# Why are $\cosh$ and $\sinh$ used in solving this case of Laplace's equation?

I'm currently trying to find the separable solution to the PDE $$u_{xx} + u_{yy} = 0$$ Subject to the boundary conditions $$u(0,y) = u(a,y) = 0$$

To attempt this, I have expressed $u(x,y) = X(x)Y(y)$ which makes the PDE $$X''(x)Y(y) + X(x)Y''(y) = 0 \\ \Rightarrow -\frac{X''}{X} = \frac{Y''}{Y} = \lambda \\ \Rightarrow X'' + \lambda X = 0$$

Solving this, using the boundary conditions gives us that the only non-trivial solution to the ODE is the oscillatory solution $$X_n (x) = \sin \left( \frac{n \pi x}{a} \right)$$ With $\lambda_n = \frac{n^2 \pi^2}{a^2}$. So far so good. I have the solutions to this question, and this is the same result.

Now comes the confusion...

We now need to find the corresponding solution in terms of $Y$ to obtain the separable solution. I calculated this as $$Y_n (y) = A_n \cos \left( \frac{n \pi y}{a} \right) + B_n \sin \left( \frac{n \pi y}{a} \right)$$ However, I have checked the mark scheme and the solution they have for $Y$ is $$Y_n (y) = A_n \cosh \left( \frac{n \pi y}{a} \right) + B_n \sinh \left( \frac{n \pi y}{a} \right)$$

I have never seen $cosh$ and $sinh$ used in the general solution to a second order ODE, and have no idea why it is being used here.

Can anyone please explain this to me?

$$-y''(x)+\lambda y (x) = 0$$
but your equation is $$y''(x)+\lambda y (x) = 0$$
• I think you have your differential equations the wrong way round: $-y''(x)+\lambda\,y(x)=0$ yields the hyperbolic basis of solutions and $y''(x)+\lambda\,y(x)=0$ the trigonometric ones. Commented May 25, 2017 at 11:51