3.3.18.1 Integrating factor that depends on $x$ only

Let

$$\begin{array}{rl}\text{(1)}& \mu M(x,y)+\mu N(x,y)\frac{dy}{dx}& =d\varphi (x,y)& =\frac{\partial \varphi }{\partial x}\frac{dx}{dx}+\frac{\partial \varphi }{\partial y}\frac{dy}{dx}\\ \text{(2)}& & =\frac{\partial \varphi }{\partial x}+\frac{\partial \varphi }{\partial y}\frac{dy}{dx}\end{array}$$

Comparing (1),(2) then

$$\begin{array}{rl}\frac{\partial \varphi }{\partial x}& =\mu M\\ \frac{\partial \varphi }{\partial y}& =\mu N\end{array}$$

The compatibility condition is $\frac{{\partial }^2\varphi }{\partial y\partial x}=\frac{{\partial }^2\varphi }{\partial x\partial y}$ then this implies

$$\begin{array}{rl}\frac{\partial }{\partial y}\left(\frac{\partial \varphi }{\partial x}\right)& =\frac{\partial }{\partial x}\left(\frac{\partial \varphi }{\partial y}\right)\\ \frac{\partial \mu M}{\partial y}& =\frac{\partial \mu N}{\partial x}\\ {\mu }_{y}M+\mu M_y& ={\mu }_{x}N+\mu N_x\\ {\mu }_{x}N& ={\mu }_{y}M+\mu M_y-\mu N_x\\ {\mu }_{x}N& ={\mu }_{y}M+\mu (M_y-N_x)\\ {\mu }_x& =\frac{{\mu }_{y}M}{N}+\frac{\mu }{N}(M_y-N_x)\end{array}$$

Assuming $\mu \equiv \mu \left(x\right)$ then ${\mu }_y=0$ and the above simplifies to

$$\begin{array}{rl}{\mu }_x& =\frac{\mu }{N}(M_y-N_x)\\ \frac{d\mu }{dx}\frac{1}{\mu }& =\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})\end{array}$$

Let $\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})=A$. If $A\equiv A\left(x\right)$ which depends only on $x$ then we can solve the above.

$$\begin{array}{rl}\frac{d\mu }{dx}\frac{1}{\mu }& =A\\ \mu & =e^{\int Adx}\end{array}$$

Let $\bar{M}=\mu M,\bar{N}=\mu N$ then the ode

$$\bar{M}(x,y)+\bar{N}(x,y)y^{\prime }=0$$

is now exact.