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I have the following ODE, which have to solved by exact ODE method. $$e^{-y}dx + e^{-x}(-e^{-y}+1)dy = 0$$ Integrating factor is given as $$ I = e^{x+y}$$ I want to understand if there is any method of obtaining integrating factor that depends on both the variables $x$ & $y$, and any sequential procedure through which one can arrive at the given integrating factor.

A rural reader
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Rohan Garg
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    Re. "methods for obtaining integrating factors", usually you can get away with guessing the form of $\mu$ (example). I've also found the method laid out here helpful in the past, but it doesn't work with every ODE. Aside from these, and determining $\mu$ "by inspection", I'm not sure there is a method that will always get you what you need. – user170231 Dec 18 '23 at 16:25

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You want to multiply both sides of the inexact ODE,

$$M(x,y) \, dx + N(x,y) \, dy = 0$$

by an integrating factor $\mu(x,y)$ that makes it exact, such that

$$\begin{align*} \frac{\partial}{\partial y} \left[\mu(x,y) M(x,y)\right] &= \frac{\partial}{\partial x} \left[\mu(x,y) N(x,y)\right] \\ \implies \frac{\mu_y}\mu M - \frac{\mu_x}\mu N &= N_x - M_y \end{align*}$$

which in this case means finding $\mu$ to satisfy

$$\begin{align*} \frac{\mu_y}\mu e^{-y} - \frac{\mu_x}\mu e^{-x} \left(1-e^{-y}\right) &= -e^{-x}\left(1-e^{-y}\right) + e^{-y} \\ \implies \frac{\mu_x}\mu \left(1-e^y\right) + \frac{\mu_y}\mu e^x &= 1 - e^y + e^x \end{align*}$$

Rearranging to the above form suggests $\dfrac{\mu_x}\mu=\dfrac{\mu_y}\mu=1$ and the solution for $\mu$ follows.

user170231
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