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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

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

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

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

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

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