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anonymous

  • one year ago

What am I doing wrong? Determine whether the given differential equation is exact. If it is exact, solve it. (If it is not exact, enter NOT.) (tanx − sinxsiny)dx + cosxcosy dy = 0 My work: pM/py= tan x-sinxsiny=-sinxcosy pM/py=cosxcosy=-cosysinx Therefore this is exact. Solve. f(x,y)=integral of (tanx-sinxsiny)dx+g(y) f(x,y)=cosxsiny-ln(cosx)+c_1+g(y) cosxsiny-ln(cosx)+c_1=c_2 Final Answer: cos(x)sin(y)-ln(cos(x))=C But my answer is wrong and I don't know why.

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  1. IrishBoy123
    • one year ago
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    looking at this form a physical sciences perspective, the big problem i think you have is that the field is full of singularities due to the fact it has a tan(x) in there. so even though you get the mixed partials to be equal, and you solve for a potential function, the potential function is not a potential function. if that means anything to you, read on here http://mathinsight.org/conservative_vector_field_determine

  2. IrishBoy123
    • one year ago
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    actually, a more mathsy answer might look at the fact that the underlying potential function is expressed as \(\Phi (x,y,z) = const\). it is from this fact that we can arrive at exact differential equations, because it follows that \[d\Phi = \Phi_x dx + \Phi_y dy + \Phi_z dz = 0\] and that's where you get the test for the mixed partials and exactness however, the singularities means you cannot claim that \(\Phi (x,y,z) = const\) in the first place in case i am leading you astray, let me tag a few people @ganeshie8 @freckles @Loser66

  3. phi
    • one year ago
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    on the other hand, wolfram likes your answer http://www.wolframalpha.com/input/?i=%28tanx+%E2%88%92+sinxsiny%29dx+%2B+cosxcosy+dy+%3D+0

  4. IrishBoy123
    • one year ago
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    OK, first of all, my bad in my first post: the tans are in the DE and not the potential, but the point still holds. in the first partial they are still signalling singularities in the underlying potential function so the claim \(\Phi = const\) cannot be made without restrictions on the domain being made also. so, for example, what happens when , say, \(x = \pi/2\) in @nelly_416 's \(cos(x)sin(y)-ln(cos(x))=C\) solution it is also interesting that the MW solution is written in a very particular way but can be reorganised to be @nelly_416 's solution. you can buy your way out of these problems by re defining the values at erstwhile singularities. so is that the problem?

  5. anonymous
    • one year ago
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    Thank you! The final answer I received was ln|secx|+cosxsiny Which turned out to be correct.

  6. IrishBoy123
    • one year ago
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    lol!!! so \(-ln(cosx) = ln|secx|\) was the problem?!?! still think this is messed up because x is allowed to be \(\pi / 2\) many times over. but great that closure has been reached, & well done for persevering! :p

  7. IrishBoy123
    • one year ago
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    out of interest, were you typing these answers into a computer?

  8. anonymous
    • one year ago
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    Yes, the homework was online.

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