Calculus 3 hw help

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  • #1
VU2
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let f(xy)=e^(2y)sin(pix), is f a solution to (fxy)^2 -fxx(fyy)=4pi^2 e^(4x)


So I found fxx and fyy and fxy, which are -pi(e^(2y))sin(pix), 4sin(pix)e^(2y), 2pi(e^(2y))cos(pix) respectively,

When i reduced everything i got e^4y(sin^2(pix))+pie^4ycos^2(pix)=pie^(4x). I am assuming it is not a solution to 4pi^2 e^(4x) because first, it doesn't add up, and second, e^(4y) is never e^(4x). Am I right, I am not sure?
 

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  • #2
haruspex
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You are right that it is not a solution, but I wonder whether it's because there is a typo in the question. Maybe it should say fxy2 -fxxfyy=4pi2 e4y
 
  • #3
VU2
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Yeah, me too. But the problem on the paper clearly states that. I'll bring it up to my professor. Thanks!
 

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