# Tenenbaum and Pollard, Problem 2.10.6

• Mar 2nd 2011, 05:25 PM
Ackbeet
Tenenbaum and Pollard, Problem 2.10.6
Problem 2.10.6 Test the following DE for exactness. If it is not exact, try to find an integrating factor. After the equation is made exact, solve by looking for integrable combinations. If you cannot find any, use the standard exact method.

$(x^{2}-y^{2}-y)\,dx+(y^{2}-x^{2}+x)\,dy=0.$

We test for exactness. Is

$-2y-1=1-2x?$

No. We try to find an integrating factor.

1. Based on the symmetry of the problem, I'm going to guess a function $h=h(xy)$. Then we have

$h(xy)(x^{2}-y^{2}-y)\,dx+h(xy)(x+y^{2}-x^{2})\,dy=0.$

The exactness condition implies

$h'(xy)(x)(x^{2}-y^{2}-y)+h(xy)(-2y-1)=h'(xy)(y)(x+y^{2}-x^{2})+h(xy)(1-2x)\quad\Rightarrow$

$h'(xy)(x^{3}-xy^{2}-xy-xy-y^{3}+x^{2}y)=h(xy)(1-2x+1+2y)\quad\Rightarrow$

$h'(xy)(x^{3}-y^{3}-2xy-xy^{2}+x^{2}y)=h(xy)(2-2x+2y).$

I'm not sure this is going anywhere, because we'd have to write all the non- $h$ stuff as a function of $xy$, which I don't think is going to happen.

2. Let's try again, this time with a function $h=h(y/x)$. The DE is

$h\!\left(\dfrac{y}{x}\right)(x^{2}-y^{2}-y)\,dx+h\!\left(\dfrac{y}{x}\right)(x+y^{2}-x^{2})\,dy=0.$

The exactness condition yields

$h'\!\left(\dfrac{y}{x}\right)\left(\dfrac{1}{x}\ri ght)(x^{2}-y^{2}-y)+h\!\left(\dfrac{y}{x}\right)(-2y-1)=

$h'\!\left(\dfrac{y}{x}\right)\left(x-\dfrac{y^{2}}{x}-\dfrac{y}{x}\right)+h\!\left(\dfrac{y}{x}\right)(-2y-1)=

$h'\!\left(\dfrac{y}{x}\right)\left(x-\dfrac{y^{2}}{x}-\dfrac{y}{x}+\dfrac{y}{x}+\dfrac{y^{3}}{x^{2}}-y\right)=

$h'\!\left(\dfrac{y}{x}\right)\left(x-\dfrac{y^{2}}{x}+\dfrac{y^{3}}{x^{2}}-y\right)=

$h'\!\left(\dfrac{y}{x}\right)\left(1-\dfrac{y^{2}}{x^{2}}+\dfrac{y^{3}}{x^{3}}-\dfrac{y}{x}\right)=
h\!\left(\dfrac{y}{x}\right)\left(\dfrac{2}{x}-2+2\,\dfrac{y}{x}\right).$

Again, I'm not seeing how the non- $h$ stuff can be written as a function of $y/x$.

3. Let's go for a simpler one: $h=h(x)$. Then we have

$h(x)(x^{2}-y^{2}-y)\,dx+h(x)(x+y^{2}-x^{2})\,dy=0.$

The exactness condition implies

$h(x)(-2y-1)=h'(x)(x+y^{2}-x^{2})+h(x)(1-2x)\quad\Rightarrow$

$h(x)(-2y-1-1+2x)=h'(x)(x+y^{2}-x^{2}).$

Again, no prospect of writing it the way I want. By symmetry, I don't think a function of the type $h=h(y)$ will work.

4. Let's try the $h=h(x/y)$ type:

$h\!\left(\dfrac{x}{y}\right)(x^{2}-y^{2}-y)\,dx+h\!\left(\dfrac{x}{y}\right)(x+y^{2}-x^{2})\,dy=0.$

Exactness implies

$h'\!\left(\dfrac{x}{y}\right)\left(-\dfrac{x}{y^{2}}\right)(x^{2}-y^{2}-y)+h\!\left(\dfrac{x}{y}\right)(-2y-1)=

$h'\!\left(\dfrac{x}{y}\right)\left(-\dfrac{x^{3}}{y^{2}}+x+\dfrac{x}{y}\right)+h\!\lef t(\dfrac{x}{y}\right)(-2y-1-1+2x)=

$h'\!\left(\dfrac{x}{y}\right)\left(x+\dfrac{x}{y}-\dfrac{x^{3}}{y^{2}}-\dfrac{x}{y}-y+\dfrac{x^{2}}{y}\right)=

$h'\!\left(\dfrac{x}{y}\right)\left(x-\dfrac{x^{3}}{y^{2}}-y+\dfrac{x^{2}}{y}\right)=

$h'\!\left(\dfrac{x}{y}\right)\left(\dfrac{x}{y}-\dfrac{x^{3}}{y^{3}}-1+\dfrac{x^{2}}{y^{2}}\right)=
h\!\left(\dfrac{x}{y}\right)\left(\dfrac{2}{y}+2-2\,\dfrac{x}{y}\right).$

Still not seeing how this could work.

Does anyone see a mistake in my computations anywhere? See any new ideas?

Thanks!
• Mar 3rd 2011, 05:56 AM
Jester
When I see your ODE this is what I try

$(x^2-y^2)(dx-dy) + xdy-ydx = 0$

so

$(x^2-y^2)(dx-dy) + x^2 d\left(\dfrac{y}{x}\right) = 0$

so

$\left(1-\dfrac{y^2}{x^2}\right)(dx-dy) + d\left(\dfrac{y}{x}\right) = 0$

so

$dx - dy + \dfrac{d\left(\dfrac{y}{x}\right)}{1-\dfrac{y^2}{x^2}} = 0.$
• Mar 3rd 2011, 06:12 AM
Ackbeet
Thanks for that. So do you think it's a fruitless task trying to find an integrating factor?
• Mar 3rd 2011, 06:17 AM
Jester
No, unless you guess the right form

i.e. $h = h(x^2-y^2)$

althought there are others.
• Mar 3rd 2011, 06:25 AM
Ackbeet
Did you mean that it is a fruitless task unless you guess the right form?
• Mar 3rd 2011, 07:45 AM
Jester
In my opinion - Yes! If your ODE is

$M dx + N dy = 0$ and is not already in exact form them you seek an integrating factor $\mu$ such that

$\dfrac{\partial}{\partial y} M \mu = \dfrac{\partial}{\partial x} N \mu$

or

$N \mu_x - M \mu_y = \left(M_y-N_x \right) \mu\;\;\;(1)$.

It is natural to assume certain form of this PDE

$\mu = \mu(x), \mu(y), \mu(xy), \mu(y/x)$ etc

but it really depends on what the PDE (1) looks like. Really to solve (1) is to solve the ODE itself.
• Mar 3rd 2011, 10:32 AM
Ackbeet
Jolly good. Thanks a bundle, as always.
• Mar 4th 2011, 09:58 AM
Ackbeet
Incidentally, I just found a way to encapsulate all five of Tenenbaum and Pollard's five guesses for integrating factors. They say that you should guess $h(x), h(y), h(xy), h(x/y),$ and $h(y/x).$ You can unify all this in one place if you just guess $h=h(x,y)=x^{n}y^{m},$ and you get more possibilities besides.

What a time-saver! I don't think this method will work for the DE in this thread, but at the very least, I won't have to trudge through guessing all five of those options above.
• Mar 4th 2011, 11:22 AM
Ackbeet
It just occurred to me that there are possibilities that my little time-saver leaves out: $\sin(x),$ for example, would be a possible integrating factor you might find if you tried $h(x),$ but you couldn't find that with $x^{n}y^{m}.$ Still, for DE's where it looks like powers of the variables will work, this could still save some time.
• Mar 4th 2011, 12:14 PM
Jester
What you could do is to assume $h = h\left(x^m y^n \right)$. This should hit all of your cases.
• Mar 4th 2011, 06:27 PM
Ackbeet
Thanks for that one!