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Theorem 3:

If $\frac{\left(\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{-}\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\right)}{\mathbf{N}}$is continuous and depends only on x, then

$渭\left(\mathbf{x}\right)\mathbf{=}\mathbf{exp}\left[鈭�\left(\frac{\left(\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{-}\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\right)}{\mathbf{N}}\right)\mathbf{dx}\right]$is an integrating factor for the equation.

If $\frac{\left(\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\mathbf{-}\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\right)}{\mathbf{M}}$is continuous and depends only on y, then

$渭\left(\mathbf{y}\right)\mathbf{=}\mathbf{exp}\left[鈭�\left(\frac{\left(\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\mathbf{-}\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\right)}{\mathbf{M}}\right)\mathbf{dy}\right]$ is an integrating factor for the equation.

Given: $\left({\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\right)\mathbf{dx}\mathbf{-}\mathbf{x}\mathbf{dy}\mathbf{=}\mathbf{0}路路路路路路\left(1\right)$

Let$\mathbf{M}\mathbf{=}{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{-}\mathbf{x}$

Then:

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

$\begin{array}{c}\frac{\left(\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{-}\frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\right)}{\mathbf{N}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{1}}{\mathbf{-}\mathbf{x}}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}\end{array}$

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

Then, find the integrating factor for y alone.

Let $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}\frac{2}{\mathbf{x}}$.

Integrate on both sides.

$\begin{array}{c}鈭�\mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}\mathbf{=}\mathbf{-}鈭�\frac{2}{\mathbf{x}}\mathbf{dx}\\ \mathbf{=}\mathbf{-}2\mathbf{ln}\left|\mathbf{x}\right|\end{array}$

Now find the value of $渭\left(\mathbf{x}\right)$.

role="math" localid="1664266938118" $\begin{array}{l}渭\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{鈭�\mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ln}\left|x\right|}\\ \mathbf{=}{\mathbf{x}}^{\mathbf{-}2}\end{array}$

Substitute the value$渭\left(\mathbf{x}\right)$ of in equation (1).

$\begin{array}{c}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}\left({\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\right)\mathbf{dx}\mathbf{-}\left({\mathbf{x}}^{\mathbf{-}\mathbf{2}}\right)\mathbf{x}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{2}}\right)\mathbf{dx}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{dy}\mathbf{=}\mathbf{0}路路路路路路\left(2\right)\end{array}$

Solve the equation (2).

Let $\mathbf{M}\mathbf{=}\frac{鈭�\mathbf{F}}{鈭�\mathbf{x}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{2}}$.

Now integrate the value to find F.

$\begin{array}{c}\mathbf{F}\mathbf{=}鈭�\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{2}}\right)\mathbf{dx}\\ \mathbf{=}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{-}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{g}\left(\mathbf{y}\right)\end{array}$

Differentiate F with respect to y.

$\begin{array}{c}\frac{鈭�\mathbf{F}}{鈭�\mathbf{y}}\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{-}1}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\\ \mathbf{=}\mathbf{N}\end{array}$

Then equalise the N values.

$\begin{array}{c}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\\ \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}0\end{array}$

Integrate on both sides with respect to y.

$\begin{array}{c}鈭�\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}鈭�\mathbf{0}\mathbf{dy}\\ \mathbf{g}\left(\mathbf{y}\right)\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\end{array}$

Substitute in F.

$\begin{array}{l}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{-}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\\ \frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{-}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{=}\mathbf{C}\\ \mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{3}}\mathbf{+}\mathbf{x}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{Cx}\\ \mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{3}}\mathbf{-}\mathbf{x}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{-}\mathbf{Cx}\end{array}$

Hence the solution is$\mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{3}}\mathbf{-}\mathbf{x}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{-}\mathbf{Cx}$