Is there a mistake here so far?

[franky2727]

question attached

 

[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.

 

[franky2727]

ok emailed

 

[Tomtom]

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

 

[HallsofIvy]

The problem is to solve
[tex]\frac{dy}{dx}= \frac{3x- y+ 1}{x+ y+ 1}[/tex]

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
[tex]dF= \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial x}dx= (x+y+1)dy+ (-3x+ y- 1)dx[/tex]
Since dF= 0, F(x,y)= constant. Do you know how to find F from
[tex]\frac{\partial F}{\partial y}= x+ y+ 1[/tex]
and
[tex]\frac{\partial F}{\partial x}= -3x+ y- 1[/tex]?

 

[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 don't have a clue where you would go after this? isn't it easier to back substitute cos I'm quite confused here

 

[franky2727]

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

 

[HallsofIvy]

If
[tex]\frac{\partial F}{\partial y}= x+ y+ 1[/tex]
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,
[tex]\frac{\partial F}{\partial x}= y+ g'(x)[/tex]
and since we know the partial derivative of with respect to x must be -3x+ y+ 1,
[tex]\frac{\partial F}{\partial x}= y+ g'(x)= -3x+ y+ 1[/tex]
we have
[tex]g'(x)= -3x+ 1[/tex]
so
[tex]g(x)= -\frac{3}{2}x^2+ x[/tex]
(I've ignored the constant of integration- which really is a constant since g is a function of x only)
Putting those together,
[tex]F= xy+ \frac{1}{2}y^2+ y-\frac{3}{2}x^2+ x[/tex]
and since dF= 0 means F= C,
[tex]xy+ \frac{1}{2}y^2+ y-\frac{3}{2}x^2+ x= C[/tex]