What Is the Laplace Transform of $\frac{\partial{\phi}}{\partial{t}} = \frac{\partial^2{\phi}}{\partial{x}^2} - \cos(x)$, where $\phi = \phi(x, t)$?

Question

I was wondering what the Laplace transform of $\dfrac{\partial{\phi}}{\partial{t}} = \dfrac{\partial^2{\phi}}{\partial{x}^2} - \cos(x)$, where $\phi = \phi(x, t)$, is?

I know the Laplace transform of $\dfrac{\partial{\phi}}{\partial{t}}$ is $s \mathcal{L}\{ \phi \} - \phi(0)$, but I'm not sure what it is for the second part, since $\dfrac{\partial^2{\phi}}{\partial{x}^2}$ is the derivative with respect to $x$ and $\cos(x)$ is a function of $x$?

I would greatly appreciate it if someone could please take the time to clarify this.

Answer 1

We get $$s \mathcal{L}\{ \phi \} - \phi(0)=\frac{d^2}{d {x^2}}\mathcal{L}\{ \phi \}-\frac{\cos x}{s}$$