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

• Drakkith
In summary, the homework statement is to find an integrating factor of the form ##x^Ay^B## and solve the equation. However, the equation is not an exact equation and one of the equations in section 2 must be a function of only a single variable, x for the first or y for the 2nd. 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. After solving the linear system, a and b are found to be 1 and 1.
Drakkith
Mentor

## 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.

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.

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.

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.

Buffu
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...

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

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.

SqueeSpleen said:
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...

SqueeSpleen said:
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?

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##.

Alright. Thanks!

Drakkith said:

## 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 ?

Drakkith said:
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.
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.

Buffu said:
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.

Buffu said:
Also isn't the DE homogenous ?

Not that I can see.

SqueeSpleen said:
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 XAYB and then solve for A and B. Trust me when I say that I had no idea what to do.

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

Drakkith said:
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.

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

Buffu said:
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.

Drakkith said:
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)##.

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

Drakkith said:
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##.

SqueeSpleen
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...

Drakkith said:
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.

SqueeSpleen
Drakkith said:
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?

SqueeSpleen
Drakkith said:
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.

Buffu said:
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.

Buffu

## 1. What is an integrating factor?

An integrating factor is a function used in differential equations to make them easier to solve. It is multiplied by the original equation to transform it into an exact equation, which can then be solved using standard techniques.

## 2. How do I find an integrating factor?

The method for finding an integrating factor depends on the form of the differential equation. In general, you can multiply the equation by a function of the independent variable or by a function of the dependent variable and its derivatives. You can also use the method of undetermined coefficients or the method of variation of parameters.

## 3. Why do we need an integrating factor?

Integrating factors are used to simplify the process of solving differential equations. They help transform the equation into a more manageable form, making it easier to find a solution. Without an integrating factor, some differential equations would be nearly impossible to solve.

## 4. Can any equation have an integrating factor?

No, not all differential equations have an integrating factor. The equation must be in a specific form, such as an exact equation or a linear equation with constant coefficients. If the equation does not meet these criteria, an integrating factor cannot be found.

## 5. How do I know if I have found the correct integrating factor?

If you have correctly found the integrating factor, the transformed equation should be an exact equation. This means that the partial derivatives of the original equation with respect to the independent and dependent variables should be equal. You can also check your solution by plugging it back into the original equation to see if it satisfies the differential equation.

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