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Given that, $\left({\mathbf{x}}^3\mathbf{-}\mathbf{y}\right)\mathbf{dx}\mathbf{+}\mathbf{x}\text{鈥�}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\text{鈥夆赌�}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{3}鈰�鈰�鈰�鈰�鈰�鈰�\left(1\right)$

Let us check whether the given equation is exact or not.

Then, $\mathbf{M}\mathbf{=}{\mathbf{x}}^3\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{x}$.

Differentiate the value of M and N.

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

So, the given equation is not exact. Then, find the special integrating factor,

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

Find the value of $渭\left(x\right)$.

Multiply x^-2 in equation (1) on both sides.

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

Now again check whether the founded equation is exact or not.

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

Differentiate the value of M and N.

$\begin{array}{l}\frac{鈭�\mathbf{M}}{鈭�\mathbf{y}}\mathbf{=}\mathbf{-}{\mathbf{x}}^{-2}\\ \frac{鈭�\mathbf{N}}{鈭�\mathbf{x}}\mathbf{=}\mathbf{-}{\mathbf{x}}^{-2}\end{array}$

Therefore, the founded equation is exact.

Now, let us assume $\mathbf{M}\mathbf{=}\frac{鈭�\mathbf{F}}{鈭�\mathbf{x}}\mathbf{=}\mathbf{x}\mathbf{-}{\mathbf{yx}}^{-2}$.

Integrate on both sides.

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

Differentiate the F with respect to y.

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

Equalize the values of N.

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

Integrate on both sides.

data-custom-editor="chemistry" $\begin{array}{c}鈭�\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}鈭�0\text{鈥�}\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}{\mathbf{C}}_1\end{array}$

Substitute in the equation of F.

$\begin{array}{c}\frac{{\mathbf{x}}^2}{2}\mathbf{+}\frac{y}{x}\mathbf{+}{\mathbf{C}}_1\mathbf{=}\mathbf{0}\\ \frac{{\mathbf{x}}^2}{2}\mathbf{+}\frac{y}{x}\mathbf{=}\mathbf{C}鈰�鈰�鈰�鈰�鈰�鈰�\left(2\right)\end{array}$

So, the solution is found.

Given that, $\mathbf{y}\left(1\right)\mathbf{=}\mathbf{3}$.

Then, x = 1 and y = 3

Substitute the value in equation (2) to get the value of C.

$\begin{array}{c}\frac{{\mathbf{x}}^2}{2}\mathbf{+}\frac{y}{x}\mathbf{=}\mathbf{C}\\ \frac{1}{2}\mathbf{+}\frac{3}{1}\mathbf{=}\mathbf{C}\\ \frac{1+6}{2}\mathbf{=}\mathbf{C}\\ \mathbf{C}\mathbf{=}\frac{7}{2}\end{array}$

Substitute the value of C in equation (2).

$\begin{array}{c}\frac{{\mathbf{x}}^2}{2}\mathbf{+}\frac{y}{x}\mathbf{=}\frac{7}{2}\\ \frac{y}{x}\mathbf{=}\frac{7}{2}\mathbf{-}\frac{{\mathbf{x}}^2}{2}\\ \mathbf{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^3}{2}\mathbf{+}\frac{7x}{2}\end{array}$

So, the solution is$\mathbf{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^3}{2}\mathbf{+}\frac{7x}{2}$