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Solution to the wave equation?

  1. Nov 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Is the function
    y(x,t) = Ae(−β2x2−2βxt−t2) + Be(−x2+2αxt−α2t2)
    a solution to the wave equation
    2y / ∂t2 = v2 (∂2y / ∂x2)

    2. Relevant equations

    2y / ∂t2 = v2 (∂2y / ∂x2)




    3. The attempt at a solution
    I have found the solution through finding the partial derivatives (∂2y / ∂t2 and ∂2y / ∂x2) of the function. I got that it is a solution if α=-1 and β=1. However, this was a very long winded process of taking derivatives and am wondering if there is another way to solve it. I notice now that if I put α=-1 and β=1 into the original equation, y(x,t) = Ae(−β2x2−2βxt−t2) + Be(−x2+2αxt−α2t2) , then the exponents are the same. Am I suppose to notice this right away instead of taking the derivatives? Thanks
     
  2. jcsd
  3. Nov 11, 2015 #2

    Orodruin

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    Yes, you are likely supposed to notice it directly. What is the general form of a solution to the wave equation?
     
  4. Nov 12, 2015 #3
    General form is : y(x,t) = Aei(kx-ωt) = A[cos(kx-ωt) + isin(kx-ωt)]

    So if I notice the exponents are the same under the condition α=-1 and β=1 how does that exactly translate to it being a solution of the wave equation?
     
  5. Nov 12, 2015 #4

    Orodruin

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    No, this is not the most general form. It is just a plane wave solution.
     
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