Doesn't the sinh function solve the DE X'' = G X, where G > 0? You get sin's out of Y'' = - G Y. Or are you more concerned about the stuff inside the argument of the sinh function?
Let U(x,y)=X(x)Y(y), which satisfies the partial equation,
d^2u/dx^2 + d^2u/dy^2 = 0 [This is the Laplace equation]
Given U(0,y)=U(a,y) = U(x,0) = 0 and U(x,b)=g(x)
Show that U(x,y) = Sigma (from n=1 to infinity) Cn Sin(n*pi*x/a)sinh(n*pi*y/a)
Where n,a and b are constants.
My trouble is not how I start it but where does my Sinh(n*pi*y/a) come from??
I got two equations they are X''(x)-G*X(x)=0 and Y''(y)+G*Y(y)=0
Now Ignore solving the first equation because I already got CnSin(...)
My problem is Y''(y)+G*Y(y)=0 and getting some how sinh(n*pi*y/a)
Keep in mind that G=(n*pi/a) which is my eigenvalue.
No, you get Sin("stuff") from X''=G*X, and you get sinH("stuff") from the Y''=-G*Y....
I already got the "stuff" inside the argument, I don't know where the sinh <-- this comes from?? I know sinh can be represented in terms of e^x-e^-x
but I can't seem to get it ???
Well, I can't say I fully understand your solution process, but if you got at the right answer, more power to you.
A couple of comments: if G > 0, then the solution to X'' = G X is sinh functions, not sin functions. If I take two derivatives of the sinh function, I get right back to the sinh function, with no sign changes. On the other hand, if I take two derivatives of the sin function, I get back to the sin function, but with a sign change.
Another comment is that hyperbolic trig functions solve X'' = G X just as well as exponentials do. In fact, they are equivalent, and you can use whichever you find more convenient.
You're welcome for whatever help I could provide. Have a good one!
Thanks for the input, yeah my problem was to bring sinh(..) involved, the trick was to replace 0 with k, just for the moment and then replace it back to 0 so that I won't have to evaluate e^(..)*0.
But its all good, I went through all G<0 G>0 and G=0.
I'd love for you to help me with this question.