Recent content by Spoony

Ok so; f(t) = e^{at}(Ae^{ibt} + Be^{-ibt}) now; do i use a direct substitution for u_1 and u_2 (ie just rearanging the e^(ibt) and e^(-ibt) to be in terms of u_1 and u_2 or is it a bit more sneaky than that?) (I checked all my lecture notes as well just to make sure i hadnt missed this, and...

Sorry do you mean factor out the; e^{at} instead of e^{a} ?

OH unless you also call for cos to be rewritten in terms of exponentials?

so; sin(t) = \frac{e^{it} - e^{-it}}{2i} (rewrote it because i wasnt sure of the 0) Now; f(t) = A(cos(t\sqrt{n^{2}\pi^{2}-1}) + \frac{e^{it\sqrt{n^{2}\pi^{2}-1}} - e^{-it\sqrt{n^{2}\pi^{2}-1}}}{2})) + B(cos(t\sqrt{n^{2}\pi^{2}-1}) - \frac{e^{it\sqrt{n^{2}\pi^{2}-1}} -...

LOL hello fellow uni dude, you finished this yet? I think I've cracked it now but not too sure of all the final steps. EDIT: keep the name of our uni private please, reasons of privacy and this could be, but i highly doubt it viewed as cheating (im of the mind that I am being helped with and...

can the cosntants c_1 and c_2 be complex here? because i think that's what I am struggling with, also do i end up with; f(t) = e^{-t}(Ae^{+(n^{2}\pi^{2}-1)^{1/2}} + Be^{-(n^{2}\pi^{2}-1)^{1/2}} ) Leading to; f(t) = e^{-t} (c_1 cos bt + c_2 sin bt) c_1 = A+B and c_2 = i(A-B) b =...

Do end up with f(t) = Ae^{-1+(1-n^{2}\pi^{2})^{1/2}} + Ae^{-1-(1-n^{2}\pi^{2})^{1/2}} Or is there individual constants for the positive and minus square root bits?

edit: wrong

why would i need sines and cosines btw? would there be more steps before i have the solution to theta?

but then it wouldn't satisfy m^{2} + 2m + n^{2}\pi^{2} = 0 as 2m would be a complex number... literally just occoured to me that m^2 will then have a complex part to cancel out the 2m part.. d'oh! cheers mate :D

aye true, but you know that 4*(n^2)*(pi^2) is greater than 4 for all n > 0 (n is an integer) except when n = 0 but if n = 0 then this is the only solution which leads to g(x) = 0 which then leads to theta = 0 and u = -(0.5)x + 2, ie u is only a function of x then and not of t.

but m can't take imaginary values as the value for 2m would give an imaginary part, and -n^2 pi^2 is a real number

the lowest point of m^2 + 2m is 1 and n is an integer therefore -n^2 pi^2 > 1 for all n... unless m can take imaginary values... oh hell i haven't been stupid enough to look at that and think "no value of m can surely satisfy that equation m^2 is always positive.." if this what it is i want to...

Anyoen have any ideas? I'll take anything atm.. I've come to a stand still on this one

Homework Statement I have the damped wave equation; u_{tt} = 4 u_{xx} -2 u_{t} which is to be solved on region 0 < x < 2 with boundary conditions; u(0,t) = 2, u(2,t) = 1. i must; 1) find steady state solution u_{steady}(x) and apply boundary conditions. 2) find \theta(x,t)...