# Proving a property of homogeneous equation that is exact

The following question was given to us in an exam:

If $0=M dx + N dy$ is an exact equation, in addition to the fact that $\frac{M}{N} = f\Big(\frac{y}{x}\Big)$ is homogeneous, then

$xM_x + yM_y = (xN_x + yN_y)f$.

Now I had absolutely no idea how to prove this question. I tried doing $M = Nf$ and taking derivatives and multiplying by $x$ or $y$, and you get the required R.H.S. but with the extra term $N(\frac{-f_x}{x} + \frac{f_y}{x})$ added. How does one approach a question like that??

I have never encountered a question like that, not even when solving for different types of integrating factors to get an exact equation or when working with a homogeneous equation.

Anyone got any ideas? Please don't post a complete solution.

Thanks.

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You had the right idea. If $M(x,y) = N(x,y) F(x,y)$, you have $x M_x + y M_y = x (N F)_x + y (N F)_y = (x N_x + y N_y) F + N (x F_x + y F_y)$. Now if $F(x,y) = f(y/x)$, you have $F_x = - f'(y/x) \frac{y}{x^2}$ and $F_y = \frac{f'(y/x)}{x}$, so $x F_x + y F_y = 0$. Note that the statement that the equation is exact was a red herring.

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No need to suppose $f$ homogeneous? – lhf Apr 27 '11 at 18:33
@lhf Hi this was given to us in a differential equations exam, no one could solve this question! Oh and could you explain to me why when you do $F_x$, or $F_y$, you get just $f'(y/x)$ and not the partial derivative of this with respect to either $y$ or $x$? I think that if I had got that, I would have been able to solve the question because when I did it not of the terms cancelled off. – user38268 Apr 27 '11 at 22:11
@lhf: "homogeneous" is the $F(x,y) = f(y/x)$ part. @D Lim: This is the chain rule. $\frac{\partial}{\partial x} f(g(x,y)) = f'(g(x,y)) g_x(x,y)$, etc. – Robert Israel Apr 28 '11 at 7:34

I guess you need to use Euler's theorem on homogeneous functions.

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Still don't get how I can use that for the problem. – user38268 Apr 27 '11 at 15:04
I don't even know the degree of homogeneity of $f$, how can I even attempt to solve it? – user38268 Apr 27 '11 at 15:09

So the solution should be:

As $\frac{M}{N} = f\Big(\frac{y}{x}\Big)$, this means that the degree of homogeneity of $M$ and $N$ must be equal.

So $xM_x + yM_y = aM$, and $xN_x + yN_y = aN$, by euler's homogeneity theorem where $a$ is the degree of homogeneity of $M$ and $N$.

So dividing the first by the second of these equations, and one should get

$xM_x + yM_y = \frac{M}{N} (xN_x + yN_y) = f\Big(\frac{y}{x}\Big)(xN_x + yN_y )$.

Is that correct?

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