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anonymous

  • 5 years ago

Looking for the solution to equations of the form y''+y=0

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  1. anonymous
    • 5 years ago
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    sinx, cosx?

  2. anonymous
    • 5 years ago
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    If y'' is proceded by a constant?

  3. anonymous
    • 5 years ago
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    y=Csinx and y=Ccosx......anybody?

  4. anonymous
    • 5 years ago
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    There is an e^ax factor before the trig functions, where a is a root of the characteristic equation. Should be something like y = ce^ax sinx + ce^ax cosx.

  5. anonymous
    • 5 years ago
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    That rings a bell, happen to know of a good tutorial/reference?

  6. anonymous
    • 5 years ago
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    http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

  7. anonymous
    • 5 years ago
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    hope that helps.

  8. anonymous
    • 5 years ago
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    Great stuff, cheers matey.

  9. anonymous
    • 5 years ago
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    cheers

  10. anonymous
    • 5 years ago
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    this is the generic equation, but since \[y=e^{ax}[C_{1}\cos(x)+C_{2}\sin(x)]\rightarrow a=0\] this will be you final equation \[y=C_{1}\cos(x)+C_{2}\sin(x)\]

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