Question: In Problems , solve the equation.$\left(\mathbf{4}{\mathbf{xy}}^3\mathbf{-}\mathbf{9}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{xy}}^2\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{-}\mathbf{6}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{x}}^2\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Given that,$\left(\mathbf{4}{\mathbf{xy}}^3\mathbf{-}\mathbf{9}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{xy}}^2\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{-}\mathbf{6}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{x}}^2\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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

Then, $\mathbf{M}\mathbf{=}\mathbf{4}{\mathbf{xy}}^3\mathbf{-}\mathbf{9}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{xy}}^2\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{-}\mathbf{6}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{x}}^2\mathbf{y}$.

Differentiate the value of M and N.

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

$\begin{array}{c}\frac{\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}}{N}\mathbf{=}\frac{\mathbf{12}{\mathbf{xy}}^2\mathbf{-}\mathbf{18}\mathbf{y}\mathbf{+}\mathbf{8}\mathbf{xy}\mathbf{-}\mathbf{6}{\mathbf{xy}}^2\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{xy}}{\mathbf{3}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{-}\mathbf{6}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{x}}^2\mathbf{y}}\\ \mathbf{=}\frac{\mathbf{6}{\mathbf{xy}}^2\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{+}\mathbf{4}\mathbf{xy}}{\mathbf{3}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{-}\mathbf{6}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{x}}^2\mathbf{y}}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{y}\left(3\mathrm{xy}-6+2x\right)}{\mathbf{xy}\left(3\mathrm{xy}-6+2x\right)}\\ \mathbf{=}\frac{2}{x}\end{array}$

Find the value of $\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \frac{2}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{ln}\left|x\right|}\\ \mathbf{=}{\mathbf{x}}^2\end{array}$

Multiply x² in equation (1) on both sides.

$\left(\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^3\mathbf{-}\mathbf{9}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^2\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{-}\mathbf{6}{\mathbf{x}}^3\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^4\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

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

$\mathbf{M}\mathbf{=}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^3\mathbf{-}\mathbf{9}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{-}\mathbf{6}{\mathbf{x}}^3\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^4\mathbf{y}$

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{12}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{-}\mathbf{18}{\mathbf{x}}^2\mathbf{y}\mathbf{+}\mathbf{8}{\mathbf{x}}^3\mathbf{y}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{12}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{-}\mathbf{18}{\mathbf{x}}^2\mathbf{y}\mathbf{+}\mathbf{8}{\mathbf{x}}^3\mathbf{y}\end{array}$

Therefore, the founded equation is exact.

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^3\mathbf{-}\mathbf{9}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^2$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^3\mathbf{-}\mathbf{9}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^2\right)\mathbf{dx}\\ \mathbf{=}{\mathbf{x}}^4{\mathbf{y}}^3\mathbf{-}\mathbf{3}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Differentiate the F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\mathbf{3}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{-}\mathbf{6}{\mathbf{x}}^3\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^4\mathbf{y}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalize the values of N.

$\begin{array}{c}\mathbf{3}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{-}\mathbf{6}{\mathbf{x}}^3\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^4\mathbf{y}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{3}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{-}\mathbf{6}{\mathbf{x}}^3\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^4\mathbf{y}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}0\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\int 0\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}{\mathbf{C}}_1\end{array}$

Substitute in the equation of F.

$\begin{array}{c}{\mathbf{x}}^4{\mathbf{y}}^3\mathbf{-}\mathbf{3}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}{\mathbf{C}}_1\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^4{\mathbf{y}}^3\mathbf{-}\mathbf{3}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{=}\mathbf{C}\end{array}$

So, the solution is ${\mathbf{x}}^4{\mathbf{y}}^3\mathbf{-}\mathbf{3}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{=}\mathbf{C}$