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Integrating Factor for Closed but not Exact Differential Form

finding an integrating factor for the nonlinear non-autonomous ODE $ (xy)y'+yln y - 2xy = 0 $Exact Equations and Integrating FactorFinding Integrating factorIntegrating factor for a non exact differential formIntegrating factor not making ODE exactDeriving the integrating factor for exact equations.Finding integrating factor for non-exact differential equation $(4y-10x)dx+(4x-6x^2y^{-1})dy=0$.A formula for homogeneous differential form integrating factorExact Differential Equation Integrating FactorHow to find the integrating factor for this ODE?

0

$begingroup$

(I kind-of skipped ODEs in undergrad, so my knowledge is a little sketchy; please excuse me if this question is elementary.)

(In addition, this problem is a bit of a sketch; I'll flesh it out with calculations in a day or so.)

I've been looking over the theory of integrating factors, and they appear to focus on making the resulting differential form $muomega$ closed, not necessarily exact.

This appears to work fine most of the time. For instance, if I take the ODE

$$-y dx + x dy = g(t) dt$$

the left-hand side is neither closed nor exact. I recall reading in $textit{Advanced Calculus}$ by Buck https://smile.amazon.com/Advanced-Calculus-SECOND-Creighton-Buck/dp/B000OG0SRK/ that the differential equation for an integrating factor is called a "Pfaffian"; if that is the case and my back-of-the-envelope calculations from last night are correct [I'll flesh this out with calculations in a day or so], the Pfaffian is

$$xfrac{partial mu}{partial x} + yfrac{partial mu}{partial y} = -2mu$$

This appears to admit the solution integrating factor

$$mu(x,y) = frac{1}{xy}$$

and so the ODE has the implicit solution

$$left(frac{y}{x}right)^{xy} = Aexpleft(xyint frac{g}{xy} dtright)$$

(Maybe one would want to take $g(t) equiv 0$ ?)

The situation is different with the closed but not exact differential form (left-hand side of the equation)/ODE

$$frac{-y}{x^2+y^2} dx + frac{x}{x^2+y^2} dy = g(t) dt$$

The Pfaffian is [again, I'll flesh this out with calculations in a day or so]

$$frac{partial mu}{partial x}x + frac{partial mu}{partial y}y = 0$$

which doesn't appear to admit a solution, as nearly as I can figure.

Does this second ODE admit an integrating factor and I'm doing something wrong? More generally, if one has a closed but not exact differential form, is it possible to find an integrating factor that makes the differential form exact?

Thanks in advance.

ordinary-differential-equations differential-forms integrating-factor

share|cite|improve this question

edited Mar 19 at 3:11

Jeffrey Rolland

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These differential equations fall under the umbra of "implicit ODEs" (at least, they generally have implicit solutions). The variables x and y are themselves functions of t in the paradigm.
  $endgroup$
  – Jeffrey Rolland
  Mar 19 at 2:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that every closed form is locally exact ... Solving the differential equation is usually a local question; topology will dictate whether you can get a global integrating factor.
  $endgroup$
  – Ted Shifrin
  Mar 19 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin Mmmm, so, "in a flow box" (to use terminology from Abraham and Marsden smile.amazon.com/Foundations-Mechanics-Ams-Chelsea-Publishing/…), that is to say, in a rectangle with respect to t, x, and y in which all functions and their partial derivatives are continuous, the second ODE actually is exact, and we have an implicit solution; very nice, thank you.
  $endgroup$
  – Jeffrey Rolland
  Mar 19 at 19:19
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

(I kind-of skipped ODEs in undergrad, so my knowledge is a little sketchy; please excuse me if this question is elementary.)

(In addition, this problem is a bit of a sketch; I'll flesh it out with calculations in a day or so.)

I've been looking over the theory of integrating factors, and they appear to focus on making the resulting differential form $muomega$ closed, not necessarily exact.

This appears to work fine most of the time. For instance, if I take the ODE

$$-y dx + x dy = g(t) dt$$

the left-hand side is neither closed nor exact. I recall reading in $textit{Advanced Calculus}$ by Buck https://smile.amazon.com/Advanced-Calculus-SECOND-Creighton-Buck/dp/B000OG0SRK/ that the differential equation for an integrating factor is called a "Pfaffian"; if that is the case and my back-of-the-envelope calculations from last night are correct [I'll flesh this out with calculations in a day or so], the Pfaffian is

$$xfrac{partial mu}{partial x} + yfrac{partial mu}{partial y} = -2mu$$

This appears to admit the solution integrating factor

$$mu(x,y) = frac{1}{xy}$$

and so the ODE has the implicit solution

$$left(frac{y}{x}right)^{xy} = Aexpleft(xyint frac{g}{xy} dtright)$$

(Maybe one would want to take $g(t) equiv 0$ ?)

The situation is different with the closed but not exact differential form (left-hand side of the equation)/ODE

$$frac{-y}{x^2+y^2} dx + frac{x}{x^2+y^2} dy = g(t) dt$$

The Pfaffian is [again, I'll flesh this out with calculations in a day or so]

$$frac{partial mu}{partial x}x + frac{partial mu}{partial y}y = 0$$

which doesn't appear to admit a solution, as nearly as I can figure.

Does this second ODE admit an integrating factor and I'm doing something wrong? More generally, if one has a closed but not exact differential form, is it possible to find an integrating factor that makes the differential form exact?

Thanks in advance.

ordinary-differential-equations differential-forms integrating-factor

share|cite|improve this question

edited Mar 19 at 3:11

Jeffrey Rolland

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These differential equations fall under the umbra of "implicit ODEs" (at least, they generally have implicit solutions). The variables x and y are themselves functions of t in the paradigm.
  $endgroup$
  – Jeffrey Rolland
  Mar 19 at 2:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that every closed form is locally exact ... Solving the differential equation is usually a local question; topology will dictate whether you can get a global integrating factor.
  $endgroup$
  – Ted Shifrin
  Mar 19 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin Mmmm, so, "in a flow box" (to use terminology from Abraham and Marsden smile.amazon.com/Foundations-Mechanics-Ams-Chelsea-Publishing/…), that is to say, in a rectangle with respect to t, x, and y in which all functions and their partial derivatives are continuous, the second ODE actually is exact, and we have an implicit solution; very nice, thank you.
  $endgroup$
  – Jeffrey Rolland
  Mar 19 at 19:19
  
  
  

  
  
  
  

add a comment | 

$begingroup$

(I kind-of skipped ODEs in undergrad, so my knowledge is a little sketchy; please excuse me if this question is elementary.)

(In addition, this problem is a bit of a sketch; I'll flesh it out with calculations in a day or so.)

I've been looking over the theory of integrating factors, and they appear to focus on making the resulting differential form $muomega$ closed, not necessarily exact.

This appears to work fine most of the time. For instance, if I take the ODE

$$-y dx + x dy = g(t) dt$$

the left-hand side is neither closed nor exact. I recall reading in $textit{Advanced Calculus}$ by Buck https://smile.amazon.com/Advanced-Calculus-SECOND-Creighton-Buck/dp/B000OG0SRK/ that the differential equation for an integrating factor is called a "Pfaffian"; if that is the case and my back-of-the-envelope calculations from last night are correct [I'll flesh this out with calculations in a day or so], the Pfaffian is

$$xfrac{partial mu}{partial x} + yfrac{partial mu}{partial y} = -2mu$$

This appears to admit the solution integrating factor

$$mu(x,y) = frac{1}{xy}$$

and so the ODE has the implicit solution

$$left(frac{y}{x}right)^{xy} = Aexpleft(xyint frac{g}{xy} dtright)$$

(Maybe one would want to take $g(t) equiv 0$ ?)

The situation is different with the closed but not exact differential form (left-hand side of the equation)/ODE

$$frac{-y}{x^2+y^2} dx + frac{x}{x^2+y^2} dy = g(t) dt$$

The Pfaffian is [again, I'll flesh this out with calculations in a day or so]

$$frac{partial mu}{partial x}x + frac{partial mu}{partial y}y = 0$$

which doesn't appear to admit a solution, as nearly as I can figure.

Does this second ODE admit an integrating factor and I'm doing something wrong? More generally, if one has a closed but not exact differential form, is it possible to find an integrating factor that makes the differential form exact?

Thanks in advance.

ordinary-differential-equations differential-forms integrating-factor

share|cite|improve this question

edited Mar 19 at 3:11

Jeffrey Rolland

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217

$endgroup$

share|cite|improve this question

edited Mar 19 at 3:11

Jeffrey Rolland

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217

share|cite|improve this question

edited Mar 19 at 3:11

Jeffrey Rolland

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217

asked Mar 18 at 20:49

Jeffrey RollandJeffrey Rolland

22217