In Problems 7-12, solve the equation.

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

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:

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

Let $\mathbf{M}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}$$\mathbf{M}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}$

Then:

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

So, we obtain the given equation is exact.

Solve the equation (2).

Let $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{3}}\mathbf{+}\mathbf{1}$.

Now integrate the value to find F.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{2}{\mathbf{xy}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\right)\mathbf{dx}\\ \mathbf{=}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^3\mathbf{+}\mathbf{g}\left(\mathbf{y}\right)\end{array}$

Differentiate F with respect to y.

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

Then equalise the N values.

$\begin{array}{c}3{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^2\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}3{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^2\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\\ \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\end{array}$

Integrate on both sides with respect to y.

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

Substitute in F.

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

Hence the solution is${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{3}}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{ln}\left|\mathbf{y}\right|\mathbf{=}\mathbf{C}$