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

[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.

 

[saltydog]

How do I go about solving

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

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

It's separable right?

 

[lurflurf]

It's separable right?

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

 

[Icebreaker]

Ok, I got it, thanks for the help.

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

 

[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.

 

[saltydog]

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.

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.