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anonymous
 4 years ago
Is \(y=Ae^{xt^2}\) a wave?
I'm testing it by using the wave equation:\[\frac{\partial ^2}{\partial x^2}=\frac{1}{u^2} \frac{\partial ^2}{\partial t^2}\]So I currently have\[\frac{\partial ^2}{\partial x^2}=Ae^{xt^2}\]and\[\frac{\partial ^2}{\partial t^2}=Ae^{xt^2} [2+4t^2]\]Normally, a wave has the form \(kx\omega t\), and the velocity can be calculated from that, but this one has a \(t^2\), which confuses me. Any thoughts?
anonymous
 4 years ago
Is \(y=Ae^{xt^2}\) a wave? I'm testing it by using the wave equation:\[\frac{\partial ^2}{\partial x^2}=\frac{1}{u^2} \frac{\partial ^2}{\partial t^2}\]So I currently have\[\frac{\partial ^2}{\partial x^2}=Ae^{xt^2}\]and\[\frac{\partial ^2}{\partial t^2}=Ae^{xt^2} [2+4t^2]\]Normally, a wave has the form \(kx\omega t\), and the velocity can be calculated from that, but this one has a \(t^2\), which confuses me. Any thoughts?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Those derivatives should all be with respect to y also.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They are derivatives of y, but not with respect to y. In the wave equation, u is a constant, i.e. the speed of propagation of the wave. Do your results suggest that your function satisfies the wave equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Aye, I meant derivatives of y. Early morning head I suppose. If the two sides were equal, the velocity squared would need to be \(2+4t^2\), which is not constant, so I suppose it cannot be a wave.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I suppose it would also have an imaginary velocity.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mhmm, it's not a wave as defined by the wave equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I shouldn't say "as defined by the wave equation".... it suffices to say that the function you are given does not satisfy the onedimensional wave equation :)
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