- #1
giacomh
- 36
- 0
xdy-ydx=(x2+y2)2(xdx+ydy)
(Hint: consider d(x2+y2)2)
d(x2+y2)2
d(arctan(y/x))=xy|-y/(x2+y2)
The answer is arctan(y/x)-(.25(x2+y2)2)=C
I correctly solved other problems of this type, but this is the hardest one in the problem set and I have no idea how to solve it. This isn't a homework I have to turn in, I'm just trying to fully understand this concept, so I would really appreciate it if someone would just help me work through this. I don't understand how d(x2+y2)2 is relevant to anything, and I'm assuming the denominator of d(arctan(y/x)) somehow plays a role in 'cancelling' the (x2+y2)2? I guess I'm mostly confused because it seems like you have to use two integrating factors?
Thanks!
(Hint: consider d(x2+y2)2)
Homework Equations
d(x2+y2)2
d(arctan(y/x))=xy|-y/(x2+y2)
The Attempt at a Solution
The answer is arctan(y/x)-(.25(x2+y2)2)=C
I correctly solved other problems of this type, but this is the hardest one in the problem set and I have no idea how to solve it. This isn't a homework I have to turn in, I'm just trying to fully understand this concept, so I would really appreciate it if someone would just help me work through this. I don't understand how d(x2+y2)2 is relevant to anything, and I'm assuming the denominator of d(arctan(y/x)) somehow plays a role in 'cancelling' the (x2+y2)2? I guess I'm mostly confused because it seems like you have to use two integrating factors?
Thanks!