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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 ${\mathbf{y}}{\mathbf{\equiv }}{\mathbf{0}}$.