# Dif Eq

1. Sep 21, 2005

### Icebreaker

How do I go about solving

x^2y' - cos (2y) = 1

This is unlike anything I've seen so far. Or so I think.

Last edited by a moderator: Sep 21, 2005
2. Sep 21, 2005

### saltydog

It's separable right?

3. Sep 21, 2005

### lurflurf

Yes
as a matter of personal taste I enjoy the use of the identity
1-cos(2y)=2(sin(y/2))^2

4. Sep 21, 2005

### Icebreaker

Ok, I got it, thanks for the help.

I got y = arctan ( c = 2/x)

Last edited by a moderator: Sep 21, 2005
5. Sep 21, 2005

### Icebreaker

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.

Last edited by a moderator: Sep 21, 2005
6. Sep 22, 2005

### saltydog

Regarding:

$$\frac{dy}{dx}=\sqrt{4x+2y-1}$$

How about squaring it for starters. Then let:

$$p=\frac{dy}{dx}$$

Differentiate it to get an ODE in p. Then separate variables.