Inexact differential equation

In order to solve the equation, we need to transform it into an exact differential equation. In order to do that, we need to find an integrating factor ${\displaystyle \mu }$ to multiply the equation by. We'll start with the equation itself. ${\displaystyle M\,dx+N\,dy=0}$, so we get ${\displaystyle \mu M\,dx+\mu N\,dy=0}$. We will require ${\displaystyle \mu }$ to satisfy ${\textstyle {\frac {\partial \mu M}{\partial y}}={\frac {\partial \mu N}{\partial x}}}$. We get

${\displaystyle {\frac {\partial \mu }{\partial y}}M+{\frac {\partial M}{\partial y}}\mu ={\frac {\partial \mu }{\partial x}}N+{\frac {\partial N}{\partial x}}\mu .}$

After simplifying we get

${\displaystyle M\mu _{y}-N\mu _{x}+(M_{y}-N_{x})\mu =0.}$

Since this is a partial differential equation, it is mostly extremely hard to solve, however in some cases we will get either ${\displaystyle \mu (x,y)=\mu (x)}$ or ${\displaystyle \mu (x,y)=\mu (y)}$, in which case we only need to find ${\displaystyle \mu }$ with a first-order linear differential equation or a separable differential equation, and as such either

${\displaystyle \mu (y)=e^{-\int {{\frac {M_{y}-N_{x}}{M}}\,dy}}}$

or

${\displaystyle \mu (x)=e^{\int {{\frac {M_{y}-N_{x}}{N}}\,dx}}.}$

• Tenenbaum, Morris; Pollard, Harry (1963). "Recognizable Exact Differential Equations". Ordinary Differential Equations: An Elementary Textbook for Students of Mathematics, Engineering, and the Sciences. New York: Dover. pp. 80–91. ISBN 0-486-64940-7.

• A solution for an inexact differential equation from Stack Exchange
• a guide for non-partial inexact differential equations at SOS math