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Q2.6 - 9E

Expert-verified
Found in: Page 71

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

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

See the step by step solution

## General form of Homogeneous equation

If the right-hand side of the equation $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)$can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

## Evaluate the given equation

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

By Evaluating,

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

## Substitution method

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

Then $y=vx$

By Differentiating,

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\phantom{\rule{0ex}{0ex}}\mathbf{v}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\phantom{\rule{0ex}{0ex}}{\mathbf{v}}^{\mathbf{2}}\mathbf{=}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{dx}$

Now integrating on both sides,

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

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

Therefore, Homogeneous equation for the given equation is ${\mathbf{y}}{\mathbf{=}}\frac{-\mathbf{x}}{\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}\mathbf{C}}$ and localid="1655121736301" ${y}{=}{0}$.