In Problems 7-12, solve the equation.

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

Theorem 3:

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

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

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

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

Given:

$\left(\mathbf{2}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}\left({\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(1\right)$

Let:

$\mathbf{M}\mathbf{=}2\mathbf{xy}\mathbf{,}\mathbf{N}\mathbf{=}{\mathbf{y}}^2\mathbf{-}3{\mathbf{x}}^2$

Then,

$\begin{array}{c}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}2\mathbf{x}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{-}6\mathbf{x}\end{array}$

Then, substitute the values to find it.

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

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

Then, find the integrating factor for y alone.

Let

$\mathbf{P}\left(\mathbf{y}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\mathbf{y}}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{P}\left(\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{-}\int \frac{\mathbf{4}}{\mathbf{y}}\mathbf{dy}\\ \mathbf{=}\mathbf{-}\mathbf{4}\mathbf{ln}\left|\mathbf{y}\right|\end{array}$

Now find the value of $\mu \left(\mathbf{y}\right)$.

$\begin{array}{c}\mu \left(\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{y}\right)\mathbf{dy}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{ln}\left|\mathbf{y}\right|}\\ \mu \left(\mathbf{y}\right)\mathbf{=}{\mathbf{y}}^{\mathbf{-}\mathbf{4}}\end{array}$

Multiply the value of $\mu \left(\mathbf{y}\right)$ in equation (1).

$\begin{array}{c}{\mathbf{y}}^{\mathbf{-}\mathbf{4}}\left(\mathbf{2}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}{\mathbf{y}}^{\mathbf{-}\mathbf{4}}\left({\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \mathbf{2}{\mathbf{xy}}^{\mathbf{-}\mathbf{3}}\mathbf{dx}\mathbf{+}\left({\mathbf{y}}^{\mathbf{-}\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{-}\mathbf{4}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(2\right)\end{array}$

Solve the equation (2).

Let $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{-}\mathbf{3}}$.

Now integrate the value to find F.

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

Differentiate F with respect to y.

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

Then equalize the N values.

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

Integrate on both sides with respect to y.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\int {\mathbf{y}}^{\mathbf{-}\mathbf{2}}\mathbf{dy}\\ \mathbf{g}\left(\mathbf{y}\right)\mathbf{=}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\end{array}$

Substitute in F.

$\begin{array}{c}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{-}\mathbf{3}}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{-}\mathbf{3}}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{C}\end{array}$

Hence the solution is , $\mu \mathbf{=}{\mathbf{y}}^{\mathbf{-}\mathbf{4}},{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{-}\mathbf{3}}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{C}$.