# Finding integrading factor for differential equation

1. Mar 17, 2012

### Lily@pie

1. The problem statement, all variables and given/known data
The differential equation:

(1-y2) dx + (1-x2) dy = 0

the integrating factor is of the form μ(x,y) = f(xy). Find f

2. The attempt at a solution

I have tried many times to get a function of (xy) such that the LHS of the differential equation to be exact.

I have tried the integrating factor 1/[(1-y^2)(1-x^2)] and it works..... but it is definitely not a function of the term (xy)...

Any hints on how should I find it? Or is there any formal way to find it?

2. Mar 17, 2012

### vela

Staff Emeritus
You know that $f(xy)(1-y^2)\,dx + f(xy)(1-x^2)\,dy$ is supposed to be an exact differential, so what does that tell you about how $M(x,y) = f(xy)(1-y^2)$ and $N(x,y) = f(xy)(1-x^2)$ are related?

3. Mar 17, 2012

### Lily@pie

This means that the partial differentiation My(x,y)=Nx(x,y).... right?

4. Mar 17, 2012

### vela

Staff Emeritus
Yes. If you do that, you'll eventually end up with a differential equation for f(xy) you can solve.

5. Mar 17, 2012

### Lily@pie

That is actually the part where I am stuck. The differential equation will be a partial differential equation right? But I hv not learnt that before... Or it is the same as the ordinary differential equation?

6. Mar 17, 2012

### vela

Staff Emeritus
No, you should end up with an ODE. Show us your work.

7. Mar 17, 2012

### Lily@pie

My(x,y)=Nx(x,y)

fy(x,y)(1-y2)-2yf(xy) = fx(x,y)(1-x2)-2xf(x,y)

Is the above correct? As I don't know how to continue from here onwards.

8. Mar 17, 2012

### vela

Staff Emeritus
You need to evaluate the partial derivatives. Think of it as f(u) where u=xy. What's $\partial f/\partial x$ equal to?

9. Mar 17, 2012

### Lily@pie

will ∂f/∂x be (df/du)y?

10. Mar 17, 2012

### vela

Staff Emeritus
Right, which is f'(xy)y.

11. Mar 17, 2012

### Lily@pie

fy(x,y)(1-y2)-2yf(xy) = fx(x,y)(1-x2)-2xf(x,y)

f'(xy)(x-xy2)-2yf(xy) =f'(xy)(y-x2y)-2xf(x,y)

f'(xy)+$\frac{2x-2y}{x-xy^{2}-y+x^{2}y}$f(xy)=0

Is this the ODE that I need to solve to find f(xy)? It feels like I'm getting something but I don't know how to start

12. Mar 17, 2012

### vela

Staff Emeritus
Simplify.

13. Mar 17, 2012

### Lily@pie

Oh, so i get

f'(xy) + $\frac{2}{1+xy}$f(xy)=0

So I'll get f(xy)=$\frac{1}{(1+xy)^{2}}$

YAY!!! Then I found the integrating factor...

Thanks A LOT!!!

14. Mar 17, 2012

### Lily@pie

For my own practice, I tried to solve the differential equation

(1-y$^{2}$) dx + (1-x$^{2}$) dy = 0

And the solution I got is

$\frac{y^{2}-1}{y(1+xy)}$+$\frac{1}{y}$=D
where D is some arbitrary constant

Is this correct?

15. Mar 17, 2012

### tiny-tim

Hi Lily@pie!
(I can see how you got it, but it only works for the dx, not the dy )

nooo …

the solution obviously is going to be symmetric in x and y …

so what's the simplest symmetric numerator you can think of?

16. Mar 17, 2012

### Lily@pie

What do u mean by it only works for the dx and not dy?

What do u mean by symmetric numerator? y=x?

17. Mar 17, 2012

### HallsofIvy

No, that's not what "symmetric in x and y" means. It means that if you were to swap "x" and "y" you would get exactly the same formula. For example, both $x^3+ y^3$ and $x^2+ 3xy+ y^2$ is "symmetric in x and y". $x^3+ y^2$ and $x^2+ 3x+ y^2$ are not.

18. Mar 17, 2012

### Lily@pie

Ohhh.... so the simplest one will be something like x+y?

But i still don't understand what does it mean by the solution only works for dx and not dy...

19. Mar 17, 2012

### vela

Staff Emeritus
I think your solution is correct, but you can simplify it a bit by putting everything over a common denominator.

20. Mar 18, 2012

### Lily@pie

Okay.. Thanks a lot =)