leonne said:
Homework Statement
Well the problem is a electromagnetism physics problem to find potential in a cube.
Homework Equations
The Attempt at a Solution
using Laplace question and combining X(x)Y(y)Z(y) to it
we get
1/X(d2X/dx2 + 1/Y(d2Y/dy2 + 1/Y(d2Y/dy2=0
Than we see its in the forum of f(x)+ g(y)+ h(z)=0 that they must be constant for this to be true so
1/X(d2X/dx2=c1 1/Y(d2Y/dy2=c2
1/Y(d2Y/dy2=c3
so c1+c2+c3=0 They go about saying like how c3 is pos while c1 ,c2 is negative
( btw do u know why its that? some other problems has c1 pos while c2 c3 neg does it matter?)
That does NOT follow from what you give here. It might follow from the boundary conditions you are given for the problem.
Than they have c1=-k2 c2=-l2 than c3=(k+l)2
btw where did the k and l come from is it just some constant they picked?
Basically, yes. Since a square is always positive, writing c1 and c2 as "[itex]-k^2[/itex]" and "[itex]-l^2[/itex]"
emphasizes that they are negative. Also, it simplifies the solution to the "characteristic equation".
now the plug in for the c we found and get like (d2X/dx2=-k2 X...
Than they solve this ODE which is the part i am lost. They just go from this to having
X(x)=Asin(ky)+bcos(ky) why is that? also in another problem when X was pos it was Aekx+Be-kx why is it like this?
The "characteristic equation" for the differential equation ay"+ by+ c= 0 is [itex]ar^2+ br+ c= 0[/itex]. If that has two distinct real solutions, say [itex]r_1[/itex] and [itex]r_2[/itex], then the general solution is [itex]y(t)= Ce^{r_1t}+ De^{r_2t}[/itex]. If it has two complex roots, a+ bi and a- bi, then the general solution is [itex]y(t)= e^{at}(Ccos(bt)+ Dsin(bt))[/itex]. The two different kinds of solutions are related through [itex]e^{iat}= cos(at)+ i sin(at)[/itex]
For this prblem the pos z was in the same format as that. Should i just accept that when its pos to use e^ and when neg to use the cos sine?
Unfortunately, one of the basic rules of physics/mathematics is "you always learn the mathematics you need for physics the
next year!". It looks to me like this course is using differential equations intensively and differential equations
should have been a prerequisite for it.
All I can say is that if you are in a course in which you are apparently expected to solve
partial differential equations and you have not yet taken Ordinary Differential Equations, you are going to have to accept a lot of things without really understanding them. You might try looking at
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/lecture-notes/