Question

Hey, can someone make this equation [cos (x) cos(y) –cot (x)] dx – [sin(x) sin (y)] dy=0 into the form dy/dx + P(x) y = Q(x)?

Answer 1

looks like some sort of diff of angles cos formula

Answer 2

cant be done :|

Answer 3

look at it for a few seconds and then wolframad it

its not linear

Answer 4

Well, that was fun .... got know idea where it went, but it was fun :)

[cos (x) cos(y) –cot (x)] dx – [sin(x) sin (y)] dy = 0

[cos (x) cos(y) –cot (x)] dx = [sin(x) sin (y)] dy


[cos (x) cos(y) –cot (x)] dy
------------------ = ---
[sin(x) sin (y)] dx


cos(x)cos(y) cot(x) dy
---------- - --------- = ---
sin(x)sin(y) sin(x)sin(y) dx


1 dy
cot(x) (cot(y) - ---------- ) = ---
sin(x)sin(y) dx


cos(y) 1 dy
cot(x) ( ----- - ---------- ) = ---
sin(y) sin(x)sin(y) dx


sin(x)cos(y) 1 dy
cot(x) ( ---------- - ----------) = ---
sin(x)sin(y) sin(x)sin(y) dx


cot(x) dy
------------------ ( sin(xy) - sin(y)cos(x) - 1) = ---
cos(x)cos(y) - cos(xy) dx

Answer 5

haha that's really cool. anyway, i am going to approach my instructor if it's really possible. thanks for the replies everyone!