Homogeneous equation; Initial Value

[Exocer]

Homework Statement

Given,

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

Homework Equations

v=y/x

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 homework to get to.

 

[hp-p00nst3r]

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

 

[Dick]

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

 

[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..

 

[Dick]

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

 

[Exocer]

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

lol. I appreciate the honesty

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?

 

[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.

 

[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 ?

 

[hp-p00nst3r]

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

 

[Dick]

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 ?

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

 

[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

Strangely enough, the answer in the book reads:

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

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

 

[Exocer]

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

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

 

[Exocer]

ah i think i see it.as you stated above, (1/u)*du is separate from (-2/u)*du correct? So i should integrate those seperately?

Edit: I see where i forgot u now.

 

[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.

 

[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!