Question

Difficult DE Question:

Determine all values of x0 and y0 such that the IVP for the the DE given below with initial condtions y(x0) = y0 is guaranteed to have a unique solution. Then, confirm the prediction by finding and analyzing the general solution

Answer 1

\[\frac{ dy }{ dx } = \frac{ 1+\cos(x) }{2-\sin(y) }\]

Answer 2

The uniqueness theorem should apply, for part one, check if the R.H.S and the derivative of the right hand side w.r.t. y are continuous on a region containing the IC.

Answer 3

How can I automatically tell if this is continuous:
http://www.wolframalpha.com/input/?i=d%2Fdy+of+(1%2Bcos(x))%2F(2-sin(y))

It just seems to make things a bit more complicated.

Answer 4

let \[f(x,y) := \frac{1+\cos(x)}{2-\sin(y)}\] then f(x,y) can only fail to be cont. if ?

Answer 5

remember, sin and cos are continuous functions.

Answer 6

I don't understand. Am I being daft?
It fails to be continuous where this denominator would equate to 0.
However, arcsin(2) isn't a thing.

Answer 7

exactly! but \[|\sin(x)|<1\]

Answer 8

so it's continuous

Answer 9

now \[f_y\]

Answer 10

fy?
Partial with respect to y?

Answer 11

Yeah, so the fy is always continuous. This means we'll have a general solution wherever. Let's tell for uniqueness by doing a partial w.r.t y. Correct?

Answer 12

Would I have to do this partial? It doesn't seem fun to do by hand. Is there any trick?

Answer 13

sry, was afk.

Answer 14

Yes, partial wrt y.

Answer 15

Ugh, well, if somehow with divine intervention and lots of algebra, I obtained this, I would see that this is well continuous because the denominator is never going to be 0. (see Wolfram Alpha link above).
I would state there is a unique solution.

Answer 16

Correct?

Answer 17

yep. Is it really that messy to do?

Answer 18

Actually no, I should have just attempted it.

Answer 19

But how does this general solution of \[2y + \cos(y)=x+\sin(x)\] show anything?

Answer 20

that's how it is sometimes, and others, it's messy!

Answer 21

Solved simply by separating and integrating with respect to variable.

Answer 22

I'm not sure, it's exist and is smooth.

Answer 23

the original analysis gives all the important answers.

Answer 24

You're the best with Differential Equations!
Thanks.

Answer 25

np