# Method of Characteristics: Prove the u and v is a solution

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• October 25th 2010, 06:46 AM
Ackbeet
Correct. So your argument for $ku$ being a solution is complete, because all the steps are reversible.

Quote:

$c(x,y)u^2+c(x,y)v^2$
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?
• October 25th 2010, 07:25 AM
bugatti79
Quote:

Originally Posted by Ackbeet
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!
• October 25th 2010, 10:04 AM
Ackbeet
Right. You're welcome. Have a good one!
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