Hyperbolic or exponential solutions to differential equation I have spent the last couple weeks in my Fourier Analysis course to solve PDEs with the method of separations of variables. However, I have come up with something that annoys me and I can't really explain it. Let me show an example.
I have this problem here
$$u_{xx}+u_{yy}=0 \\
u(0,y) = u(1,y) = 0 \\
u(x,0) = u(x,1) = \frac{x^3-x}{6}$$
So I separate the variables and get these ODEs
$$
X''(x)+ \lambda^2 X(x) = 0 \\
Y''(y)- \lambda^2 Y(y) = 0
$$
and these are simple to solve. The first one is just $X(x)= A\cos(\lambda x)+B \sin(\lambda x)$ and now is the confusing part. For me the solution to the second equation have always been $Y(y)=Ce^{- \lambda y}+ De^{ \lambda y}$ but the book have suddenly started to use the hyperbolic equations $\cosh$ and $\sinh$, why is that? Is it something I am missing?
 A: The awesome thing about hyberbolic (trig) functions is how they can be represented as sums of exponentials (and vice versa). Recall that $\cosh(u) = \dfrac{e^u + e^{-u}}{2}$ and $\sinh(u) = \dfrac{e^u - e^{-u}}{2}$. It's just a matter of how you want to indicate the constants that come about from your initial conditions.
For example, since you're used to the solution $Y(y) = Ce^{\lambda y} + De^{-\lambda y}$. Here's how you could express the same solution with hyperbolic trig functions (namely $\cosh$ and $\sinh$). Let $C = \frac{A+B}{2}$ and $D=\frac{A-B}{2}$. Then
$$ 
Ce^{\lambda y} + De^{-\lambda y} \\ 
= \frac{A+B}{2}e^{\lambda y} + \frac{A-B}{2}e^{-\lambda y} \\ 
= \frac{A}{2}e^{\lambda y} + \frac{B}{2}e^{\lambda y} + \frac{A}{2}e^{-\lambda y} - \frac{B}{2}e^{-\lambda y} \\ 
= \frac{A}{2}(e^{\lambda y} + e^{-\lambda y}) + \frac{B}{2}(e^{\lambda y} - e^{-\lambda y}) \\ 
= A \frac{e^{\lambda y} + e^{-\lambda y}}{2} + B \frac{e^{\lambda y} - e^{-\lambda y}}{2} \\ 
= A\cosh(\lambda y) + B \sinh(\lambda y).
$$
Again, notice that the only thing that changed was really how you defined the constants with a relation from $C,D$ to $A,B$. I stress that these are direct results from your initial (or given) conditions.

As far as your question from the comments, "Why write it in that way instead of just the exponentials?", it is really a matter of conveniently denoting the properties of a solution. In an analogous fashion, even your familiar sines and cosines are merely just one way of expressing solutions to almost identical differential equations.
Since $\cos(u) = \dfrac{e^{iu}+e^{-iu}}{2}$ and $\sin(u) = \dfrac{e^{iu}+-e^{-iu}}{2i}$ (where $i^2 = -1$), it depends on the context of your problem or possibly just your preference of where to write a solution as $P\cos(x)+Q\sin(x)$ or in the form $Ue^{ix}+Ve^{-ix}$. It is straightforward to relate the constants $P,Q$ to $U,V$ in the same way that $A,B$ and $C,D$ were related for the hyperbolic trig case.

There are several examples that I'd love to present to show why expressing your solutions as a sum of hyperbolic trig functions. There are properties of hyperbolic trig functions that are so closely related to the familiar circular trig functions that manipulating them can be very natural (as opposed to working with a sum of 2 exponentials).
For example, let's say your solution was expressed as $y(x) = C\cosh(mx)+D\sinh(mx)$. How would you find the $n$-th derivative of $y(x)$? Since $\frac{d}{dx}[\cosh{mx}] = m\sinh(mx)$ and $\frac{d}{dx}[\sinh{mx}] = m\cosh(mx)$, it's very easy to express the derivatives of $y(x)$:
$$ 
y(x) = y^{(0)}(x) = C\cosh(mx)+D\sinh(mx) \\
y'(x) = y^{(1)}(x) = Cm\sinh(mx) + Dm\cosh(mx) \\
y''(x) = y^{(2)}(x) = Cm^2\cosh(mx)+Dm^2\sinh(mx) \\
y'''(x) = y^{(3)}(x) = Cm^3\sinh(mx) + Dm^3\cosh(mx) \\
\vdots
$$
A: I think that
the reason that
hyperbolic functions are used
is that they satisfy
many identities similar to the ones
satisfied by the trig functions.
An example is
$\cosh^2(x)-\sinh^2(x)
= 1
$.
