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Elementary differential equations classes usually cover exact differential equations. These are equations of the form: $$M(x,y)+N(x,y)y'=0 \qquad \mathrm{such\;that} \qquad \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

We know that in many cases if one cooks up the right integrating factor $I(x,y)$, we can multiply through the equation and get a new equation $I(x,y)M(x,y)+I(x,y)N(x,y)y'=0$ which is exact (i.e. $\frac{\partial (IM)}{\partial y} = \frac{\partial (IN)}{\partial x}$) and has the same solutions.

Also, we know that finding $I$ (in general) is quite hopeless since this boils down to solving the first order partial differential equation: $\frac{\partial (IM)}{\partial y} = \frac{\partial (IN)}{\partial x}$ for $I$.

To find $I$ one usually makes some simplifying assumption (like $I(x,y)$ is a monomial or $I$ does not depend on $x$ or $y$). In such cases finding $I$ reduces to algebra or solving an easier first order ODE.

Of course, $I=0$, always makes the equation exact (but this introduces solutions which are not solutions to the original equation). Also, it's not too hard to come up with examples which have no integrating factors which are monomials or functions of $x$ or $y$ alone (the special assumptions don't always pan out).

My question: Is it true (under some assumptions -- like $M$ and $N$ are analytic or something) that there always exists some $I$ such that $IM+INy'=0$ is exact (and has the same set of solutions)?

If so, could you provide a reference or two? Preferably something better than "look at DeRham cohomology...blah blah...foliations." A fairly elementary reference accessible to someone without extensive differential topology background would be nice (if it exists).

If this is not true, are there reasonable assumptions one can make so that it is true?

I'm no DEs expert (I've taught a few introductory courses but this is outside my realm of expertise). By brother brought up this question a few weeks ago and it's been bothering me. He says he's seen the answer "Yes" claimed in some text (but that text was written for engineers so who knows whether the author meant "integrating factors always exist" or "integrating factors exist for the equations we care about").

Thanks in advance!

Edit: Thanks to Julian for his answer. Here's a more detailed version of what he posted...

Given a differential equation, $M(x,y)+N(x,y)y'=0$. Let's assume $M$ and $N$ have continuous first partials. In addition assume $M(x,y)$ and $N(x,y) \not=0$ [If $N(x,y)=0$, this is not a DE. If $M(x,y)=0$ we just have trivial constant solutions].

Then the equation is equivalent to $y'=-M(x,y)/N(x,y)$. Call on the fundamental existence/uniqueness theorem and get a solution $F(x,y,C)=0$. If $F_y=0$, then $y$ does not appear in $F$ (which is absurd) so $F_y \not=0$.

Fix a constant $C$, by the chain rule, since $F(x,y,C)=0$ we have $F_x(x,y,C)+F_y(x,y,C)y'=0$. Therefore, $y' = -F_x/F_y$ (recall $F_y \not=0$). But $y'=-M/N$. Therefore, $-F_x/F_y=-M/N$ and so $F_x/M = F_y/N$ (recall $M$ and $N$ are non-zero). Let $I(x,y) = F_x(x,y)/M(x,y) = F_y(x,y)/N(x,y)$ (this is our integrating factor).

Then $I(x,y)M(x,y) + I(x,y)N(x,y)y' = 0$ which is $\frac{F_x}{M}M+\frac{F_y}{N}Ny'=0$ and so $F_x(x,y)+F_y(x,y)y'=0$ (which is an exact equation).

Therefore, we can (in theory) always find (except in a trivial case) an integrating factor to make a non-exact first order ODE into an exact one.

Also, note: I have not been precise about where things are non-zero, but we don't really need these functions and partials to be non-zero everywhere just in the regions we care about.

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  • $\begingroup$ I suppose you could try scavenging through existence theorems for PDEs. $\endgroup$
    – anon
    Jan 23, 2012 at 15:47
  • $\begingroup$ I certainly thought this question was interesting :) $\endgroup$
    – James S. Cook
    Aug 19, 2012 at 0:23
  • $\begingroup$ I had this exact (no pun intended) same question. Here is my intuitive comment/thoughts which forces a 'yes': All DE's of any order, if they are valid DE's in the sense that they have a solution (just as I can randomly write down an algebraic equation which might or might not have a solution), have to spring from a $0$th order DE (which a $0$th order DE is just an equation - but not just any equation with discrete solutions like discrete $(x,y)$ points as solutions - but equations which define an implicit curve - because you can only take derivatives of functions). A 1st order DE, if its valid $\endgroup$
    – DWade64
    Oct 27, 2018 at 16:26
  • $\begingroup$ has to come from a $0$th order DE. The only way to get from a $0$th order DE $R(x,y) = c$ to a valid 1st order DE is to do $\frac{d}{dx}R(x, f(x)) = 0$. Which just turns into an exact differential equation. So since all valid 1st order DE's come from $0$th order DE's, and the only way to get from a 0th order to a 1st order is to take a derivative - you always form an exact equation $\endgroup$
    – DWade64
    Oct 27, 2018 at 16:29
  • $\begingroup$ Thank you for the edit - it was very useful and insightful to me. Getting from non-exact to exact, always with multiplication, was the question (and say not square rooting randomly $M$). I guess now my question would be why does multiplying an equation by a function not change the solution set? How can we be sure multiplying through by $F_x/M$ doesn't add solutions to the original problem? Is it because any valid differential equation has to be exact? $\endgroup$
    – DWade64
    Oct 27, 2018 at 16:48

1 Answer 1

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The answer is in fact yes (under usual conditions for existence and uniqueness). But to write an equation in exact form is a problem of the same difficulty as solving it; in fact, once in exact form it is already solved.

If the solution of the equation is in the form $F(x,y,C)=0$, solve for $C$ to obtain $G(x,y)=C$. Taking derivatives with respect to $x$ we get $$ \frac{\partial G}{\partial x}(x,y)+\frac{\partial G}{\partial y}(x,y)\,y'=0\ , $$ which is equivalent to the exact form $$ M(x,y)\,dx+N(x,y)\,dy=0 $$ with $$ M=\frac{\partial G}{\partial x},\qquad N(x,y)=\frac{\partial G}{\partial y}\ . $$

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  • $\begingroup$ This isn't what I am asking. I want to know whether there is an integrating factor $I$ which will turn the ODE into an exact equation. Given $M+Ny'=0$, multiply by $I$ and get $IM+INy'=0$. Can we guarantee that $IM=G_x$ and $IN=G_y$ for some $G$? So that $I=G_x/M=G_y/N$? Equivalently, can $G$ be found so that $G_x/M=G_y/N$? $\endgroup$
    – Bill Cook
    Jan 23, 2012 at 18:29
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    $\begingroup$ As I said, the answer is yes. Given the equation $M+N\,y'=0$, from the general solution you can find $G(x,y)$ such that $G_xdx+G_ydy=0$, or equivalently $y'=-G_x/G_y$. From the equation it folows that $-M/N=-G_x/G_y$. that is, $G_x/M=G_y/N$. $\endgroup$
    – Julián Aguirre
    Jan 23, 2012 at 23:05
  • $\begingroup$ Thanks for the forceful rebuttal - I was being too dense. Now I see. By the way, I don't believe you need to "solve for $C$". Dealing with $F$ directly should work. Fix a choice for $C$, then $F(x,y,C)=0$ implies $F_x+F_y y' =0$ (chain rule). So $I=F_y/N=F_x/M$ works as the integrating factor. I guess this uses assumptions: $F_y \not=0$, $M \not=0$, and $N \not=0$ (which isn't too surprising). THANKS!!! $\endgroup$
    – Bill Cook
    Jan 24, 2012 at 2:43

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