Integrating Factor for Closed but not Exact Differential Form The Next CEO of Stack Overflowfinding 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?
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Integrating Factor for Closed but not Exact Differential Form
The Next CEO of Stack Overflowfinding 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?
$begingroup$
(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 $textitAdvanced Calculus$ 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
$$xfracpartial mupartial x + yfracpartial mupartial y = -2mu$$
This appears to admit the solution integrating factor
$$mu(x,y) = frac1xy$$
and so the ODE has the implicit solution
$$left(fracyxright)^xy = Aexpleft(xyint fracgxy dtright)$$
(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
$$frac-yx^2+y^2 dx + fracxx^2+y^2 dy = g(t) dt$$
The Pfaffian is [again, I'll flesh this out with calculations in a day or so]
$$fracpartial mupartial xx + fracpartial mupartial yy = 0$$
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
asked Mar 18 at 20:49
Jeffrey RollandJeffrey Rolland
22217
$endgroup$
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 $textitAdvanced Calculus$ 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
$$xfracpartial mupartial x + yfracpartial mupartial y = -2mu$$
This appears to admit the solution integrating factor
$$mu(x,y) = frac1xy$$
and so the ODE has the implicit solution
$$left(fracyxright)^xy = Aexpleft(xyint fracgxy dtright)$$
(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
$$frac-yx^2+y^2 dx + fracxx^2+y^2 dy = g(t) dt$$
The Pfaffian is [again, I'll flesh this out with calculations in a day or so]
$$fracpartial mupartial xx + fracpartial mupartial yy = 0$$
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
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 $textitAdvanced Calculus$ 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
$$xfracpartial mupartial x + yfracpartial mupartial y = -2mu$$
This appears to admit the solution integrating factor
$$mu(x,y) = frac1xy$$
and so the ODE has the implicit solution
$$left(fracyxright)^xy = Aexpleft(xyint fracgxy dtright)$$
(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
$$frac-yx^2+y^2 dx + fracxx^2+y^2 dy = g(t) dt$$
The Pfaffian is [again, I'll flesh this out with calculations in a day or so]
$$fracpartial mupartial xx + fracpartial mupartial yy = 0$$
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
asked Mar 18 at 20:49
Jeffrey RollandJeffrey Rolland
22217
$endgroup$
(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 $textitAdvanced Calculus$ 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
$$xfracpartial mupartial x + yfracpartial mupartial y = -2mu$$
This appears to admit the solution integrating factor
$$mu(x,y) = frac1xy$$
and so the ODE has the implicit solution
$$left(fracyxright)^xy = Aexpleft(xyint fracgxy dtright)$$
(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
$$frac-yx^2+y^2 dx + fracxx^2+y^2 dy = g(t) dt$$
The Pfaffian is [again, I'll flesh this out with calculations in a day or so]
$$fracpartial mupartial xx + fracpartial mupartial yy = 0$$
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
ordinary-differential-equations differential-forms integrating-factor
asked Mar 18 at 20:49
Jeffrey RollandJeffrey Rolland
22217
asked Mar 18 at 20:49
Jeffrey RollandJeffrey Rolland
22217
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
22217
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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 |
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$begingroup$
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