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