Why does the wave equation involve second partial derivatives? If it were \[\frac{\partial f}{\partial x} = \frac{1}{v}\frac{\partial f}{\partial t},\] would it yield the same set of solutions \(f\) as the standard wave equation?

Take a sample wave equation and try the derivation yourself

f=Ae^(i(kx-wt))

The subtlety of my question is more whether the formulation I wrote admits more solutions than the standard wave equation.

Of course...solutions to PDEs come in families dont they..in effect any linear combinations of valid solutions would satisfy this

I realize this. I don't think you understand my equation thoroughly. In clearer terms, does \[\frac{\partial f}{\partial x} = \frac{1}{v}\frac{\partial f}{\partial t},\] which is my wave equation, have any solutions \(f(x,t)\) that does not satisfy\[\frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2},\] which is the standard wave equation?

I dont know much...but I guess there must be...becz the second eqn imposes one more condition..not sure

@yakeyglee: wow, this is a good question! Did you come up with this yourself, or did your teacher point at it? Now, here's my contribution: Answer to your question is: this equation has fewer solutions than the wave equation. Here are two examples: f(x,t) = A cos [ω(t - x/v)] is solution of wave equation, but not of your first order equation. Try it! f(x,t) = A cos(ωt).cos(ωx/v) has same properties. Ok, I'll let you digest this: I hope it will lead you to more questioning, because it is not the end of it.

This was my own question. And interesting...I see that allows for the translation of waves in both directions. What if we limited ourselves to waves that only propagated in the negative direction? Would the solutions be different then?

Exactly, what you got is only one set of solution of the wave equation, namely "travelling waves in -x direction". But travelling waves in +x direction and standing waves, are also solutions of the (full) wave equation.

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