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

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

Given, $({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dx}+2\mathrm{xydy}=0$.

Evaluate it.

role="math" localid="1654940518297" $\begin{array}{rcl}({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dx}+2\mathrm{xydy}& =& 0\\ 2\mathrm{xydy}& =& -({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dx}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& -\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{2\mathrm{xy}}\\ & =& -\frac{\mathrm{x}}{2\mathrm{y}}-\frac{\mathrm{y}}{2\mathrm{x}}\\ & =& -\frac{1}{2}\left(\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}\right)\end{array}$

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

Then $\mathrm{y}=\mathrm{vx}$.

By Differentiating,

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

role="math" localid="1654941378213" $\begin{array}{rcl}\mathrm{xdx}& =& -\left(\frac{1+{\mathrm{v}}^2+2{\mathrm{v}}^2}{2\mathrm{v}}\right)\\ & =& -\frac{1+3{\mathrm{v}}^2}{2\mathrm{v}}\\ \frac{2\mathrm{v}}{1+3{\mathrm{v}}^2\mathrm{dv}}& =& -\frac{1}{\mathrm{x}}\mathrm{dx}\end{array}$

Now, integrate on both sides.

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

Integrate$\begin{array}{rcl}& & \int \frac{2\mathrm{v}}{1+3{\mathrm{v}}^2}\mathrm{dv}\end{array}$separately.

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

Then, localid="1654944579504" $$\begin{gathered} \begin{array}{rcl}& & \frac{\mathrm{dw}}{\mathrm{dv}}=6\mathrm{v} \\ \mathrm{dv}=\frac{1}{6\mathrm{v}}\mathrm{dw} \\ \end{array} \end{gathered}$$

Now,

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

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

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

Then,

localid="1654943733305" $\begin{array}{rcl}\frac{1}{3}\ln \left|1+3{\mathrm{v}}^2\right|& =& -\ln \left|x\right|+C_1\\ \frac{1}{3}\ln \left|1+3{\mathrm{v}}^2\right|+\ln \left|x\right|& =& C_1\\ \ln \left|1+3{\mathrm{v}}^2\right|+3\ln \left|x\right|& =& 3C_1\\ \ln \left|1+3{\mathrm{v}}^2\right|+\ln \left|x^3\right|& =& C_2\end{array}$

$\begin{array}{rcl}\ln \left|\left(1+3{\mathrm{v}}^2\right)\times {\mathrm{x}}^3\right|& =& {\mathrm{C}}_2\\ \left(1+3{\mathrm{v}}^2\right)\times {\mathrm{x}}^3& =& {\mathrm{e}}^{{\mathrm{C}}_2}\\ \left(1+3{\mathrm{v}}^2\right)\times {\mathrm{x}}^3& =& \mathrm{C}\end{array}$

Substitute $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}$.

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

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