Question

Use the method discussed under “Homogeneous Equations” to solve problems 9-16. $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}$

Short answer

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

Step-by-step solution

General form of Homogeneous equation

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.

Evaluate the given equation

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

Evaluate it.

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

Substitution method

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

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

By Differentiating,

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1}{3}\left(\frac{1}{\mathrm{v}}-\mathrm{v}\right)& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1}{3\mathrm{v}}-\frac{\mathrm{v}}{3}-\mathrm{v}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1-{\mathrm{v}}^2-3{\mathrm{v}}^2}{3\mathrm{v}}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1-4{\mathrm{v}}^2}{3\mathrm{v}}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}& =& \frac{1}{3\mathrm{x}}\mathrm{dx}\end{array}$

Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl}\int \frac{\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}& =& \int \frac{1}{3\mathrm{x}}\mathrm{dx}\\ \int \frac{\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}& =& \ln \left|\mathrm{x}\right|+\mathrm{C}\end{array}$

Find the value of$\begin{array}{rcl}& & \int \frac{\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}\end{array}$separately.

Let $\begin{array}{rcl}\mathrm{w}& =& 1-4{\mathrm{v}}^2\\ \frac{d\mathrm{w}}{d\mathrm{v}}& =& -8\mathrm{v}\\ \mathrm{dv}& =& -\frac{1}{8\mathrm{v}}\mathrm{dw}\end{array}$

$\begin{array}{rcl}\int \frac{\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}& =& -\int \frac{\mathrm{v}}{\mathrm{w}}\frac{1}{8\mathrm{v}}\mathrm{dw}\\ & =& -\frac{1}{8}\int \frac{1}{\mathrm{w}}\mathrm{dw}\\ & =& -\frac{1}{8}\ln \left|\mathrm{w}\right|\end{array}$

Then, $\begin{array}{rcl}\int \frac{\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}& =& -\frac{1}{8}\ln \left|1-4{\mathrm{v}}^2\right|\end{array}$.

Androle="math" localid="1655139371679" $\begin{array}{rcl}-\frac{1}{8}\ln \left|1-4{\mathrm{v}}^2\right|& =& \frac{1}{3}\ln \left|\mathrm{x}\right|+{\mathrm{C}}_1\end{array}$

Substitution method

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

$\begin{array}{rcl}-\frac{1}{8}\ln \left|1-4{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2\right|& =& \frac{1}{3}\ln \left|\mathrm{x}\right|+{\mathrm{C}}_1\\ -3\ln \left|1-4{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2\right|& =& 8\ln \left|\mathrm{x}\right|+{\mathrm{C}}_2\\ -3\ln \left|\frac{{\mathrm{x}}^2-4{\mathrm{y}}^2}{{\mathrm{x}}^2}\right|& =& 8\ln \left|\mathrm{x}\right|+{\mathrm{C}}_2\\ & & \end{array}$

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

Then,

$\begin{array}{rcl}-3\ln \left|{\mathrm{x}}^2-4{\mathrm{y}}^2\right|+3\ln \left|{\mathrm{x}}^2\right|& =& 8\ln \left|\mathrm{x}\right|+{\mathrm{C}}_2\\ -3\ln \left|{\mathrm{x}}^2-4{\mathrm{y}}^2\right|+3\ln \left|{\mathrm{x}}^2\right|-8\ln \left|\mathrm{x}\right|& =& {\mathrm{C}}_2\\ \ln \left|{\mathrm{x}}^6\right|-\ln \left|{\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3\right|-\ln \left|{\mathrm{x}}^8\right|& =& {\mathrm{C}}_2\\ \ln \left|\frac{{\mathrm{x}}^6}{{\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3}\right|-\ln \left|{\mathrm{x}}^8\right|& =& {\mathrm{C}}_2\end{array}$

$\begin{array}{rcl}\ln \left|\frac{1}{{\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3{\mathrm{x}}^2}\right|& =& {\mathrm{C}}_2\\ \frac{1}{{\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3{\mathrm{x}}^2}& =& {\mathrm{e}}^{{\mathrm{C}}_2}\\ \frac{1}{{\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3{\mathrm{x}}^2}& =& {\mathrm{C}}_3\\ {\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3{\mathrm{x}}^2& =& \frac{1}{{\mathrm{C}}_3}\\ {\left({\mathrm{x}}^2-4{\mathrm{y}}^2\right)}^3{\mathrm{x}}^2& =& \mathrm{C}\end{array}$

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