91Ó°ÊÓ

Theorem 3:

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

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

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

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

Given,

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

To find integrating factor of the form ${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$.

Multiply ${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$on equation (1).

$\begin{array}{c}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\left(\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\left(\mathbf{3}\mathbf{xy}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left(\mathbf{2}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{2}\mathbf{+}\mathbf{m}}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{2}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(2\right)\end{array}$

Let

$\mathbf{M}\mathbf{=}\mathbf{2}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{2}\mathbf{+}\mathbf{m}}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{2}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}$

Now let us consider the found equation is exact. Then,$\frac{∂\mathbf{M}}{∂\mathbf{y}}\mathbf{=}\frac{∂\mathbf{N}}{∂\mathbf{x}}$.

$\begin{array}{c}\frac{∂\mathbf{M}}{∂\mathbf{y}}\mathbf{=}\mathbf{2}\left(\mathbf{2}\mathbf{+}\mathbf{m}\right){\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{-}\mathbf{6}\left(\mathbf{1}\mathbf{+}\mathbf{m}\right){\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\\ \frac{∂\mathbf{N}}{∂\mathbf{x}}\mathbf{=}\mathbf{3}\left(\mathbf{1}\mathbf{+}\mathbf{n}\right){\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{-}\mathbf{4}\left(\mathbf{2}\mathbf{+}\mathbf{n}\right){\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\end{array}$

So, we can equalise the coefficients.

$\begin{array}{l}\mathbf{2}\left(\mathbf{2}\mathbf{+}\mathbf{m}\right)\mathbf{=}\mathbf{3}\left(\mathbf{1}\mathbf{+}\mathbf{n}\right)\\ \mathbf{4}\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{n}\\ \mathbf{2}\mathbf{m}\mathbf{-}\mathbf{3}\mathbf{n}\mathbf{=}\mathbf{-}\mathbf{1}\\ \mathbf{-}\mathbf{6}\left(\mathbf{1}\mathbf{+}\mathbf{m}\right)\mathbf{=}\mathbf{-}\mathbf{4}\left(\mathbf{2}\mathbf{+}\mathbf{n}\right)\\ \mathbf{6}\mathbf{+}\mathbf{6}\mathbf{m}\mathbf{=}\mathbf{8}\mathbf{+}\mathbf{4}\mathbf{n}\\ \mathbf{6}\mathbf{m}\mathbf{-}\mathbf{4}\mathbf{n}\mathbf{=}\mathbf{2}\end{array}$

Solve the founded equation to get the value of m and n.

$m=1$

The value of m = 1 and n = 1.

${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\mathbf{=}\mathbf{xy}$

Then,

So, the integrating factor is found.

Substitute m and n in equation (2)

$\begin{array}{c}\left(\mathbf{2}{\mathbf{xy}}^{\mathbf{2}\mathbf{+}1}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}1}{\mathbf{y}}^{\mathbf{1}\mathbf{+}1}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{x}}^{\mathbf{1}\mathbf{+}1}{\mathbf{y}}^{\mathbf{1}\mathbf{+}1}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{2}\mathbf{+}1}{\mathbf{y}}^{\mathbf{1}\mathbf{+}1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left(\mathbf{2}{\mathbf{xy}}^3\mathbf{-}\mathbf{6}{\mathbf{x}}^2{\mathbf{y}}^2\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{x}}^2{\mathbf{y}}^2\mathbf{-}\mathbf{4}{\mathbf{x}}^3{\mathbf{y}}^2\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(3\right)\end{array}$

Solve the equation (3).

Let $\mathbf{M}\mathbf{=}\frac{∂\mathbf{F}}{∂\mathbf{x}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^3\mathbf{-}\mathbf{6}{\mathbf{x}}^2{\mathbf{y}}^2$.

Now integrate the value to find F.

$\begin{array}{c}\mathbf{F}\mathbf{=}∫\left(\mathbf{2}{\mathbf{xy}}^3\mathbf{-}\mathbf{6}{\mathbf{x}}^2{\mathbf{y}}^2\right)\mathbf{dx}\\ \mathbf{=}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{3}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\left(\mathbf{y}\right)\end{array}$

Differentiate F with respect to y.

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

Then equalise the N values.

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

Integrate on both sides with respect to y.

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

Substitute in F.

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

Hence the solution is

${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{3}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{C}$