Finding integrading factor for differential equation

[Lily@pie]

Homework Statement

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?

 

[vela]

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?

 

[Lily@pie]

This means that the partial differentiation M_y(x,y)=N_x(x,y)... right?

 

[vela]

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

 

[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 learned that before... Or it is the same as the ordinary differential equation?

 

[vela]

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

 

[Lily@pie]

M_y(x,y)=N_x(x,y)

f_y(x,y)(1-y²)-2yf(xy) = f_x(x,y)(1-x²)-2xf(x,y)

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

 

[vela]

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

 

[Lily@pie]

will ∂f/∂x be (df/du)y?

 

[vela]

Right, which is f'(xy)y.

 

[Lily@pie]

f_y(x,y)(1-y2)-2yf(xy) = f_x(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

 

[vela]

Simplify.

 

[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!

 

[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?

 

[tiny-tim]

Hi Lily@pie!

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

(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? ​

 

[Lily@pie]

(I can see how you got it, but it only works for the dx, not the dy )

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

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

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

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

 

[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.

 

[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...

 

[vela]

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

 

[Lily@pie]

Okay.. Thanks a lot =)