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Integrating Factor Techniques for Exact ODE

Is there any theorem on uniqueness of integrating factor for inexact ordinary differential equations?Does nudging an exact differential equation nudge or destroy the identity integrating factor?Finding integrating factor (IF) when integrating factor IF will be a function of both $x$ and $y$finding an integrating factor for the nonlinear non-autonomous ODE $ (xy)y'+yln y - 2xy = 0 $Integration Factor in terms of $x$ and $y$ used to reduce an ODE to an exact formA line integral equation popped up when trying to derive Exact ODE integrating factor, can it be solved analytically?Solving a Non-Exact First-Order ODEAlgorithm to find an integrating factor in a first order ODE?Why do Integrating Factors Work?Exact Differential Equation Integrating FactorExact Differential Equation GeometryIntegrating Factor for Closed but not Exact Differential Form

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As explained in this answer, an inexact ODE in the form $M(x, y) dx + N(x, y) dy = 0$ can be transformed into an exact ODE via multiple integrating factors which vary from each other non-trivially (by more than a factor of a constant). This implies that an exact ODE can be transformed into a non-trivially different exact ODE by some integrating factor, even though there would be less motivation to do so.

The usual method of finding integrating factors of only $x$ or only $y$ always returns $1$ when it eats an exact ODE, so my guess is that such a non-trivial transformation can only be a function of both $x$ and $y$ explicitly. One example of this might be multiplying an exact ODE by an integrating factor in $x$ and $y$ which separates the variables, which is guaranteed to output another exact ODE. However, this technique can just as easily be applied to inexact ODEs if the variables are separable.

What I want to know is if there are any new techniques that become available to find a non-trivial integrating factor when the given ODE is already exact. By analogy, there may be some "difficulty (at least if you don't have a calculator or patience)" in finding a particular linear combination of nickels and quarters that total $$2,547,042,997.85$, but once you find one, the rest follow effortlessly via the nullspace $n*beginbmatrix1 quarter \ -5 nickelsendbmatrix$. I'm wondering if finding one exact form of an ODE can aid in finding others (aside from the trivial case of multiplying the integrating factor by a constant).

ordinary-differential-equations algorithms vector-fields differential

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asked Mar 20 at 20:02

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0

$begingroup$

As explained in this answer, an inexact ODE in the form $M(x, y) dx + N(x, y) dy = 0$ can be transformed into an exact ODE via multiple integrating factors which vary from each other non-trivially (by more than a factor of a constant). This implies that an exact ODE can be transformed into a non-trivially different exact ODE by some integrating factor, even though there would be less motivation to do so.

The usual method of finding integrating factors of only $x$ or only $y$ always returns $1$ when it eats an exact ODE, so my guess is that such a non-trivial transformation can only be a function of both $x$ and $y$ explicitly. One example of this might be multiplying an exact ODE by an integrating factor in $x$ and $y$ which separates the variables, which is guaranteed to output another exact ODE. However, this technique can just as easily be applied to inexact ODEs if the variables are separable.

What I want to know is if there are any new techniques that become available to find a non-trivial integrating factor when the given ODE is already exact. By analogy, there may be some "difficulty (at least if you don't have a calculator or patience)" in finding a particular linear combination of nickels and quarters that total $$2,547,042,997.85$, but once you find one, the rest follow effortlessly via the nullspace $n*beginbmatrix1 quarter \ -5 nickelsendbmatrix$. I'm wondering if finding one exact form of an ODE can aid in finding others (aside from the trivial case of multiplying the integrating factor by a constant).

ordinary-differential-equations algorithms vector-fields differential

share|cite|improve this question

asked Mar 20 at 20:02

user10478user10478

454413

$endgroup$

add a comment | 

$begingroup$

As explained in this answer, an inexact ODE in the form $M(x, y) dx + N(x, y) dy = 0$ can be transformed into an exact ODE via multiple integrating factors which vary from each other non-trivially (by more than a factor of a constant). This implies that an exact ODE can be transformed into a non-trivially different exact ODE by some integrating factor, even though there would be less motivation to do so.

The usual method of finding integrating factors of only $x$ or only $y$ always returns $1$ when it eats an exact ODE, so my guess is that such a non-trivial transformation can only be a function of both $x$ and $y$ explicitly. One example of this might be multiplying an exact ODE by an integrating factor in $x$ and $y$ which separates the variables, which is guaranteed to output another exact ODE. However, this technique can just as easily be applied to inexact ODEs if the variables are separable.

What I want to know is if there are any new techniques that become available to find a non-trivial integrating factor when the given ODE is already exact. By analogy, there may be some "difficulty (at least if you don't have a calculator or patience)" in finding a particular linear combination of nickels and quarters that total $$2,547,042,997.85$, but once you find one, the rest follow effortlessly via the nullspace $n*beginbmatrix1 quarter \ -5 nickelsendbmatrix$. I'm wondering if finding one exact form of an ODE can aid in finding others (aside from the trivial case of multiplying the integrating factor by a constant).

ordinary-differential-equations algorithms vector-fields differential

share|cite|improve this question

asked Mar 20 at 20:02

user10478user10478

454413

$endgroup$

share|cite|improve this question

asked Mar 20 at 20:02

user10478user10478

454413

share|cite|improve this question

asked Mar 20 at 20:02

user10478user10478

454413

asked Mar 20 at 20:02

user10478user10478

454413

asked Mar 20 at 20:02

user10478user10478

454413