Question

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

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

Short answer

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

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

Evaluate it.

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

Substitution method

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

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

By Differentiating,

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

Now, integrate on both sides.

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

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

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

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