Given a non-exact differential equation, $$M(x,y) dx + N(x,y)dy = 0,$$ an integrating factor is a function $\mu(x,y) \ne 0$ such that the equation

$$\mu Mdx + \mu N dy = 0$$ is exact.

I understand how to find integrating factors, but my only confusion is, why do they work? How do we know that the resulting function will have the same solution set as the original differential equation?

I understand that multiplying a function by another function can drastically change the behavior. So for this procedure, how do we know that solving the new differential equation will lead to potential solutions to the original one?

  • $\begingroup$ Not saying this to be annoying but, $\mu*0=0, \mu \neq 0$ $\endgroup$
    – keoxkeox
    Oct 4 '18 at 23:27
  • $\begingroup$ I edited the question. Thanks. $\endgroup$
    – math783625
    Oct 4 '18 at 23:53
  • $\begingroup$ Blindly, if a function $y = f(x)$ satisfies $M + N\frac{dy}{dx} = 0$ Then it will also satisfy $\mu (x, f(x)) M + \mu(x, f(x)) N \frac{dy}{dx} = 0$. However I think you are right. $y = f(x)$ only satisfies the differential equation on a certain domain. If the $\mu$ you throw in has a $1/(x-1)$, then suddenly, any solution to this new differential equation must have have $1$ restricted from it's domain. So my guess is that integrating factors could change the solution set ($f(x)$ defined over 2 different domains are different results). However as an answer has already stated, most integrating $\endgroup$
    – DWade64
    Oct 5 '18 at 3:12
  • $\begingroup$ factors are exponential functions. But these are just thoughts of mine, and they might be totally wrong $\endgroup$
    – DWade64
    Oct 5 '18 at 3:13

Usually the integrating factor is an exponential function and it is not zero, therefore

$$ \mu M dx +\mu Ndy =0 \iff \mu (M dx + Ndy) =0 \iff (M dx + Ndy) =0$$

Of course integrating factors are there to help solving the original equation and we better check our solutions.

If we find a solution which makes the integration factor zero, then we can check it in our original differential equation and in case that it does not satisfy our equation we disregard it.

  • $\begingroup$ Thanks for your response. My problem is that I don't understand why the procedure even leads to potential solutions to the original differential equation. $\endgroup$
    – math783625
    Oct 5 '18 at 2:48
  • $\begingroup$ We find the integrating factors such that the new equation is exact. Thus the new equation will have solutions. Since $\mu$ is not zero, these solutions are solutions to the original eqautions. $\endgroup$ Oct 5 '18 at 10:00
  • $\begingroup$ Oh, alright that makes it clear. Thanks! $\endgroup$
    – math783625
    Oct 5 '18 at 10:02

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