- #1
How do I go about solving
x^2y' - cos (2y) = 1
This is unlike anything I've seen so far. Or so I think.
x^2y' - cos (2y) = 1
This is unlike anything I've seen so far. Or so I think.
Last edited by a moderator:
Icebreaker said:How do I go about solving
x^2y' + cos (2y) = 1
This is unlike anything I've seen so far. Or so I think.
Yessaltydog said:It's separable right?
Icebreaker said:18 down, 2 more to go:
y' = (4x + 2y - 1)^(1/2)
and
(1-xy)y' + y ^2 + 3xy^3 = 0
The last one kinda looks like Riccati, but I can't get it to work.
edit: I got the second one. One more ODE remains... I don't know how to approach that one.