Question

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

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

Short answer

Homogeneous equation for the given equation is$\mathbf{y}\mathbf{=}\frac{\mathbf{-}\mathbf{x}}{\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}}$ and$\mathbf{y}\mathbf{\equiv }\mathbf{0}$ .

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, $(\mathrm{xy}+{\mathrm{y}}^2)\mathrm{dx}-{\mathrm{x}}^2\mathrm{dy}=0$.

By Evaluating,

role="math" localid="1654867974528" $\begin{array}{rcl}(\mathrm{xy}+{\mathrm{y}}^2)\mathrm{dx}-{\mathrm{x}}^2\mathrm{dy}& =& 0\\ -{\mathrm{x}}^2\mathrm{dy}& =& -(\mathrm{xy}+{\mathrm{y}}^2)\mathrm{dx}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{(\mathrm{xy}+{\mathrm{y}}^2)}{{\mathrm{x}}^2}\\ & =& \frac{\mathrm{y}}{\mathrm{x}}+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}\\ & =& \frac{\mathrm{y}}{\mathrm{x}}+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2\end{array}$

Substitution method

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

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

By Differentiating,

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \mathrm{v}+{\mathrm{v}}^2& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ {\mathrm{v}}^2& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1}{{\mathrm{v}}^2}\mathrm{dv}& =& \frac{1}{\mathrm{x}}\mathrm{dx}\\ & & \\ & & \end{array}$

Now integrating on both sides,

role="math" localid="1654868488187" $\begin{array}{rcl}\int {\mathrm{v}}^{-2}\mathrm{dv}& =& \int \frac{1}{\mathrm{x}}\mathrm{dx}\\ -{\mathrm{v}}^{-1}& =& \ln \left|\mathrm{x}\right|+\mathrm{C}\\ -\frac{1}{\mathrm{v}}& =& \ln \left|\mathrm{x}\right|+\mathrm{C}\end{array}$

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

$\begin{array}{rcl}-\frac{1}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}}& =& \ln \left|\mathrm{x}\right|+\mathrm{C}\\ -\frac{\mathrm{x}}{{\displaystyle \mathrm{y}}}& =& \ln \left|\mathrm{x}\right|+\mathrm{C}\\ \mathrm{y}& =& -\frac{\mathrm{x}}{\ln \left|\mathrm{x}\right|+\mathrm{C}},\mathrm{y}\equiv 0\end{array}$

Therefore, Homogeneous equation for the given equation is $\mathbf{y}\mathbf{=}\frac{\mathbf{-}\mathbf{x}}{\mathbf{ln}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{+}\mathbf{C}}$and $\mathbf{y}\mathbf{\equiv }\mathbf{0}$.