Use the method discussed under “Homogeneous Equations” to solve problems 9 -16.$\mathbf{ }\left(x^2+y^2\right)\mathbf{ }\mathit{d}\mathit{x}\mathbf{+}\mathbf{2}\mathit{x}\mathit{y}\mathbf{}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

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

Evaluate it.

$\begin{array}{rcl}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\right) \mathbf{dx}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{dy}& \mathbf{=}& \mathbf{0}\\ \mathbf{2}\mathbf{xydy}& \mathbf{=}& \mathbf{-}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{dx}\\ \frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{2}\mathbf{xy}}\\ & \mathbf{=}& \mathbf{-}\frac{\mathbf{x}}{\mathbf{2}\mathbf{y}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{2}\mathbf{x}}\\ & \mathbf{=}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(\frac{\mathbf{x}}{\mathbf{y}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\right)\\ & & \end{array}$

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

Then $\mathbf{y}\mathbf{=}\mathbf{vx}$

By Differentiating,

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(\frac{\mathbf{1}}{\mathbf{v}}\mathbf{+}\mathbf{v}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(\frac{\mathbf{1}}{\mathbf{v}}\mathbf{+}\mathbf{v}\right)\mathbf{-}\mathbf{v}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \mathbf{-}\frac{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}\mathbf{v}}\mathbf{-}\mathbf{vdv}& \mathbf{=}& \mathbf{xdx}\\ & & \end{array}$

Now, integrate on both sides.

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

Integrate$\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}$separately.

Let us take$\mathbf{w}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}$

Then,

$$\begin{gathered} \frac{\mathbf{dw}}{\mathbf{dv}}\mathbf{=}\mathbf{6}\mathbf{v} \\ \mathbf{dv}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}\mathbf{v}}\mathbf{dw} \end{gathered}$$

Now,

$\begin{array}{rcl}\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{w}}\frac{\mathbf{1}}{\mathbf{6}\mathbf{v}}\mathbf{dw}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\int \frac{\mathbf{1}}{\mathbf{w}}\mathbf{dw}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{w}\right|\\ & & \end{array}$

Substitute $\mathbf{w}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}$

$\begin{array}{rcl}\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{w}\right|\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right|\\ & & \end{array}$

Then,

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

$\begin{array}{rcl}\mathbf{ln}\left|\left(\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right)\mathbf{\times }{\mathbf{x}}^{\mathbf{3}}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}}\\ \left(\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right)\mathbf{\times }{\mathbf{x}}^{\mathbf{3}}& \mathbf{=}& {\mathbf{e}}^{{\mathbf{C}}_{\mathbf{2}}}\\ \left(\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right)\mathbf{\times }{\mathbf{x}}^{\mathbf{3}}& \mathbf{=}& \mathbf{C}\\ & & \end{array}$

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

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

Therefore, Homogeneous equation for the given equation is ${\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{xy}}^{\mathbf{2}}\mathbf{=}\mathbf{C}$.