First Order Equations

[Link-]

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?

[waht]

Separate the variables x and y first.

[menager31]

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?

It's a total differential equation. It's how i solved it:
10588
two or three combinations and you get it:
10589
where C is real

:surprised

[HallsofIvy]

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

[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

[menager31]

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

so where's an error?

[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.

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Link:

Why don't you show us what you did?

[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

[menager31]

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

y and x aren't constant

[AiRAVATA]

y and x aren't constant

Really? Are you serious?

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

[HallsofIvy]

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
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
makes it clear that this is a troll.

[menager31]

10592

:approve::approve::approve::approve:

[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.

[menager31]

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

\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.
ok, sorry, no I've understood

[HallsofIvy]

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!

[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

[Link-]

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.

-------

Link:

Why don't you show us what you did?

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.

[Link-]

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.

[cks]

Let y=kx,

then you find that k=-1

so, y=-x is the answer.