Can you find the integrating factor for this non-exact equation?

[Drakkith]

Homework Statement

Find an integrating factor of the form $x^Ay^B$ and solve the equation.
$(2y^2-6xy)dx+(3xy-4x^2)dy=0$

Homework Equations

$M=2y^2-6xy$
$N=3xy-4x^2$
$IF = exp(\int \frac{M_y-N_x}{N}\,dx)$
or
$IF = exp(\int \frac{N_x-M_y}{M}\,dy)$

The Attempt at a Solution

[/B]
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.

 

[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 x^Ay^B. 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 x^Ay^B. 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.

 

[Drakkith]

Hi SqueeSpleen,

The idea of multiplying the original equation by X^AY^B 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.

 

[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.

 

[Drakkith]

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...

 

[Drakkith]

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

 

[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.

 

[Drakkith]

Sorry. I sometimes commit the sin of being too lazy and I write the things as I know them instead of trying to write it in the best way possible to convey my ideas. I think it's a very common sin between mathematicians.

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...

 

[Drakkith]

This must be true for any x and y. So in particular if we take x=1 and y=0...

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

 

[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 $b$.

 

[Drakkith]

Alright. Thanks!

 

[Buffu]

Homework Statement

Find an integrating factor of the form $x^Ay^B$ and solve the equation.
$(2y^2-6xy)dx+(3xy-4x^2)dy=0$

Homework Equations

$M=2y^2-6xy$
$N=3xy-4x^2$
$IF = exp(\int \frac{M_y-N_x}{N}\,dx)$
or
$IF = exp(\int \frac{N_x-M_y}{M}\,dy)$

The Attempt at a Solution

[/B]
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.

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

 

[SqueeSpleen]

Hi SqueeSpleen,

The idea of multiplying the original equation by X^AY^B 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.

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.

 

[Drakkith]

I remember such a problem in Ross' book. Is that the you been using ?

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

Also isn't the DE homogenous ?

Not that I can see.

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.

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

 

[SqueeSpleen]

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

 

[Buffu]

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 X^AY^B and then solve for A and B. Trust me when I say that I had no idea what to do.

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

 

[Drakkith]

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

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.

 

[Buffu]

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.

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 $y^\prime = M(1,y/x)/N(1, y/x) = g(y/x)$.

 

[Drakkith]

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

 

[Buffu]

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

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 $y/x$.

 

[Drakkith]

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...

 

[Buffu]

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...

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

Maybe this is correct.

 

[Drakkith]

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

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?

 

[Buffu]

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?

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

 

[Drakkith]

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

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.