First Order Equations

1. Jun 12, 2007

I have the following equation:

(x^2-y^2-y)dx-(x^2-y^2-x)dy=0

I was trying to find the integral factor of this to make it a exact differential equation, but ended in a almost imposible integral. do anyone have any idea of how to make this?

2. Jun 13, 2007

waht

Separate the variables x and y first.

3. Aug 10, 2007

menager31

It's a total differential equation. It's how i solved it:

two or three combinations and you get it:

where C is real

:surprised

4. Aug 10, 2007

HallsofIvy

Staff Emeritus
It certainly is NOT an 'exact' equation (total differential) and the solution you give is not correct.

5. Aug 10, 2007

matness

I don't know whether there is an easier solution but i would start with
dy/dx=(x^2-y^2-y)/(x^2-y^2-x)
then divide both sides with x^2 and use u=y/x substitution

6. Aug 10, 2007

menager31

so where's an error?

7. Aug 10, 2007

AiRAVATA

$P(x,y)=x^2-y^2-y$, and $Q(x,y)=-x^2+y^2+x$, so

$$P_y=-2y-1 \neq -2x+1 = Q_x.$$

-------

Why don't you show us what you did?

Last edited: Aug 10, 2007
8. Aug 10, 2007

menager31

omg I don't want to write 5 more lines.
but if it's too hard for someone who would read it i can write

9. Aug 10, 2007

menager31

y and x aren't constant

10. Aug 10, 2007

AiRAVATA

Really? Are you serious?

Did you even bothered to calculate $\partial P/\partial y$ and $\partial Q/ \partial x$?

Last edited: Aug 10, 2007
11. Aug 11, 2007

HallsofIvy

Staff Emeritus
do you KNOW how to calculate Py and Qx?

Your remark makes it clear that you don't know how to take a partial derivative!

It isn't important, of course, that
makes it clear that this is a troll.

12. Aug 11, 2007

menager31

13. Aug 11, 2007

AiRAVATA

?????????????

$$\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$$

False => False. Your hypothesis is wrong! Go back to the book and read REALLY CAREFULLY this time.

The least you should do this time is check your answer before claiming it's correct.

14. Aug 11, 2007

menager31

ok, sorry, no I've understood

15. Aug 12, 2007

HallsofIvy

Staff Emeritus
So you firmly believe (a quote from your post) that
-2y- 1= 1- 2x? If so there is not point in trying to change your mind!

16. Sep 3, 2007

DiffEQprof

For a different view on the subject, try this book:

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&rd=1&item=120156979964&ssPageName=STRK:MESE:IT&ih=002 [Broken]

Last edited by a moderator: May 3, 2017
17. Sep 3, 2007

I couldn't resolve it.

I suggested integral factor because I founded this equation on an old book in the integral factor chapter. I will give it another try to see what happens.

18. Sep 3, 2007

OK, I tried again using the integral factors to make the equation an exact differential equation I end up with nothing again.

($$\frac{\delta M}{\delta y}$$ - $$\frac{\delta N}{\delta x}$$ )/ N

I get F(x,y) so this is not helping.

The other

($$\frac{\delta N}{\delta x}$$ - $$\frac{\delta M}{\delta y}$$) / M

Get G(x,y) nothing

and the other cases doesn't help at all because multiplying by x or y or X^2 or y^2 wouldn't end up in f(u) u=x/y or u=xy.
So maybe this book was trying to make headaches on students.

Anyway thanks for the help to all.

19. Sep 5, 2007

cks

Let y=kx,

then you find that k=-1