# Separating these equations

• Nov 25th 2009, 02:49 PM
Open that Hampster!
Separating these equations
1) $\displaystyle (x^2+30)y' = xy$

If I separated it like this:

$\displaystyle \frac{y'}{y} = \frac{x}{x^2+30}$

...can I integrate anyway and get lnY = something?

2) $\displaystyle y' = 38y^2sin(x)$

This is the same above. Getting the y and x variables, but since there's not dx can I then Integrate anyway?

3) $\displaystyle \frac{du}{dt} = 1+u+t+tu$

This one I'm just not sure how to separate.

Any help on these is appreciated. I know they're simple, but well, so am I.
• Nov 25th 2009, 03:29 PM
Jester
Quote:

Originally Posted by Open that Hampster!
1) $\displaystyle (x^2+30)y' = xy$

If I separated it like this:

$\displaystyle \frac{y'}{y} = \frac{x}{x^2+30}$

...can I integrate anyway and get lnY = something?

2) $\displaystyle y' = 38y^2sin(x)$

This is the same above. Getting the y and x variables, but since there's not dx can I then Integrate anyway?

3) $\displaystyle \frac{du}{dt} = 1+u+t+tu$

This one I'm just not sure how to separate.

Any help on these is appreciated. I know they're simple, but well, so am I.

Factor $\displaystyle 1+u+t+tu = (u+1)(t+1)$
• Nov 25th 2009, 03:31 PM
galactus
You separated fine.

Then integrate and get:

$\displaystyle \int\frac{1}{y}dy=\int\frac{x}{x^{2}+30}dx$

$\displaystyle ln(y)=\frac{1}{2}ln(x^{2}+30)+C$

e to both sides isolates y and gives us a constant $\displaystyle C_{1}$

$\displaystyle y=C_{1}\sqrt{x^{2}+30}$
• Nov 26th 2009, 11:38 AM
Open that Hampster!
Thanks for the help. Just two more quick questions:

For the second one:

$\displaystyle y' = 38y^2sin(x)$

$\displaystyle \int y^{-2}dy = \int 38sin(x)$

$\displaystyle -\frac{1}{y} = -38cos(x) + C$

$\displaystyle y = \frac{1}{38cos(x)+C}$

Is this wrong? Is there more than one solution?

And then for the third one, I'm solving for u, so:

$\displaystyle \int (u+1)du = \int (t+1)dt$
$\displaystyle \frac{u^2}{2}+u = \frac{t^2}{2}+t+C$

Not sure where to go to isolate it from here to isolate u.

Happy Thanksgiving everyone.
• Nov 26th 2009, 05:39 PM
HallsofIvy
Quote:

Originally Posted by Open that Hampster!
1) $\displaystyle (x^2+30)y' = xy$

If I separated it like this:

$\displaystyle \frac{y'}{y} = \frac{x}{x^2+30}$

...can I integrate anyway and get lnY = something?

I don't know what you mean by "integrate anyway". Yes, of course, you can integrate $\displaystyle \frac{dy}{y}= \frac{ xdx}{x^2+ 30}$.

Quote:

2) $\displaystyle y' = 38y^2sin(x)$

This is the same above. Getting the y and x variables, but since there's not dx can I then Integrate anyway?
Oh, I see. There is a "dx"! y'= dy/dx and you can separate them: $\displaystyle y^{-2}dy= 38 sin(x)dx$.

Quote:

3) $\displaystyle \frac{du}{dt} = 1+u+t+tu$

This one I'm just not sure how to separate.

Any help on these is appreciated. I know they're simple, but well, so am I.
Danny told you how to do the last one.
• Nov 27th 2009, 06:20 AM
Open that Hampster!
Quote:

Originally Posted by HallsofIvy
I don't know what you mean by "integrate anyway". Yes, of course, you can integrate $\displaystyle \frac{dy}{y}= \frac{ xdx}{x^2+ 30}$.

Oh, I see. There is a "dx"! y'= dy/dx and you can separate them: $\displaystyle y^{-2}dy= 38 sin(x)dx$.

Danny told you how to do the last one.

My post right above yours is my progress after the above two posters.
• Nov 27th 2009, 07:38 AM
dedust
Quote:

Originally Posted by Danny
Factor $\displaystyle 1+u+t+tu = (u+1)(t+1)$

using this, we get

$\displaystyle \frac{du}{(u+1)} = (t+1)dt$

integrate it and we're done.