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Show that if $\frac{M y-N x}{N}=Q_{1}$ , Where $Q$ is a function of $x$ only, then the DE $M+N y^{\prime}=0$ has an
integrating factor of the form $M=e^{\int Q(x) d x}$

$M d x+N d y=0 \quad * \mu(x, y)$

$\mu {M dx}+\mu {Ndy}=0$ Exact-Equation

then $\rightarrow(\mu M)_{y}=(\mu N)_{x}$

$\mu {y} M+\mu M y=\mu_{x} N+\mu {N x}$

$\mu {x}=\frac{d \mu}{d x}$

$\mu y=0$

$0+\mu M{y}=\frac{d \mu N}{d x}+\mu N x$

$\mu M y-\mu N x=\frac{d \mu}{d x} N$

$\frac{d \mu}{d x} N=\mu M y-\mu N x$

$\rightarrow \frac{d {\mu}}{d x} N=\mu\left(M {y}-N {x}\right)$

$* \frac{d x}{\mu}$

$\frac{d {\mu}}{\mu} N=\frac{M y-N x}{1} d x$

$\frac{d {\mu}}{\mu}=\frac{M y-N x}{N} d x$ Integrate

$\ln |\mu|=\int \frac{M y-N x}{N} d x$

$e^{\ln |\mu|}=e^{\int\frac{M y-N x}{N} d x}$

$\Rightarrow \mu=\exp^{\left(\int \frac{M y-N x}{N} d x\right)}$

$\mu=e^{\int Q(x) d x}$

Show that if $\frac{N x-M y}{M}=Q$ , Where $Q$ is a function of y only, then the DE
$M+N y^{\prime}=0$ has an integrating factor of the form $M=e^{\{Q(y) \mathrm{d} y}$

$M d x+N d y=0 \quad * \mu(x, y)$

$\mu M d x+\mu N d y=0$ Exact equation

then $(\mu M)_{y}=(\mu N)_{x}$

$\mu {y} M+\mu M y=\mu{x} N+\mu N {x}$

$\mu x=0$

$\mu y=\frac{d \mu}{d y}$

$\frac{d \mu M}{d y}+\mu M y=0+\mu N x$

$\frac{d \mu}{d y} M=\mu N x-\mu M y$

$\frac{d \mu}{d y} M=(N x-M y) \mu$

$\div M$

$\frac{d \mu}{d y}=\frac{\left(N {x}-M {y}\right) \mu}{M}$

$* \frac{d y}{\mu}$

$\frac{d y}{\mu}=\frac{\left(N {x}-M {y}\right)}{M} d y$ Integrat

$\int {\frac{d y}{\mu}}=\int {\frac{\left(N {x}-M {y}\right)}{M} d y}$

$\ln |\mu|=\int \frac{N x-M y}{M} d y$

$e^{\ln |\mu|}=e^{\int \frac{N x-M y}{M} d y}$

$\rightarrow \mu=\exp^{\int Q(y) d y}$