I'm REALLY stuck on this question as I don't really know how to begin, I understand that it has something to do with:
$$\sinh x=\frac{e^x-e^{-x}}{2}$$
I'm definitely not asking for someone to do it all for me but I've given the entire question(s) below so that you have some context, basically, I would like to be pointed in the right direction. This is my first time asking a question on a site like this so I apologise if I've missed out on some normal protocol.
The temperature distribution, θ(x), along a fin of length L is given by:
$θ(x)=Ae^{kx}+Be^{-kx}$ where A, B and k are constants and x is the distance along the fin. A and B can be evaluated from $A+B=θ_0$ and $Ae^{kL}+Be^{-kL}=θ_L$ where $θ_0$ is the temperature at the base of the fin and $θ_L$ is the temperature at length L.
Show that:
$$A=\frac{θ_L-θ_0 e^{-kL}}{e^{kL}-e^{-kL}}$$
$$B=\frac{θ_0-e^{kL}-θ_L}{e^{kL}-e^{-kL} }$$
$$θ(x)=\frac{θ_L \sinh(kx)+θ_0 \sinh[k(L-x)]}{\sinh(kL)}$$
Thank you very much in advance to anybody who can offer me a push in the right direction...
EDIT:
Thank you Conrad Turner, I've gotten this far, I promise this is the last question:
$$Ae^{kL}+Be^{-kL}=θ_L$$ $$A(e^{kL} )+θ_0-A(e^{-kL} )=θ_L$$ $$A(e^{kL} )-A(e^{-kL} )=θ_L-θ_0$$ $$A(e^{kL}-e^{-kL} )=θ_L-θ_0$$ $$A=\frac{θ_L-θ_0}{e^{kL}-e^{-kL}}$$
I'm not sure how $e^{-kL}$ come to be in the numerator.
This is a problem designed by the look of it to give you practice at algebraic manipulation.
For the first observe that $A+B=\theta_0$ means $B=\theta_0-A$. Now substitute this for $B$ in $A e^{kL}+Be^{-kL}=\theta_L$ and solve for $A$.
A similar process will deal with the second, or just substituting the first result into $A+B=\theta_0$ and rearranging should also work.
The third uses the two previous results and that $\sinh(u)=\frac{e^u-e^{-u}}{2}$