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!


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: $$\mu(x,y)=\exp\left(\int \dfrac{dx}{x}\right)\exp\left(\int\dfrac{dy}{y}\right)=xy$$

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.