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Expert-verified Found in: Page 76 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Use the method discussed under “Homogeneous Equations” to solve problems 9 - 16. $\left(\mathrm{xy}+{y}^{2}\right){\mathbf{dx}}{\mathbf{-}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{dy}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{0}}$

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

See the step by step solution

## Step 1: 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.

## Step 2: Evaluate the given equation

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

By Evaluating,

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

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

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

Therefore, Homogeneous equation for the given equation is ${\mathbf{y}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{x}}{\mathbf{ln}\mathbf{|}\mathbf{x}\mathbf{|}\mathbf{+}\mathbf{C}}$and ${\mathbf{y}}{\mathbf{\equiv }}{\mathbf{0}}$. ### Want to see more solutions like these? 