Need Help - Exact Integrating Factor?

[NeutronStar]

Here's the problem:

$$y\left( 2x-y-1 \right) dx + x\left( 2y-x-1 \right) dy = 0$$

So rewrite it as:

$$\left( 2xy-y^2-y \right) dx + \left( 2xy-x^2-x \right) dy = 0$$

Now it's in the form,…$P(x,y) dx + Q(x,y) dy = 0$

$$\frac{\partial P(x,y)}{\partial y} = 2x-2y-1$$

$$\frac {\partial Q(x,y)}{\partial x} = 2y-2x-1$$

These are not equal so we need an integrating factor,…

$$h(u) =e^{\int{F(u) du}}$$

Where: $u=xy$, and,...

$$F(u)=\frac {\frac {\partial}{\partial y} P(x,y)- \frac {\partial}{\partial x}Q(x,y)}{y Q(x,y)-x P(x,y)}$$

$$F(u)=\frac{2x-2y-1-(2y-2x-1)}{y(2xy-x^2-x)-x(2xy-y^2-y)}$$

$$F(u)=\frac{4x-4y}{2xy^2-x^2y-xy-2x^2y+xy^2+xy}$$

$$F(u)=\frac{4(x-y)}{3xy^2-3x^2y}=\frac{4(x-y)}{3xy(y-x)} =-\frac{4(x-y)}{3xy(x-y)} =-\frac{4}{3xy}$$

$$h(u) =e^{\int -\frac{4}{3}\frac {du}{u}}}=e^{ -\frac{4}{3}\ln{u}}}=e^{ \ln{(u^{-\frac{4}{3}}})}} =u^{-\frac{4}{3}}=(xy) ^{-\frac{4}{3}}$$

Except this doesn't seem to work. It's also quite different from the book answer.

Book answer for the integrating factor is,..

$$x^{-1} y^{-1}\left( x+y+1\right)^{-1}$$

So where did I go wrong?

Just for the record this problem is from the Dover Book Ordinary Differential Equations by Tenenbaum and Pollard. It's problem # 14 on page 91 in exercise 10.

 

[TenaliRaman]

I think even (x+y+1)^(-4) works as an IF.

-- AI

 

[NeutronStar]

I think even (x+y+1)^(-4) works as an IF.

-- AI

Well, that's interesting. But I have no idea how you got that IF, and I still have no clue why I got the wrong answer.

 

[TenaliRaman]

Neutron,
Sorry not to have responded to this earlier.

I got confused by the way you are finding the IF. AFAIK, there is no direct way of obtaining IF's. The method i use is still ad-hoc.

If u want i will detail the method i use.By the time, i would like to know the source of your method of finding IFs.

-- AI

 

[NeutronStar]

Neutron,
I would like to know the source of your method of finding IFs.

-- AI

I referenced the book that I am using in the opening post. It's in chapter 2 Lesson 10.

Just for the record this problem is from the Dover Book Ordinary Differential Equations by Tenenbaum and Pollard. It's problem # 14 on page 91 in exercise 10.

So I'm using the method that they use in this lesson plan. I wouldn't think that they would have included this problem if it did not fit the criteria for this method (especially since they just give the answer in the answer section with no mention of any special treatment)

I thought that this book would be popular enough that there might be other people out there using it. It's a pretty nice book actually.

 

[Hurkyl]

The formula in my book doesn't look like that.

BTW I found it much more helpful, when I took DiffEq, to work through the steps to solve problems, rather than memorize the end formula.

 

[NeutronStar]

The formula in my book doesn't look like that.

So you're saying that your book has a different formula for calculating the integrating factor?

I got this from page 87 equation (10.73) in my book:

$$F(u)=\frac {\frac {\partial}{\partial y} P(x,y)- \frac {\partial}{\partial x}Q(x,y)}{y Q(x,y)-x P(x,y)}$$

BTW I found it much more helpful, when I took DiffEq, to work through the steps to solve problems, rather than memorize the end formula.

I do too. I'm actually only on page 38 of this book working through some conic sections. I stopped studying the ODE to go back and refresh my knowledge of conic sections. My cousin is also studying this same book and he is on page 91 doing this problem actually.

I looked at it briefly to see if I could tell where he was going wrong but it appears to me that he's done everything just as the book says (as far as I can tell). So I haven't really studied lesson 10 in depth myself. When I get there I intend to study it in great detail.

To be honest I'm not clear on why he chose $u=xy$, but I can't see why any other choice would make any better sense. I really don't want to jump this far ahead of myself to study lesson 10 just yet. I prefer to work up through the book at my own pace. Thus I decided to post for help instead.

I am also studying the Schaum's outline for ODE's, and I'm taking the MIT OCW video course on ODE's (when I can get them to download!). So I have some understanding of the idea behind Exact equations how integrating factors work. I just don't see what my cousin is doing wrong in this problem.

Maybe there's a misprint in the formula on page 87 in our books?

I guess I'll have to read the chapter myself.

 

[Hurkyl]

Ah, I think the difference arises because your text introduced a change of variable.

 

[Dr Transport]

I have been following this tread from the beginning, I looked at my Schaum's Outline for Differential Equations (I can't find my copy of Bopyce and DiPrima because I recently moved and it is not unpacked yet) but I have not found the formula

$$F(u)=\frac {\frac {\partial}{\partial y} P(x,y)- \frac {\partial}{\partial x}Q(x,y)}{y Q(x,y)-x P(x,y)}$$

anywhere. I also looked at the form of the original differential equation

$$y\left( 2x-y-1 \right) dx + x\left( 2y-x-1 \right) dy = 0$$

and noticed that for the integrating funtion to be of the form noted above the Diff EQ should be of the form

$$y F(xy) dx + x G(xy) dy = 0$$, now that is not the same as above. For some reason I think that the manner you are calculating the integrating factor is not correct. I'll try o apply another way and and back to you.

 

[ReyChiquito]

the integrating factor $\mu (x,y)$ must satisfy the equation

$$P\mu_y-Q\mu_x+(P_y-Q_x)\mu=0$$

 

[TenaliRaman]

Rey Chiquito's equation is the crux of the method i use ...

The method goes like this,
Let z be some functions of x and y.
Find T(x,y) where
$$T(x,y) = \frac{Q_x - P_y}{Pz_y - Qz_x}$$
The function z is arbitrarily chosen. Some of the standards normally applicable are,
z = x
z = y
z = x+y
z = x+y+1
z = xy(x+y)
z = xy(x+y+1)

Once u found T(x,y).
Try to express T(x,y) in terms of z only. (If its not possible , then try another z function). Once u are able to express T(x,y) in terms of z alone, your function becomes T(z).

The integration factor is then given by,
$$\mu(z) = e^{\int T(z) dz}$$

Its not hard to prove (albeit not rigorously) why this works.

For your example, if u put z = x+y+1, you will get the IF i got. If u sub in z = xy(x+y+1) then u should prolly get the IF the author of the book is hinting at.

-- AI

 

[ReyChiquito]

then, does your IF works?

The only problem i see here is that the equation for $\mu (x,y)$ as i wrote it has a unique solution (in this case). If the IF is not unique, then neither the solution to the DE right?

 

[TenaliRaman]

Rey,
I haven't really checked. Let me see, if i find time this evening i will try to solve it and see if my IF works.

-- AI

 

[cronxeh]

Im not all sure if this is right but here goes:

$$y\left( 2x-y-1 \right) dx + x\left( 2y-x-1 \right) dy = 0$$

$$y\left P( x,y \right) dx + x\left Q( x,y \right) dy = 0$$

$$y\left P( x,y \right) dx = -x\left Q( x,y \right) dy$$

$$\frac {\left P( x,y \right) dx} {(-x)} = \frac {\left Q( x,y \right) dy} {y}$$

$$\int {\frac {\left P( x,y \right) dx} {(-x)} } = \int { \frac {\left Q( x,y \right) dy} {y} }$$

$$y + x = \frac {\ln{\mid x \mid } (y + 1) + \ln { \mid y \mid } (x+y)} {2}$$


I'm not sure about this one guys.. any takers?