Recent content by 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.

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 separately? Edit: I see where i forgot u now.

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

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...

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 ?

lol. I appreciate the honesty :smile: 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?

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...

Homework Statement Given, (y+2)dx + y(x+4)dy = 0, y(-3) = -1Homework Equations v=y/xThe 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)...