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ODE y''+b^2 y=0 where b is vector

  1. Apr 14, 2012 #1
    please help me with this task. I'm wondering what is the right result.
    I have a ODE
    [itex]y'' - b^2 y =0[/itex]

    also the result should be
    [itex]y=C e^{\pm bx}[/itex]

    but what is the result when b is vector?
    [itex]\vec b=(b_x, b_y)[/itex]

    is this the result?
    [itex]y=C e^{\pm \vec{b}x}[/itex]

    or this?
    [itex]y=C e^{\pm |b| x}[/itex]

    how to solve it?
    thanks for help!
  2. jcsd
  3. Apr 14, 2012 #2


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    Science Advisor

    If b is a vector, then what do you mean by "[itex]b^2[/itex]"? The dot product? Then coefficient is the square of the length of the vector.
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