1

Suppose we have the following problem: $$(y-xy^2)dx+(x+x^2y^2)dy=0 \label{1}\tag{$*$}$$ If we try to find an integrating factor $\mu$ of a single variable (I.e., either $\mu(x)$ or $\mu(y)$) we will have a contradiction. Meaning, the integrating factor is a function of two variables, namely, $\mu(x,y)$.

There has been a lot of theory finding it in a "general case". The majority of the techniques are quite advanced and/or quite complicated. However, searching the internet, I came across the method of integrating factor, described for example in this Q&A.

But I want to understand "why" these sequence of steps to find $\mu(x,y)$ works.

For instance, at the start, we let $M=0$ so that we can find $y_0(x)$ then we let $N=0$ so that we can find $x_0(y)$. My first question is: why is the subscript "0" used? More importantly, why do we let $M=0$ and $N=0$?

Secondly, they find $f(x)$ and $g(y)$ using these formulae, that I can derive them; however, this depends on eliminating $x$ or $y$ via the substitution $y_0(x)$ or $x_0(y)$, but again don't know why it works.

Lastly, the relation that they satisfy: $$ \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = N\cdot f(x,y)-M\cdot g(x,y)\label{2}\tag{$***$} $$

Why there is not a "2" in front of the left-hand side?

I.e., we know: $$ f(x) = \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N} $$ and $$ g(y) = -\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M} $$

Consequently, $$N\cdot f(x) = \frac{\partial M}{\partial y} -\frac{\partial N}{\partial x}$$

and $$-M\cdot g(y) = \frac{\partial M}{\partial y} -\frac{\partial N}{\partial x} $$

Adding both equations gives: $$ 2\frac{\partial M}{\partial y} -2\frac{\partial N}{\partial x} = N\dot f(x)-M\cdot g(y)$$

But in \eqref{2}, there is no "2" on the left-hand side?.

My last question is, where did: $\mu(x,y) = e^{\int f(x) dx + \int g(y) dy}$ come from? I.e., how did he go from the above relation to finding the integrating factor $\mu(x,y)$?

An example where it does not work. $$y dx +(x+3x^3y^4)dy = 0 $$

Here are my steps (following the method): let $$ \begin{split} M = 0 & \implies y_0(x)=0\\ N = 0 & \implies x_0(y)=\sqrt{\frac{1}{-3y^4}} \end{split} $$ Then $$ f(x) = \frac{-9x^2y^4}{x+3x^3y^4} = \frac{0}{x}=0 $$ as $y(x)=0$, and $$ g(y) =\frac{-3}{y} $$ Hence, $$ \mu = e^{{\int 0dx + \int\frac{-3}{y}dy}} =\frac{1}{y^3}$$

Which is not the correct integrating factor, according to the book, and I had tried it and did not work. Sorry for the long post and thank you in advance.

Daniele Tampieri
  • 9,727
Nero
  • 395
  • Note: the link of the paper the answerer placed does not direct to anything related to the method. – Nero Sep 12 '23 at 06:14
  • The solution is $\frac{x y^2-x y \ln (x)-1}{x y}=C$ – Cesareo Sep 12 '23 at 08:04
  • Hi Cesareo, yes this is the correct solution, thanks. But my concern was why the method for finding the integrating factor of 2 variables work. – Nero Sep 12 '23 at 08:21
  • Your many questions "why" that method works are too unfocussed. Admire the tricks by Kwame Asante and practice them by applying them to other examples. Check that the solutions found solve the ODE. Sometimes math is an art. – Kurt G. Sep 12 '23 at 09:39
  • It works because it works, change a coefficient or an exponent, or even worse, add some term, and it stops working. Textbook examples are (usually) designed to have a solution, which imposes structure on the equation, which can lead to multiple "tricks" being successful. – Lutz Lehmann Sep 12 '23 at 09:43
  • @Kurt G., yes, I already practiced them by applying them to 3 examples. I agree, indeed a substantial part of mathematics is an "art". However, I believe it is crucial to have, if not a proof, an intuition about why we are doing a particular "step". That is, I am not asking for a rigorous proof nor a rough proof; just the motivation behind the step. Excuse my lack of knowledge, but I do not think the "method" used in Kwame asante's answer is a "trick", I think it is somewhat more general, which is exactly what I am asking if it is. I don't get the part about "Unfocussed". I wrote all details. – Nero Sep 12 '23 at 10:13
  • @Lutz Lehmann, yeah I get your point. But a group of problems having the same "trick" is no more called a "trick", rather a "Technique" or "Method" for that "form of problems". For instance, some DEs are solved using exact equations "idea"; hence we have found the theory behind it that it works only if mixed partials are equal, and later came the trick of integrating factors, but they solved a huge class of problems, hence the textbooks do not call it anymore a "trick", rather it has a whole chapter in introductory DE books about it. And the above method does not seem to be "specific" either. – Nero Sep 12 '23 at 10:17
  • At least it is not so clear. Maybe I am wrong; thats why I am asking.. – Nero Sep 12 '23 at 10:17
  • For instance, I just tried it on a similar example with M = y only, and the method does not work, although the question looks identical.. At least one should know when to use the trick? Which is not answered? – Nero Sep 12 '23 at 10:28
  • Focus, focus, focus. Ditch the examples where the method works (we believe that). Edit your question and show the detailed steps of an example where you believe it does not work. – Kurt G. Sep 12 '23 at 10:42
  • Ok noted. Will add these, and won't add the examples in the future. – Nero Sep 12 '23 at 10:47
  • This script is discussing a few cases where the "tricks" work and where they don't. It seems like even for some innocently looking inexact ODEs it can be a tremendous task to find the integrating factor esp. when it depends on $x$ and $y,.$ – Kurt G. Sep 12 '23 at 14:30
  • Awesome, this pretty much summarizes all the main cases, and the one I was looking for seems to not have a general way. Thank you very much @KurtG. Since it is a comment, I can not make this an "answer". – Nero Sep 12 '23 at 15:46

2 Answers2

2

You might render the differential equation thusly: $$ y(1-xy)dx+[x+(xy)^2]dy =0, $$ suggesting a possible substitution $u=xy$. Recognizing that $(x)dy+(y)dx=du$, we have as an intermediate step $$ du-(yu)dx+u^2dy=0, $$ suggesting we should further eliminate $dx$ getting an equation with $y$ and $u$ as variables. Thus $$ dx=d(u/y)=\dfrac{du}{y}-\dfrac{(u)dy}{y^2} $$ and after substituting and combining terms with the same differential: $$ \begin{split} du-(yu)\left(\dfrac{du}{y}-\dfrac{(u)dy}{y^2}\right)+u^2dy &= 0,\\ (1-u)du+[u^2(1+y)/y]dy &= 0. \end{split} $$ Now the variables are completely separared by dividing by $u^2$ (thus the original integrating factor turned out to be $1/(x^2y^2)$) and the rest is routine integration.

Daniele Tampieri
  • 9,727
Oscar Lanzi
  • 39,403
  • That is an amazing way to solve it! But I think it is specific to this "problem" and the substitution will not always be obvious like u=xy; whereas, the method used by Kwame's seems to work in most cases except few (just added one, which I don't understand why it does not work).

    All that I am trying to say is, when to use Kwame's method and when not and so on; as it seems to be a general result, which I am not quite sure of. But about solving this DE, there are multiple ways, but the way you solved it, is one of the best I saw :); thanks!

     – Nero Sep 12 '23 at 11:10
2

To find an integrating factor $\mu$ for $$ y\,dx +(x+3x^3y^4)\,dy = 0 $$ we expand the exactness condition $$ \partial_y(\mu y)=\partial_x(\mu x+3\mu x^3y^4) $$ to $$ (\partial_y\mu)\,y+\mu =(\partial_x\mu)\,(x+3x^3y^4)+\mu+9\mu x^2y^4\,. $$ This is a linear first order PDE for $\mu(x,y)\,:$ $$ 0=-(\partial_y\mu)\,y+(\partial_x\mu)\,(x+3x^3y^4)+9\mu x^2y^4 $$ The solutions could hardly be uglier. I refrain from transcribing the output from Wolfram Alpha.

A better method for this type of ODE is it to step into Oscar Lanzi's footsteps and try the ansatz $$ \mu(x,y)=x^\alpha\, y^\beta\,. $$ This leads to \begin{align} 0&=-x^\alpha \beta\,y^\beta+\alpha\,x^{\alpha-1}y^\beta(x+3x^3y^4)+9\,x^{\alpha+2} y^{\beta+4}\\[2mm] &=-\beta\,x^\alpha y^\beta+\alpha\,x^\alpha y^\beta+3\,\alpha\,x^{\alpha+2}y^{\beta+4}+9\,x^{\alpha+2} y^{\beta+4}\,. \end{align} It is quickly seen that $$\alpha=\beta=-3$$ solves this equation leading to a usable integrating factor.

Kurt G.
  • 14,198
  • You asked for the general solution, so WA went big; but you needed only a specific factor and using the method in my answer we easily find $1/(xy)$. So might you consider a way to pluck a simple particular solution from the partial differential equation? – Oscar Lanzi Sep 13 '23 at 15:54
  • @OscarLanzi Thanks for joining the discussion. Trying to see if $\mu=1/(xy)$ is what we want. Seems not?. – Kurt G. Sep 13 '23 at 16:22
  • Upon furthervreview, I shoukd have said $1/[(xy)^3]$. Can we try again? – Oscar Lanzi Sep 13 '23 at 20:07
  • Bang on!. Awesome. – Kurt G. Sep 14 '23 at 01:32
  • Well, if we want to try $\mu(x,y) =x^{\alpha}y^{\beta}$ then the following method immediately tells us so (and finds the factor): $$\frac{1}{\mu(z)}\frac{d\mu}{dz}=\frac{M_y-N_x}{yN-xM}=g(z)$$ In words, if the RHS is a function of z, where z=xy, then we can solve the above DE to get the integrating factor, as you have found for this particular question. But, my question was regarding whether there was a general method to find the integrating factor without trying it is "xy" or powers of it etc.. But, as you wrote in the answer, it is a 1st ord. PDE, which can't be solved "easily". Thanks. – Nero Sep 14 '23 at 09:22