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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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    aah.. I see....2012-06-15

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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 $\mu = \frac{1}{x^{m+1}(M(1,u)+uN(1,u)}=\frac{1}{x\cdot x^m(M(1,u)+y\cdot x^m N(1,u)}=\frac{1}{xM(x,y)+yN(x,y)}$

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    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.2012-06-15