2
$\begingroup$

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.

3 Answers 3

1

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?

2

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.

  • 0
    @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.2011-04-28
1

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

  • 0
    I don't even know the degree of homogeneity of $f$, how can I even attempt to solve it?2011-04-27