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Math Help - Method of Characteristics: Prove the u and v is a solution

  1. #16
    A Plied Mathematician
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    Correct. So your argument for ku being a solution is complete, because all the steps are reversible.

    c(x,y)u^2+c(x,y)v^2
    Actually, you'd get c(x,y)(u^{2}+2uv+v^{2}). So the proof for the u+v case won't end up with zero, because you can't get zero out of this messier expression. Make sense?
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  2. #17
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Correct. So your argument for ku being a solution is complete, because all the steps are reversible.



    Actually, you'd get c(x,y)(u^{2}+2uv+v^{2}). So the proof for the u+v case won't end up with zero, because you can't get zero out of this messier expression. Make sense?
    O yes as it would be (u+v)^2. I guess both unknown functions u and v would have to be 0 for that term to go to 0. Thanks Ackbeet!
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  3. #18
    A Plied Mathematician
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    Right. You're welcome. Have a good one!
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