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How do I go about finding the integrating factor for the equation:

$$3x + \frac{6}{y} + (\frac{x^2}{y} + \frac{3y}{x})\frac{dy}{dx}=0$$

can I find it using the following method?

$\frac{N_x - M_y}{xM-yN}$ ? if yes, then I didnt manage to find the correct one, the solution says the integrating factor should by xy ?

any tips/advice solutions on how I can find the integrating factor to this problem? thanks in advance!

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The general form for your differential equation is $$M(x,y)dx+N(x,y)dy=0$$ Thus we identify: $$M=3x+\dfrac{6}{y},\quad N=\dfrac{x^2}{y}+\dfrac{3y}{x}$$ We then compute the derivatives: $$M_x=3,\quad M_y=-\dfrac{6}{y^2},\quad N_x=\dfrac{2x}{y}-\dfrac{3y}{x^2},\quad N_y=-\dfrac{x^2}{y^2}+\dfrac{3}{x}$$ The first thing to check, for exactness, is whether $M_y=N_x$. Clearly this isn't the case.

Next, we look at the following expression and check if it is a function of $x$ only: $$\dfrac{M_y-N_x}{N}=\dfrac{-6x^2-2x^3y+3y^3}{x^2y^2}\cdot\dfrac{xy}{x^3+3y^2}\neq f(x)$$ Therefore the integrating factor is not a function only of $x$. We check similarly for $y$: $$\dfrac{N_x-M_y}{M}=\dfrac{6x^2+2x^3y-3y^3}{x^2y^2}\cdot\dfrac{y}{3xy+6}\neq f(y)$$ Therefore the integrating factor is not a function only of $y$. We finally check for a function of both $x$ and $y$: $$\dfrac{N_x-M_y}{xM-yN}=\dfrac{6x^2+2x^3y-3y^3}{x^2y^2}\cdot\dfrac{xy}{2x^3y+6x^2-3y^3}=\dfrac{1}{xy}=f(x,y)$$ Therefore, the integrating factor will be: $$\mu(x,y)=\exp\left(\int \dfrac{dx}{x}\right)\exp\left(\int\dfrac{dy}{y}\right)=xy$$

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  • $\begingroup$ wow thanks! I managed to get the first expression: $\endgroup$ – yeyyeat May 18 '15 at 21:00
  • $\begingroup$ (6x^2 +2x^3y-3y^3)/(x^2*y^2), but why do you multiply it with the other expression? $\endgroup$ – yeyyeat May 18 '15 at 21:00
  • $\begingroup$ @yeyyeat You mean, why do I have two fractions multiplied? The first one is the numerator, brought together as one fraction, and the second one is the inverse of the denominator. $\endgroup$ – Demosthene May 18 '15 at 21:02
  • $\begingroup$ oh ok I see the first one is N_x-M_y and the second one is xM-yN inverse right? $\endgroup$ – yeyyeat May 18 '15 at 21:04
  • $\begingroup$ Yes, exactly :) $\endgroup$ – Demosthene May 18 '15 at 21:05
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The first step in solving such a differential equation is checking if it is exact.

To do so, first write the differential equation in standard form, that is

$$Mdx + Ndy = 0$$

Then check if $N_x = M_y$

If this is not the case, you must check with of

$$\frac{M_y - N_x}{-M}$$ $$\frac{M_y - N_x}{N}$$

will give you a function of x or y alone.

The one you pick will be the integrating and I trust you know how to do the rest.

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  • $\begingroup$ didnt get a function of x or y alone on any of the two formulas..? $\endgroup$ – yeyyeat May 18 '15 at 18:50
  • $\begingroup$ Working it out, give me a few. $\endgroup$ – Kevin Zakka May 18 '15 at 18:59
  • $\begingroup$ It's not working out for me, can you double check your initial differential equation? $\endgroup$ – Kevin Zakka May 18 '15 at 19:08
  • $\begingroup$ its not the correct one, the first term should be 3x + 6/y, but thats the one I have been using and it didnt work out for me.. $\endgroup$ – yeyyeat May 18 '15 at 19:11
  • $\begingroup$ fixed the equation now. $\endgroup$ – yeyyeat May 18 '15 at 19:12

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