# Inexact Differential Equations: how to know if μ will be a function of x or y instead of guessing?

Here is an inexact differential equation: $$(3xy+y^2)+(x^2+xy)y'=0$$ The integrating factor is $μ=x$.

My question is: instead of guessing that $μ$ will be a function of one of the variables and then manually working it out on that assumption, is there a quick check method or rule of thumb for knowing whether the integrating factor will end up being a function of $x$ or $y$?

For instance, If I had guessed that for this equation that $μ$ was going to be a function of $y$ I would've wasted my time. How can I avoid this?

Additionally, is there a situation where an integrating factor for a differential equation MUST be a factor of $x$ and $y$?

• AFAIK there's no universal method. Finding an integrating factor amounts to solving a PDE which is generally harder than solving the original problem. – Giuseppe Negro Jun 17 '14 at 11:04

I would like to address your last question: "Is there a situation where an integrating factor must be a function of x and y?"

Consider Sophus Lie's integrating factor which, because it is a function of M and N, will almost always be a function of x and y (assuming that M and/or N is a function of x and y). The only trick is finding an invariant infinite continuous group transformation (a Lie group, one that preserves the structure of the smooth manifold) but given your DEQ this isn't difficult. Use X=$\lambda$x and Y=$\lambda^\beta$y. This is a family of groups where the unitary transformation requires $\lambda=1$, and you are looking for a specific group whose image does not alter your DEQ. It might move the points on the solutions curves around, but the solutions themselves do not change. By using Newtonian notation and noting that $\dot{y}=\frac{dy'}{dx'}=\frac{\lambda^\beta dy}{\lambda dx}$ you have $$3\lambda x\lambda^\beta y+\lambda^{2\beta}y^2+(\lambda^2 x^2+\lambda x\lambda^\beta y)\lambda^{\beta -1}\dot{y}=0$$ For invariance the $\lambda$ terms must cancel out. This requires $\beta=1$. Lie's integrating factor is $$\mu=\bigg[\bigg(\frac{\partial{X}}{\partial\lambda}\bigg)_{\lambda=1}M+\bigg(\frac{\partial{Y}}{\partial\lambda}\bigg)_{\lambda=1}N\bigg]^{-1}=(xM+\beta yN)^{-1}=(4x^2y^2+2xy^2)^{-1}$$ Multiplying by $\mu$ turns your DEQ into a perfect differential. I tried this on your equation and ran into some fairly nasty integrations, so I decided to take an easier path.

A much simpler approach is to take advantage of Lie's theorem 4.4.1 from his Differential Invariant paper of 1884 which states that any DEQ invariant to a Lie group can be rewritten as a function of the stabilizers of that group. Since any function of group stabilizers is also a group stabilizer, the DEQ itself becomes a group stabilizer and may then be solved by using the Lie algebra between the stabilizers which is also in the kernel of the map. The stabilizers for the family of groups we are using is $$S_1=\frac{y}{x^\beta}=\frac{y}{x}$$ and $$S_2=\frac{\dot{y}}{x^{\beta -1}}=\dot{y}$$ There are an infinite number of these stabilizers, but we only need the first two. The rest follow the pattern $S_3=\frac{\ddot{y}}{x^{\beta -2}}$, $S_4=\frac{\dddot{y}}{x^{\beta-3}}$, etc. (To see that these are group stabilizers just apply the Lie group we are using to them and you will notice that they don't change. Please note, this isn't a commonly known use for stabilizers, but Sophus Lie was a genius and over a hundred years after his death many of his ideas remain relatively untouched. This is just one of them.) If we divide the entire DEQ by $x^2$, we get $$3\frac{y}{x}+\bigg(\frac{y}{x}\bigg)^2+\dot{y}+\frac{y}{x}\dot{y}=0$$ which becomes $$3S_1+S_1^2+S_2+S_1S_2=0$$ This works because the stabilizers form an embedded subspace in the solution manifold. At singularities, saddle points and along separatrices in the direction field of the Lie plane of stabilizers, the Lie algebra between the stabilizers is particularly simple: $S_2=\beta S_1$, $S_3=\beta(\beta -1)S_1$, $S_4=\beta(\beta-1)(\beta-2)S_1$, and so on. For $\beta=1$ your DEQ becomes $$3S_1+S_1^2+S_1+S_1^2=0$$ and discarding the trivial solution $y=0$, $$S_1=-2=\frac{y}{x}$$ Therefore a special solution to your DEQ is $y=-2x$. This is easy to verify.
To find a more general solution, assume that $y_2=y(x)v(x)=-2xv$. Differentiating and plugging into your equation leads to the equation $$\frac{4}{x}\bigg(\frac{v^2-v}{1-2v}\bigg)=\frac{dv}{dx}$$ This is a fairly easy integral to solve. Again discarding the trivial solution where $v(c=0)=0$
$$v=\frac{1+\sqrt{1+\frac{4C}{x^4}}}{2}$$ This results in a general solution to your DEQ (which is somewhat labor-intensive to check through substitution). $$y_2=-2xv=-x-\sqrt{x^2+\frac{4c}{x^2}}$$ Note that if C=0 the general solution collapses to the special solution, as expected.

• interesting answer, do you happen to have a link to the paper of Lie you mention? Also, how do you identify the "family of groups" for the given ODE? – James S. Cook Jun 21 '14 at 14:18
• Sophus Lie's Differential Invariant Paper of 1884 can be difficult to find, and I don't know any link to it. I can send you a paper of my own aimed at a college sophomore level, or I can refer you to my old professor, Dr. Lawrence Dresner. His book, Similarity Solutions of Nonlinear Partial Differential Equations, deals more with PDE's, but it's a good source. Peter Olver's Application of Lie Groups to Differential Equations is an excellent source. If you want my basic paper on the subject, send me an email at michael.pugh@chattanoogastate.edu – atomteori Jun 21 '14 at 14:28
• As for the "family of groups," that is the big problem. I have investigated a variety of simple groups, but finding the right group is currently a hit-or-miss proposition. Peter Olver's book outlines a way to find an invariant Lie group for any given DEQ, but it's a bit over my head and I'm currently working to understand it. – atomteori Jun 21 '14 at 14:33
1. As it is said in the comment there is generally no universal method to find an integrating factor.

2. To determine whether the integrating factor is a function of $x$ or $y$ only is quite straightforward. Consider the equation $$M(x,y)dx+N(x,y)dy=0.$$ If the expression $$(M'_y-N'_x)\over N$$ depends on $x$ only then $\mu=\mu(x)$. If the expression $$(N_x'-M_y')\over M$$ depends on $y$ only then $\mu=\mu(y)$.

3. Problem 1: Prove the formulas above.
4. Problem 2: (which answers your other question) Show that if $$N_x-M_y\over xM-yN$$is a function of $xy$ only, then the differential equation has an integrating factor and find it.