Is there a mistake here so far?

1. Aug 17, 2008

franky2727

question attached

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2. Aug 17, 2008

Tomtom

Mail it to my on wing-tales@gmail.com (without the - between wing and tales). Let me have a look.

You might want to include the topic of your problem in the title.

3. Aug 17, 2008

franky2727

ok emailed

4. Aug 17, 2008

Tomtom

Could you write the starting problem here? The first sentence is a little hard to read.

5. Aug 17, 2008

HallsofIvy

Staff Emeritus
The problem is to solve
$$\frac{dy}{dx}= \frac{3x- y+ 1}{x+ y+ 1}$$

What franky2727 has done is to set u= 3x- y+1 and v= x+ y+ 1 so that, from the equation, y'= u/v. Then from the first equation u'= 3- y'= 3- u/v. I frankly don't see how adding more variables is going to help, especially since he then defines z= u/v so he winds up with a total of 5 variables!

franky2727, rewrite your equation as (x+y+ 1)dy = (3x-y+1)dx so that (x+y+1)dy- (3x-y+1)dx= (x+y+1)dy+ (-3x+ y- 1)dx= 0. Since (x+y+1)x= 1= (-3x+y-1)y, that is an exact equation. That is, there exist F(x,y) so that
$$dF= \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial x}dx= (x+y+1)dy+ (-3x+ y- 1)dx$$
Since dF= 0, F(x,y)= constant. Do you know how to find F from
$$\frac{\partial F}{\partial y}= x+ y+ 1$$
and
$$\frac{\partial F}{\partial x}= -3x+ y- 1$$?

6. Aug 17, 2008

franky2727

erm, no not a clue the way my lecturer did it was to add u,v, and z and then back substitute when he got to the end but it seemed long winded and hard. i see what your doing adding DF to emiminate but i dont have a clue where you would go after this? isnt it easier to back substitute cos i'm quite confused here

7. Aug 17, 2008

franky2727

oh and also just out of curiosity have i logged out that last line correctly?

8. Aug 17, 2008

HallsofIvy

Staff Emeritus
If
$$\frac{\partial F}{\partial y}= x+ y+ 1$$
the, integrating with respect to y (treating x as if it were a constant)
g(x)[/itex]
Notice that the "constant of integration" here may be a function of x because we are treating x as a constant. Differentiating that with respect to x,
$$\frac{\partial F}{\partial x}= y+ g'(x)$$
and since we know the partial derivative of with respect to x must be -3x+ y+ 1,
$$\frac{\partial F}{\partial x}= y+ g'(x)= -3x+ y+ 1$$
we have
$$g'(x)= -3x+ 1$$
so
$$g(x)= -\frac{3}{2}x^2+ x$$
(I've ignored the constant of integration- which really is a constant since g is a function of x only)
Putting those together,
$$F= xy+ \frac{1}{2}y^2+ y-\frac{3}{2}x^2+ x$$
and since dF= 0 means F= C,
$$xy+ \frac{1}{2}y^2+ y-\frac{3}{2}x^2+ x= C$$