# Homework Help: Homogeneous equation; Initial Value

1. Sep 15, 2008

### Exocer

1. The problem statement, all variables and given/known data
Given,

(y+2)dx + y(x+4)dy = 0, y(-3) = -1

2. Relevant equations
v=y/x

3. The attempt at a solution

I've been REALLY struggling with homogeneous equations for some reason...I just don't understand them all.

so far i've tried two things.

(1)dx -(y)dy
----- -------
(x+4) = (y+2)

Not even sure what else to do from here. I tried getting dy/dx and got

y+2 dy
---- = -----
y(x+4) dx

Thanks in advanced for any assistance... I've been looking at this problem for hours and have other hw to get to.

2. Sep 15, 2008

### hp-p00nst3r

I think the idea here is to use separable equations. There should be a section in your math book on this concept.

3. Sep 15, 2008

### Dick

Take your first form and integrate both sides. You've 'separated' the variables.

4. Sep 15, 2008

### Exocer

thanks guys.

Dick,
Glad to know i was on the right track. Just realized it isn't homogeneous.

As far as integrating:

dx/(x+4), I get lx|x+4|

how do i go about integrating

y
----- dy ?
y+2

Probably a dumb question, but I took a semester break between this and calc II, so lots of areas are rusty :(

Simply couldn't get it into du/u form..

5. Sep 15, 2008

### Dick

It is dumb. :) Sorry, you are just out of practice. Try substituting u=y+2.

6. Sep 15, 2008

### Exocer

lol. I appreciate the honesty :rofl:

u = y + 2

gives me

(U-2)
-------du
U

Which i guess can be broken up into

U
---- = 1
U

and -2/U

now all i have to integrate is

du
--------
U

bringing the -2 outside of the integral, correct?

7. Sep 15, 2008

### Dick

Sure. You might just write that as ((u-2)/u)*du=(1-2/u)*du=1*du-2*(1/u)*du. It's a little easier that trying to place the '-----' correctly. Just use parentheses. Or preferably tex if you dare.

8. Sep 15, 2008

### Exocer

i really appreciate the help.

so this is where i'm at.
ln |x+4| = 2*ln |Y+2| + ln C

which, IIRC gives

X+4 = (Y+2)^2 + C ?

9. Sep 15, 2008

### hp-p00nst3r

and since you are given a certain x y value from the question, you can sub those in to solve for the constant c

10. Sep 15, 2008

### Dick

You missed a part. The integral of 1-2/u is u-2ln(|u|). What happened to the u?

11. Sep 15, 2008

### Exocer

Ok great! Thanks again guys for the help.

I'll do that and double check my answer with the one in the book.

plugging in the x and y values. C = 0

x + 4 = (y + 2)^2*e^(-y+1)

I understand how they got everything up until the e with exponent.

12. Sep 15, 2008

### Exocer

hmmm im taking a minute to look at it now. not sure where i messed up.

13. Sep 15, 2008

### Exocer

ah i think i see it.

as you stated above, (1/u)*du is seperate from (-2/u)*du correct? So i should integrate those seperately?

Edit: I see where i forgot u now.

Last edited: Sep 15, 2008
14. Sep 15, 2008

### Exocer

Dick, p00nst3r, thanks for not doing the problem for me, and helping me think things through. I hate it when people do all the work for me lol.

I'll be checking back before I go to bed.

15. Sep 15, 2008

### hp-p00nst3r

No problem. Glad to have helped!
I was actually considering doing the question for you and then I realized that I should guide you to the answer instead.
Good thing I didn't do it for you!