Use the method discussed under “Homogeneous Equations” to solve problems 9-16.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}$

If the right-hand side of the equation $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)$can be expressed as a function of the ratio $\frac{\mathbf{y}}{\mathbf{x}}$alone, then we say the equation is homogeneous.

Given, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}$

Evaluate it.

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}\\ \frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}\\ & \mathbf{=}& \frac{\mathbf{x}}{\mathbf{3}\mathbf{y}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}\mathbf{x}}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\left(\frac{\mathbf{x}}{\mathbf{y}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{x}}\right)\\ & & \end{array}$

Let us take $v=\frac{y}{x}$

Then,$y=vx$

By Differentiating,

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{\mathbf{1}}{\mathbf{3}}\left(\frac{\mathbf{1}}{\mathbf{v}}\mathbf{-}\mathbf{v}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\end{array}$

$\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{3}\mathbf{v}}\mathbf{-}\frac{\mathbf{v}}{\mathbf{3}}\mathbf{-}\mathbf{v}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{\mathbf{1}\mathbf{-}{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{3}\mathbf{v}}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{3}\mathbf{v}}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{\mathbf{v}}{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}\mathbf{x}}\mathbf{dx}\end{array}$

Now, integrate on both sides.

$$\begin{gathered} \int \frac{\mathbf{v}}{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\int \frac{\mathbf{1}}{\mathbf{3}\mathbf{x}}\mathbf{dx} \\ \int \frac{\mathbf{v}}{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{1}} \end{gathered}$$

Find the value $\int \frac{\mathbf{v}}{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}$of separately.

Let,

Then, $\int \frac{\mathbf{v}}{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{ln}\left|\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}\right|$

And $\mathbf{-}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{ln}\left|\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{\mathbf{2}}\right|\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{1}}$

Substitute $\mathbf{v}\mathbf{=}\frac{\mathbf{y}}{\mathbf{x}}$

$\begin{array}{rcl}\mathbf{-}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{ln}\left|\mathbf{1}\mathbf{-}\mathbf{4}{\left(\frac{\mathbf{y}}{\mathbf{x}}\right)}^{\mathbf{2}}\right|& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{-}\mathbf{3}\mathbf{ln}\left|\mathbf{1}\mathbf{-}\mathbf{4}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\right|& \mathbf{=}& \mathbf{8}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\\ \mathbf{-}\mathbf{3}\mathbf{ln}\left|\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\right|& \mathbf{=}& \mathbf{8}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\end{array}$

Where $\mathbf{ln}\left(\frac{\mathbf{M}}{\mathbf{N}}\right)\mathbf{=}\mathbf{ln}\mathbf{M}\mathbf{-}\mathbf{ln}\mathbf{N}$

Then,

$\begin{array}{rcl}\mathbf{-}\mathbf{3}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right|\mathbf{+}\mathbf{3}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{2}}\right|& \mathbf{=}& \mathbf{8}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\\ \mathbf{-}\mathbf{3}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right|\mathbf{+}\mathbf{3}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{2}}\right|\mathbf{-}\mathbf{8}\mathbf{ln}\left|\mathbf{x}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}}\\ \mathbf{ln}\left|{\mathbf{x}}^{\mathbf{6}}\right|\mathbf{-}\mathbf{ln}\left|{\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}\right|\mathbf{-}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{8}}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}}\\ \mathbf{ln}\left|\frac{{\mathbf{x}}^{\mathbf{6}}}{{\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}}\right|\mathbf{-}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{8}}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}}\end{array}$

role="math" localid="1655200256535" $\begin{array}{rcl}\mathbf{ln}\left|\frac{\mathbf{1}}{{\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}}\\ \frac{\mathbf{1}}{{\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}}& \mathbf{=}& {\mathbf{e}}^{{\mathbf{C}}_{\mathbf{2}}}\\ \frac{\mathbf{1}}{{\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}}& \mathbf{=}& {\mathbf{C}}_{\mathbf{3}}\\ {\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{3}}}\\ {\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}& \mathbf{=}& \mathbf{C}\end{array}$

Therefore, Homogeneous equation for the given equation is ${\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\right)}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{C}$.