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
pM/py= tan x-sinxsiny=-sinxcosy
Therefore this is exact.
f(x,y)=integral of (tanx-sinxsiny)dx+g(y)
Final Answer: cos(x)sin(y)-ln(cos(x))=C
But my answer is wrong and I don't know why.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
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
on the other hand, wolfram likes your answer
Not the answer you are looking for? Search for more explanations.
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?
Thank you! The final answer I received was ln|secx|+cosxsiny
Which turned out to be correct.
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!
out of interest, were you typing these answers into a computer?