Thanks for looking at my question. I'm working through/self-studying the second edition of Partial Differential Equations: An Introduction by Walter A. Strauss. 

On page three, example two, he says

"Solve the PDE $u\_{xx} + u = 0$. Again, its really an ODE with an extra variable y. We know how to solve the ODE, so the solution is $u = f(y)cos(x) + g(y)sin(x)$, where again $f(y)$ and $g(y)$ are two arbitrary functions of $y$. You can easily check this formula by differentiating twice to verify that $u\_{xx} = -u$."

What I don't understand is how he gets $u = f(y)cos(x) + g(y)sin(x)$.

He says it's basically just an ODE with an extra variable y, but I'm not quite seeing that. I was able to understand Example 1 before it and Example 3 after it, and I can sort of see that this PDE is similar to the ODE form $y'' + y = 0$, but it's just been a hot minute since I've solved an ODE like this. I see it's homogenous, and one could use the method of integrating factors for it, but since this is a PDE I'm not sure how to solve this. My initial guess of $C\_{1}e^{r\_{1}t} + C\_{2}e^{{r\_2}t}$ didn't work, so I'm not sure how they got what they got for $u$. I get that when you integrate with respect to $x$ the constant you get is a function of $y$, but that's all I understand about this problem. Could someone show me how Walter got his solution for $u$, please? Thanks.