# Thread: is my solution correct?

1. ## is my solution correct?

Hi, i attempted this question, but not sure whether it is correct or not..

using separation of variables:

X''(x)=S X(x) , Y''(y)=-S Y(y), where S is separation constant

becos X'(0)=0, X'(1)=0, so using the boundary value table i get:

S=(npi)^2

X=cos(n pi x) n=0,1,2....

Y=Acos*square root of S * y + Bsin* square root of S * y

since Y(0)=0

so Y=Bsin(n pi y)

so X(x)Y(y)=Bcos(n pi x)* sin(n pi y)

i know it's not finished, but is it correct from here? cos i dont feel quite sure about it....thank you~

2. Originally Posted by lollol
Hi, i attempted this question, but not sure whether it is correct or not..

using separation of variables:

X''(x)=S X(x) , Y''(y)=-S Y(y), where S is separation constant

becos X'(0)=0, X'(1)=0, so using the boundary value table i get:

S=(npi)^2
No, that's impossible. If X is a trig function (and it must be for its derivative to be 0 at 0 and 1), then S must be negative: you must mean $\displaystyle S= -(n\pi)^2$.

X=cos(n pi x) n=0,1,2....

Y=Acos*square root of S * y + Bsin* square root of S * y
and now, because S is negative, -S is positive. $\displaystyle Y"= n\pi Y$ so Y is exponential or hyperbolic: I would use Y(y)= C cosh(y)+ D sinh(y) because then Y(0)= C cosh(0)+ D sinh(0)+ C(1)+ D(0)= C= 0.

since Y(0)=0

so Y=Bsin(n pi y)

so X(x)Y(y)=Bcos(n pi x)* sin(n pi y)

i know it's not finished, but is it correct from here? cos i dont feel quite sure about it....thank you~

3. thanks hallsofivy, from ur solution, i would get Y=Dsinh(n pi y), and for X, since

. so it involves complex number after substituting S into the equation,
therefore X=Acos(n pi x) + Bsin(n pi x)

does it seem right?

4. Yes, you had X correct all along: X(x)= A cos(n pi x).

Your entire solution will be a sum such as $\displaystyle \sum_n C cos(n\pi x)Sinh(\pi y)$.

5. ## hi

sry.. but i think i am stuck again..while i continue with my work, i got An=0.. i dont quite understand why i got that, so i think may be i am wrong? please see the attachment, thank you