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$\frac{鈭�\mathbf{F}}{鈭�\mathbf{x}}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{M}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)鈥�鈥�鈥�鈥�鈥�\mathbf{and}鈥�鈥�鈥�鈥�\frac{鈭�\mathbf{F}}{鈭�\mathbf{y}}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{N}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)$

Given, $\left(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$.

Evaluate it.

$\begin{array}{l}\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\end{array}$

Compare the given equation with general form of separable and linear equation.

So, the given equation is neither separable nor linear.

Given, $\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Let $\mathbf{M}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{xy}\mathbf{-}\mathbf{1}$

Then,

$\begin{array}{c}\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\\ \frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\mathbf{=}\mathbf{y}\end{array}$

So, $\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}鈮�\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}$

Therefore, the given equation is not exact.

If $\frac{\left(\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{-}\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\right)}{\mathbf{N}}$ then the given function is x alone.

If$\frac{\left(\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\mathbf{-}\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\right)}{\mathbf{M}}$then the given function is y alone.

Then, substitute the values to prove it.

$\begin{array}{c}\frac{\left(\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{-}\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\right)}{\mathbf{N}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{y}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\frac{\frac{\mathbf{1}\mathbf{-}\mathbf{xy}}{\mathbf{x}}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\mathbf{-}\frac{\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}{\mathbf{x}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\end{array}$$\begin{array}{c}\frac{\left(\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{-}\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\right)}{\mathbf{N}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{y}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\frac{\frac{\mathbf{1}\mathbf{-}\mathbf{xy}}{\mathbf{x}}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\mathbf{-}\frac{\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}{\mathbf{x}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\end{array}$

So, we obtain an integrating factor that is a function of x alone.

Hence, the given equation is having an integrating factor that is a function of x alone.