# Integrating Factor: How to solve it

Verify that: $$\frac12(Mx+Ny)d(\ln(xy))+\frac12(Mx-Ny)d(\ln(x/y))=Mdx+Ndy$$

Hence show that, if the de $Mdx+Ndy=0$ is homogenous, then $Mx+Ny$ is an integrating factor unless $Mx+Ny=0$

Note: Verification is trivial, hence nothing much to be done there, but I couldnt solve the second part of the question "Hence..." so for the completeness of the problem I added it. Further on, isnt the statement " $Mdx+Ndy=0$ is homogenous " superfluous as RHS is already zero, so why add the word homogenous. Perhaps I am being pedantic? And lastly I would like to have some hints in solving the INTEGRATING Factor part.

EDIT: My approach I approached like this: I multiplied the function $Mx+Ny$ to both sides of the equation $Mdx+Ndy=0$ and tried to show, that $d(u(x,y))=0$ but I couldnt prove it.

Soham

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I think that in this context "homogeneous" might mean that $M$ and $N$ are homogeneous polynomials in $x$ and $y$, and of the same degree. E.g., $$(2x^2+3xy+4y^2)dx+(5x^2-6xy-7y^2)dy=0$$ –  Gerry Myerson Jun 14 '12 at 23:53
aah.. I see.... –  Soham Jun 15 '12 at 5:16

The statement of the question is not entirely correct. In fact the integrating factor for equation $$Mdx+Ndy=0$$ where $M$ and $N$ are homogeneous functions of both $x$ and $y$ (i.e. $M(x,y)=x^m M(1,\frac{y}{x})$, see Gerry Myerson's comment for an example) will be $$\mu = \frac{1}{Mx+Ny}$$ in order to ascertain this divide both sides of your equality by $Mx+Ny$: $$\frac12d(\ln(xy))+\frac12\frac{Mx-Ny}{Mx+Ny}d(\ln(x/y))=\frac{Mdx+Ndy}{Mx+Ny}$$ $$\frac12d(\ln(xy))+\frac12\frac{M(x,y)\frac{x}{y}-N(x,y)}{M(x,y)\frac{x}{y}+N(x,y)}d(\ln(x/y))=\frac{Mdx+Ndy}{Mx+Ny}$$ Using homogeneity: $$\frac12d(\ln(xy))+\frac12\frac{M(\frac{x}{y},1)\frac{x}{y}-N(\frac{x}{y},1)}{M(\frac{x}{y},1)\frac{x}{y}+N(\frac{x}{y},1)}d(\ln(x/y))=\frac{Mdx+Ndy}{Mx+Ny}$$ On the LHS variables are separated, so it is effectively an exact differential (you can let $\frac{x}{y}=e^t$ to complete the form. Therefore, $\mu$ as given above is an integrating factor.
Constructive proof goes in a somewhat similar way. Let $u=\frac{x}{y}$. Then again, using homogeneity (assuming $M$ and $N$ are homogeneous of the order $m$): $$M(x,y)=M(x,ux)=x^m M(1,u)$$ Similarly $$N(x,y)=x^m N(1,u)$$ Now $$dy=udx+xdu$$ Inserting in the original equation we obtain $$x^m (M(1,u)+uN(1,u))dx+x^{m+1}N(1,u)du$$In order to separate variables we must divide both sides by $x^{m+1}(M(1,u)+uN(1,u)$. However $$\mu = \frac{1}{x^{m+1}(M(1,u)+uN(1,u)}=\frac{1}{x\cdot x^m(M(1,u)+y\cdot x^m N(1,u)}=\frac{1}{xM(x,y)+yN(x,y)}$$
Couple of questions: 1. In the last step how did you arrive at the value of $\mu$ 2. Can you talk a bit more how the LHS is separated? 3. Does an exact de always need an equation in the form of $Idx+Jdy=0$ ? [To be frank, I am stunned at this question. I could easily solve the questions on integrating factor from my text, but when given this sum I couldnt. Not only that, I didnt get a clue how to go about it] –  Soham Jun 15 '12 at 6:02
Sure. 1) effectively, you have here an equation of the form: $$P(x)Q(u)dx+R(x)S(u)du=0$$ which is obviously separated: $$\frac{P(x)}{R(x)}dx+\frac{S(u)}{R(u)}du=0$$ Note thatt solving any separable equation amounts to multiplying it by an integrating factor. 2. LHS can be cast in the form $$d(A(xy))+B(\ln(x/y))d(ln(x/y))$$ which is an exact differential.3. This particular IF is specific to this form of equation. Personally, I never use it and treat IF problems on a case by case basis. –  Valentin Jun 15 '12 at 9:51