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?
you use sin and cosine for a differential equation like this
$$ -y''(x)+\lambda y (x) = 0 $$
but your equation is $$ y''(x)+\lambda y (x) = 0 $$
can you spot the diffente sign in the second derivative ?