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anonymous
 3 years ago
show that the expression u = sin (xvt) describing a sinusoidal wave, satisfies the wave equation. Show that in general u = f(x  vt) and u = f(x + vt) satisfies the wave equations, where f is any function with a second derivative
anonymous
 3 years ago
show that the expression u = sin (xvt) describing a sinusoidal wave, satisfies the wave equation. Show that in general u = f(x  vt) and u = f(x + vt) satisfies the wave equations, where f is any function with a second derivative

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0slow down buddy . you wrote u=f(xvt) twice

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i wrote once, u = f(x  vt) and u = f(x + vt)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The wave equation is: \[\frac{\partial^2 u}{\partial t^2} = v^2\frac{\partial^2 u}{\partial x^2}\] In your case, \(u = \sin(x  vt)\) And \[u_{tt} = \sin(x  vt) * v^2 \] \[u_{xx} = \sin(x  vt)\] So since \(u_{tt} = v^2 u_{xx}\), u satifies the wave equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good job @Meepi i appreciate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, \(u_{tt} = \sin(x  vt) * v^2\) and \(u_{xx} = \sin(x  vt)\) the v^2 is from the chain rule when you derive with respect to t in \(u_{tt}\) since \(u_{xx} = \sin(x  vt)\), \(u_{tt} = u_{xx} * v^2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isn't it \[\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\] ??? you typed \(\frac{ ∂^{2}u }{ ∂t^{2} } = v^{2} \frac{ ∂^{2}u }{ ∂x^{2} } \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought \(\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ ∂^{2}u }{ ∂x^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂t^{2} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's alright your text book probably has the partial derivatives swapped

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ur allright :), cool thanks :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Also, for the general case, just use the chain rule as well to get \[u_{xx} = f''(x + vt)\] \[u_{tt} = f''(x + vt) * v^2\] Then \[u_{xx} = \frac{u_{tt}}{v^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Same can be done for u = f(x  vt)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok.., i got it now :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wanna help me with Legendre series ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Haven't really done much of that but I can give it a shot :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0go to http://openstudy.com/study#/updates/5172d293e4b0f872395bc5be
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