Question

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

Answer 1

int(x-y)dx and try to derive it back down to (x+y)

Answer 2

\[\int(x-y)dx\]
\[{x^2 \over2} -yx +g(y)\]
\[Dy(\frac{x^2}{2}-yx +g(y))=(x+y)\]

Answer 3

what do you mean by "Dy"?

Answer 4

Dy is another symbol for 'derivative with respect to' ... well, in this case its with respect to y

Answer 5

i dont quite follow, sorry

Answer 6

\[Dy(\frac{x^2}{2}-yx+g(y))=-x+g'(y)\]
\[-x+g'(y) = x+y\] solve for g'(y) maybe?

Answer 7

When you say Dy, are you referring to partial derivatives?

Answer 8

i spose thats a way to view it yes.

Answer 9

u need to find the solution?

Answer 10

\[Dy =\frac{d}{dy}(F(x,y))\]

Answer 11

ok

Answer 12

where do i go from here?

Answer 13

g'(y) = 2x +y

\(\int g'(y)dy=\int(2x+y)dy\)
\[g(y) = 2xy + \frac{y^2}{2}+C\]
right?

Answer 14

now test this to see if we get our results:

F(x,y) = \(\frac{x^2}{2}-yx+2xy+\frac{y^2}{2}+C\)

Answer 15

\[F(x,y)=x^2/2+y^2/2+xy+C\]

Answer 16

where do i go from here?

Answer 17

depends on what the problem is asking for ... you never realy stated what you needed from this; and we might be missing a y' from that

Answer 18

"Integrate the following homogeneous equation"

Answer 19

well, derive what we got and see if it gets you back to the original

Answer 20

I currently have F(x,y), x, y and C in my answer, how do i get rid of the F(x,y)?

Answer 21

F(x,y) is just the name of the answer ... it equals the equation.

Answer 22

i realise that, but surely the equation itself should contain an equals sign?

Answer 23

no. what we are calling an equation is more accurately denoted as a expression that is equivalent to the function of 2 variables (x,y).

Answer 24

the answer will be the 'expression' once we determine if its correct :)

Answer 25

but surely as my original differential equation contained an equals sign, then it should also contain an equals sign when intergrated?

Answer 26

the integration of 0 goes to a constant if anything .... so the +C absorbs it in a way ...

Answer 27

am still confused sorry... surely our answer should be something like:
\[x^2/2+y^2/2+xy+C=?\]

Answer 28

place a C over there if you want .... but its rather pointless

Answer 29

am still confused.
surely if we look at a simple example of:
\[y'=x^2\]then when we integrate that we get
\[y=x^3/3+c\]
why are we left with an expression as opposed to an equation when we integrate:
\[(x-y)dx+(x+y)dy=0\]?

Answer 30

ok ... lets use that same example but conform it to our problem

(x^2)dx - dy = 0

this goes to:

F(x) = x^3/3 + C; now why would you get rid of the F(x) and introduce an = sign into the mix?

Answer 31

Surely it would go to:
\[x^3/3-y+c=0\]which is the same as
\[y=x^3/3+c\]?

Answer 32

sorry if i'm being slow here...

Answer 33

exactly....

y = F(x) in this case ... the +C absorbs any and all constants that are involved in the problem into one nice little package.

C = {C1 + C2 + C3 + C4 +...+ C{n-1} + C{n} }

Answer 34

so in our original question, surely that would make our F(x,y) equal to z?

Answer 35

yes...

Answer 36

but is it not possible to give an answer that would only contain x and y?

Answer 37

you seem to be jumping all around the board here.

can you think of a way to ask what is on your mind right now in a more clear and concise manner?

Answer 38

I just want to know, step by step if possible, how to answer:
"Integrate the following homogeneous equation:
\[(x-y)dx+(x+y)dy=0\]"

Answer 39

there is no one way to solve this ... assuming it can be solved.

what methods do you know of?

Answer 40

It can definitely be solved, it was a question on a past exam paper

Answer 41

i included the assuming part becasue not all integrands can be solved. such as \(\int e^{-x^2}\)dx

Answer 42

there is 'seperation of variables' which does no good in this instance since we cant seperate x and y into a nice package

Answer 43

there is another method I know of, which I am trying to work on this where you integrate one portion and try to derive it into the other .

Answer 44

ok

Answer 45

M(x,y)dx + N(x,y)dy = 0

int(M(x,y))dx to get a function: F(x,y) +C; where C = some function of y, they tend to name it g(y). Becasue it is a function of y only, it vanishes in the partial derivative with respect to x.

Now we derive this down with respect to y to equate it with what we have for the N(x,y)dy part

Answer 46

not to butt in (because i certainly don't know how to do this) but does it say "integrate" or "solve"?
you have a first order ode so it is not clear what "integrate" would mean. especially since there is a zero on one side of the equal sign.

just asking...

Answer 47

there are other methods, of which i cant recall the details of at the momnet :)

Answer 48

@satellite73 it says integrate

Answer 49

okay

Answer 50

at the moment, M(x,y)dx = (x-y)dx and integrates up to:

x^2/2 - yx + g(y) ; we need to derive this with respect to y to get it down to and equate it with N(x,y)dy: (x+y)dy in this case

Answer 51

ok

Answer 52

i get:

-x + g'(y) = x+y

g'(y) = 2x +y .... since we now know what g' is; we integrate this with respect to y to find g(y) and include it into our integral solution

Answer 53

g(y) = 2xy + y^2/2 to which:

F(x,y) = x^2/2 -xy +2xy +y^2/2 +C ....

the only problem is, i think there is a chain rule involved in this becasue I cant seem to derive it back down to our original problem

Answer 54

if you don't mind me asking, are you a student/lecturer, and to what level?

Answer 55

or ... ive intergrated wrong by introducing a simple mistake somplace

Answer 56

im a college student in my 2nd year

Answer 57

ok thanks

Answer 58

there is a method of substituting z = y/x but i dont recall the details of it, or even if that part is accurate :)

Answer 59

what do you think of this:
http://www.mathhelpforum.com/math-help/f59/first-order-differential-equation-184051.html#post664406
?

Answer 60

thats the method i cant recall to well :)

Answer 61

this might be better for following the solution :)

http://www.youtube.com/watch?v=UpLQUGBznE4

Answer 62

thanks, will have a look

Answer 63

understand now, thanks :-)

Answer 64

im understanding it better meself :) youre welcome