- #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.

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

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

Last edited by a moderator:

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- Thread starter Icebreaker
- Start date

- #1

Icebreaker

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

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

Last edited by a moderator:

- #2

saltydog

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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.

It's separable right?

- #3

lurflurf

Homework Helper

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Yessaltydog said:It's separable right?

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

- #4

Icebreaker

Ok, I got it, thanks for the help.

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

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

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- #5

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.

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.

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- #6

saltydog

Science Advisor

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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.

Regarding:

[tex]\frac{dy}{dx}=\sqrt{4x+2y-1}[/tex]

How about squaring it for starters. Then let:

[tex]p=\frac{dy}{dx}[/tex]

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

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