 * e-ax2-ikx * e-ax2+ikx = C1/2 * e-2ax2 Now ∫e-2ax2 dx = [1/√(2a)]pi using the gaussian integral here The constant was (2a/pi) so the whole thing becomes √(2a/pi) * √(1/2a) * pi The √2a cancels and we're...

Yes. In this case we replace the (-i) with i which when squared and ignoring the constant ends up giving me e-2ax2cos(2kx) ?

If I just had the e-2ax2 term, I can see I would need to use the identity ∫e-x2 dx = pi But I don't know how to change that given this term is multiplied by cos(2kx) in my problem I can't find any gaussian integral that relates e and cos. Im sure its just a manipulation problem but I am not...

Homework Statement Normalize the wave function ψ(x,0) = C1/4 * ea(x2)-ikx a and k are positive real constantsHomework Equations ∫|ψ|2dx = 1The Attempt at a Solution Now, my maths is a little weak, so I'm struggling a little bit here. The constant is easy to deal with in all aspects of...