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henpen

  • 2 years ago

Mathematically, prove that there are there no arbitrary constants required for the particular solution of \[Ly=f(x)\].

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  1. UnkleRhaukus
    • 2 years ago
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    what is \(L\) ?

  2. henpen
    • 2 years ago
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    Linear DE operated on...

  3. henpen
    • 2 years ago
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    I know that it's obvious, but I've yet to encounter a formal proof.

  4. UnkleRhaukus
    • 2 years ago
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    i can not see the differential equation

  5. henpen
    • 2 years ago
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    Oh, it's just the general sign for\[Ly=\sum_{i=0}^{i=n}a_i\frac{d^i}{dx^i}y=0\]

  6. henpen
    • 2 years ago
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    Or is this only provable for more specific DE?

  7. UnkleRhaukus
    • 2 years ago
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    i thought there were usually as many constants in the solution as the order of the DE

  8. henpen
    • 2 years ago
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    Yes, but they're all in the complimentary solution.

  9. henpen
    • 2 years ago
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    I've not come across 'roots' with regards to DE, but I suppose so, yes.

  10. UnkleRhaukus
    • 2 years ago
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    actually the roots are in the complementary solution

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