# Homework Help: Finding an Integrating Factor

1. Jun 18, 2017

### Drakkith

Staff Emeritus
1. The problem statement, all variables and given/known data
Find an integrating factor of the form $x^Ay^B$ and solve the equation.
$(2y^2-6xy)dx+(3xy-4x^2)dy=0$

2. Relevant equations
$M=2y^2-6xy$
$N=3xy-4x^2$
$IF = exp(\int \frac{M_y-N_x}{N}\,dx)$
or
$IF = exp(\int \frac{N_x-M_y}{M}\,dy)$
3. The attempt at a solution

From the solutions in the back of my book I know that the integrating factor is supposed to be $xy$, but I can't figure out how to find it.

First, I found the partial derivatives of M and N:
$M_y=4y-6x$
$N_x=3y-8x$

Since the two don't match, the equation isn't an exact equation.
To find the integrating factor, one of the latter two equations in section 2 must be a function of only a single variable, x for the first or y for the 2nd.

$\frac{M_y-N_x}{N}=\frac{4y-6x-3y+8x}{3xy-4x^2}=\frac{y+2x}{x(3y-4x)}$
I don't see any way to simplify this and get a single variable.

$\frac{N_x-M_y}{M}=\frac{3y-8x-4y+6x}{2y^2-6xy}=\frac{-y-2x}{2y(y-3x))}$
Again, I don't see any way to simplify this down to get one variable.

If neither of these can be simplified to a single variable I feel I'm at an impasse.

2. Jun 18, 2017

### SqueeSpleen

You need an integration factor to put this equation in a way that's an exact differential equation.
If I remember well, a neccesary condition to a PDE to be exact was
$$\dfrac{ \partial M }{ \partial y } = \dfrac{ \partial N }{ \partial x }$$
So you have to use the integration factor to "force it"
You know the integration factor is of the form of xAyB. So you have to do
$$\dfrac{ \partial }{ \partial y } \left( (x^{A}y^{B} ( 2y^{6}-6xy ) \right) =\dfrac{ \partial }{ \partial x } \left( (x^{A}y^{B} ( 3xy-4x^{2} ) \right)$$
I did the following in the calculator:
Evaluated the expression I showed to you. Divided by xAyB. And I arrived to
$$( 4a-6b+2 ).x + ( -3a+2b+1 ).y =0$$
From which is easy to get a linear system, and (a,b)=(1,1) is a solution of that system.

3. Jun 18, 2017

### Drakkith

Staff Emeritus
Hi SqueeSpleen,

The idea of multiplying the original equation by XAYB never even occurred to me. From there, I'm able to get to your last equation, but unfortunately I have no idea how to get to a linear system of equations from there. How do you go from a single equation to more than one?

Thank you.

4. Jun 18, 2017

### SqueeSpleen

Well, given that polynomials are a linear space, I mean
$$\{ 1,x,x^{2}, ..., x^{n} \}$$
are linearly independent.
This is also valid in more variables, I mean
$$\{ 1,x,y,x^{2},xy,y^{2},.. \}$$
You can do it because both x and y are independent variables, I mean there is no dependence between them.

Then the equation
$$( 4a-6b+2 ).x + ( -3a+2b+1 ).y =0$$
Is true if and only if both 4a-6b+2=0 and -3a+2b+1=0 are true. So you got a linear system whose unknowns are a and b. From this the rest is linear algebra.

5. Jun 18, 2017

### Drakkith

Staff Emeritus
Sorry, I'm not familiar with most of the math terminology you just used and I haven't taken linear algebra yet. I'll have to take your word that the equation is true if 4a-6b+2=0 and -3a+2b+1=0.

Man, I really feel like I'm missing some background math here...

6. Jun 18, 2017

### Drakkith

Staff Emeritus
Okay, after solving that linear system everything worked like a charm. Thanks!

7. Jun 18, 2017

### SqueeSpleen

Sorry. I sometimes commit the sin of being too lazy and hasty to take my time and I write the things as I'm most comfortable instead of trying to write it in the best way possible to convey my ideas to the person I'm writing to. I have an easier time doing this while talking because the feedback is immediate. Anyway I think it's a very common sin between mathematicians and I should try not to do it too often.

Well. Let's think x and y as variables and forget the vector space jargon.
$$( 4a-6b+2 ).x + ( -3a+2b+1 ).y =0$$
This must be true for any x and y. So in particular if we take x=1 and y=0 we have
$$4a-6b+2 = ( 4a-6b+2 ).1 + ( -3a+2b+1 ).0 =0$$
Also, if we take $x=0$ and $y=1$ we have
$$-3a+2b+1 = ( 4a-6b+2 ).0 + ( -3a+2b+1 ).1 =0$$
So we arrived to both $4a-6b+2=0$ and $-3a+2b+1=0$.

Now you can solve it using methods to solve systems of linear equations.

8. Jun 18, 2017

### Drakkith

Staff Emeritus
Plus you really don't know my level of knowledge on the subject.
But yeah, I keep running into notation or terminology that I'm not familiar with which makes it much more difficult to follow the proofs and explanations in my book. Very frustrating...

9. Jun 18, 2017

### Drakkith

Staff Emeritus
So did you just choose an 'arbitrary' number for X and Y, picking 1 and 0 because it makes it easy?

10. Jun 18, 2017

### SqueeSpleen

You can chose anyone, I needed to pick one of them equal to zero in order to delete the other equation*. And picking 1 was because it was easy, but any nonzero number was as good a 1.
* This wasn't really necessary. you can pick $(x,y)=(c_{1},c_{2})$ and later $(x,y)=(d_{1},d_{2})$ such that there isn't any real number $\lambda$ such that $( \lambda c_{1}, \lambda c_{2}) = (d_{1},d_{2})$ because this would make you have "two times the same equation" in some sense, and you need 2 equations to find both $a$ and $b$.

11. Jun 18, 2017

### Drakkith

Staff Emeritus
Alright. Thanks!

12. Jun 18, 2017

### Buffu

I remember such a problem in Ross' book. Is that the you been using ?
Also isn't the DE homogenous ?

13. Jun 18, 2017

### SqueeSpleen

Well, I'm not very fond of differential equations so I may be wrong but... wasn't an integrating factor something you had to multiply by in order to have convert your equation to an exact differential equation? I mean, the original idea behind the integration factor.

14. Jun 18, 2017

### Drakkith

Staff Emeritus
Nope. I'm using Fundamentals of Differential Equations, by Nagle, Saff, and Snider. Eighth Edition.

Not that I can see.

That's right. I just wouldn't have thought to multiply the original equation by XAYB and then solve for A and B. Trust me when I say that I had no idea what to do.

15. Jun 18, 2017

### SqueeSpleen

Oh, sorry. I honestly thought it had might have another meaning and the "equivalence" had to be proven.

16. Jun 18, 2017

### Buffu

Why this is not homogenous ? Isn't $M(tx, ty) = t^2M(x, y)$ and same for $N(x,y)$ ?

17. Jun 18, 2017

### Drakkith

Staff Emeritus
It looks like it, but I don't know what that has to do with homogeneity. My book as a different criteria, namely that if the right side of $\frac{dy}{dx}=f(x,y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone then the equation is homogeneous. I haven't been able to make the right side a ratio of $\frac{y}{x}$ though.

18. Jun 18, 2017

### Buffu

Both of them aren't equivalent ? Like if $M(tx, ty) = t^n M(x, y)$ then let $t = 1/x$, $M(1, y/x) = (1/x)^n M(x,y)$, likewise for $N$ and then susbtitute it in $y^\prime = M(x,y)/N(x,y)$, which gives $y^\prime = M(1,y/x)/N(1, y/x) = g(y/x)$.

19. Jun 18, 2017

### Drakkith

Staff Emeritus
I'm sorry Buffu, but I can't follow what you're doing.

20. Jun 18, 2017

### Buffu

Sorry I hurried it.

Let $M(tx,ty) = t^n M(x,y)$ and same for $N$.

Then $\require{cancel} \displaystyle \dfrac {dy}{dx} = f(x,y) = \dfrac {M(x,y)}{N(x,y)} = {t^{-n}M(tx, ty) \over t^{-n} N(tx, ty)} = {M(tx, ty) \over N(tx. ty)}$

Let $t = 1/x$

Then $\displaystyle \frac{dy}{dx} = {M(1, y/x) \over N(1, y/x)}$, which is a function in variable $y/x$.

21. Jun 18, 2017

### Drakkith

Staff Emeritus
That looks fine to me, but I can't seem to make it work in this problem. I just end up with $$\frac{dy}{dx}=\frac{-2x^{-1}y^2+6y}{3y-4x}$$
I don't see any way to get that into the right form. I've checked my math twice, but I can't guarantee that I haven't made an error.

Edit: Actually I think I may have gone too far in my simplification. Stand by...

22. Jun 18, 2017

### Buffu

$(2y^2-6xy)dx+(3xy-4x^2)dy=0 \implies -\dfrac{dy}{dx} = \dfrac{2y^2 - 6xy}{3xy - 4x^2} = \dfrac{(y^2)(2 - 6(x/y) )}{(x^2)(3(y/x) - 4)} = \dfrac{y^2}{x^2} \dfrac{2 - 6(y/x)^{-1}}{3(y/x) - 4}$

Maybe this is correct.

23. Jun 18, 2017

### Drakkith

Staff Emeritus
Yes, I went too far in my simplification. Backing up a step I end up with $$\frac{dy}{dx}=\frac{-2x^{-2}y^2+6x^{-1}y}{3x^{-1}y-4}$$
Which is all a function of $\frac{y}{x}$

So now that we have that out of the way, why did we do this?

24. Jun 18, 2017

### Buffu

We can substitute $y = kx$ to solve this.

25. Jun 18, 2017

### Drakkith

Staff Emeritus
Oh, yeah, of course.

Also, I just found a single sentence in my book saying that one test for homogeneity is to replace $x$ by $tx$ and $y$ by $ty$. Then the equation $\frac{dy}{dx}=f(x,y)$ is homogenous if and only if $f(tx,ty)=f(x,y)$. Too bad they don't use that in any of the examples. I just happened to stumble across while re-reading.