Question

(x+y)dx-xdy=0

Answer 1

(x+y)dx-xdy=0
(x+y)dx=xdy
dx/dy=x/x+y

lol whats ur qs ?

Answer 2

exact equation?

Answer 3

like M = x+ y N = -x
My = 1 Nx = -1
gotta use a ....... wow I'm so tired... I think you need to find the integrating factor.. or go through substitution but you don't wanna do that because it's long as *(@#*(#@UIO@U

Answer 4

y = xu
y' = x du + u dx

Answer 5

(x+y)dx-xdy=0
xdx+ydx-xdy = 0
xdx+xudx-x(xdu+udx) = 0
xdx+ xudx-x^2du-xudx=0

Answer 6

xdx-x^2du+xudx-xudx = 0
xdx-x^2du = 0
-x^2 du = -xdx

Answer 7

du = -x/-x^2 dx

Answer 8

du = 1/x dx

Answer 9

u = ln x + c

Answer 10

oops forgot one more subsitution ...
u = y/x
y/x = lnx +C
y=x(lnx+C)
y = xlnx + Cx.

hmm this problem looks familiar...

Answer 11

woo got it correct!
http://www.wolframalpha.com/input/?i=%28x%2By%29dx-xdy%3D0

Answer 12

yeah it does look familiar....this was one of the problems in my homework from last semester O_o

Answer 13

first order differential equations. I remember those.. substitution/ homogeneous is the hardest part of the course right next to slope fields.

Answer 14

i think all kind of this equation r the same hmm
just know how to do it once then all examples wont be different

Answer 15

it's not a separable equation... it needs all y's to be on the left and all x's on the right so dx/dy=x/x+y is out.

Answer 16

linear equations have the form y' +P(x)y = Q(x)
that wasn't it either...
the equation was either exact (maybe) or substitution/homogeneous

Answer 17

i know :D
but its not enough to mintion the equation
its also good to mintion what u need in it
and ur answer is correct

Answer 18

yeah it's easier than this http://math.stackexchange.com/questions/656806/determining-which-pairs-of-quantified-statements-are-equivalent that's for sure

Answer 19

luckily I'm almost done... just need to figure out the last one. I can translate it, but I just don't know for sure if it's equivalent.

Answer 20

this is ur qs ?
Determining which pairs of quantified statements are equivalent

Answer 21

yeah

Answer 22

For all x, P(x) if and only if Q(x) is the left
For all x in P(x) if and only if all x in Q(x).

Answer 23

logic is for the birds.

Answer 24

hmmm if and only if is like one condition.