# Thread: Separable equations (variables)

1. ## Separable equations (variables)

Out of these equations: which one(s) are the variables not separable? Now I think II and III can be separated but I can not. Is this right? The way I came to this conclusion is by simply testing each one. I could not figure out a way to separate the variables for I. but for II and III I could. I might just be completely missing something for I and maybe it can be separate, maybe the other two can't and I just did them wrong! I'm not sure. As I said I couldn't figure out a way to separate I, but for II I had: d^2y/y*dy= 3x+dx after separating the variables and for III I had: dx/x=y^2/y-dy I'm not confident at all about how I separated them, it just doesn't seem right.

2. ## Re: Separable equations (variables)

You are incorrect in your reasoning above. That's all I'm telling you.

3. ## Re: Separable equations (variables)

Differential equations have to be first-order to be separable, don't they?

- Hollywood

4. ## Re: Separable equations (variables)

Originally Posted by hollywood
Differential equations have to be first-order to be separable, don't they?

- Hollywood
So then I and III are separable?

5. ## Re: Separable equations (variables)

Originally Posted by canyouhelp
So then I and III are separable?
Stop Guessing!!

I is trivially separable.

since you believe it is separable please write III in separated form for us.

6. ## Re: Separable equations (variables)

Originally Posted by romsek
Stop Guessing!!

I is trivially separable.

since you believe it is separable please write III in separated form for us.
For III I had: dx/x=y^2/y-dy ..but I think I separated it wrong :/

7. ## Re: Separable equations (variables)

Originally Posted by canyouhelp
For III I had: dx/x=y^2/y-dy ..but I think I separated it wrong :/
$\dfrac{dy}{dx}=y^2+yx = y(y+x)$

$\dfrac{dy}{y}=(y+x)~dx$

how are you going to separate that?

You're not.

8. ## Re: Separable equations (variables)

Originally Posted by romsek
$\dfrac{dy}{dx}=y^2+yx = y(y+x)$

$\dfrac{dy}{y}=(y+x)~dx$

how are you going to separate that?

You're not.
Oh, okay. Thanks!
So is it impossible to separate #2 because it's not first-order?

9. ## Re: Separable equations (variables)

Originally Posted by romsek
$\dfrac{dy}{dx}=y^2+yx = y(y+x)$

$\dfrac{dy}{y}=(y+x)~dx$

how are you going to separate that?

You're not.
So my answer should be II and III? because they both cannot be separated. Or can II be separated? I tried to above, but that was probably just really wrong.

10. ## Re: Separable equations (variables)

I think we've decided that II is second-order, so can not be separated. And III is first-order, so the question at least makes sense, but it can't be separated. So only I can be separated.

In your original post, you marked I only, but your argument was for II and III, the exact opposite.

- Hollywood

11. ## Re: Separable equations (variables)

Originally Posted by hollywood
I think we've decided that II is second-order, so can not be separated. And III is first-order, so the question at least makes sense, but it can't be separated. So only I can be separated.

In your original post, you marked I only, but your argument was for II and III, the exact opposite.

- Hollywood
Yea, sorry. I miss-read the question thinking it was asking which ones were separable. But thank you!