G=c3Cosh(ky) + c4 sinh (ky); G(b) = 0 how does this simplify to…

when k=n*pi/a, and applying the boundary condition G(b)=0 the book simplifies to this expression, I dont know how to get G=c3Cosh(ky) + c4 sinh (ky) to this thing below

G=c2Cosh(n*pi*y/a)- c2[Cosh(n*pi*b/a)/sinh(n*pi*b/a)] *sinh (n*pi*y/a)

Question

G=c3Cosh(ky) + c4 sinh (ky); G(b) = 0 how does this simplify to…

when k=n*pi/a, and applying the boundary condition G(b)=0 the book simplifies to this expression, I dont know how to get G=c3Cosh(ky) + c4 sinh (ky) to this thing below

G=c2Cosh(n*pi*y/a)- c2[Cosh(n*pi*b/a)/sinh(n*pi*b/a)] *sinh (n*pi*y/a)

anonymous · Answer

i'm confused... you're using $Y$ and $y$ as separate functions?

anonymous · Answer

rewrite the problem without using the same letter and all the weird @s :p

dan815 · Answer

not sure how they are simplfying to this.. is it some trig identity?

dan815 · Answer

|dw:1375526179503:dw|

anonymous · Answer

ah ok let's see... first off $G(b)=0$ means:$$c_3\cosh kb+c_4\sinh kb=0\c_4\sinh kb=-c_3\cosh kb\c_4=-c_3\frac{\cosh kb}{\sinh kb}$$this appears to be what they did... looks like you mixed up $c_3,c_4$ though?

dan815 · Answer

OH!! U DOG

dan815 · Answer

thankyou so much

anonymous · Answer

er mixed up $c_2,c_3$ -- anyways if you substitute $k=n\pi/a$ it looks the sameish

anonymous · Answer

hehe that was tricky... I immediately thought some identity too