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Answers without the blur. Sign up and see all textbooks for free! Q2.6 - 10E

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.${ }\left(\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\right){ }{\mathbf{dx}}{\mathbf{+}}\left(\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\right){\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

Homogeneous equation for the given equation is $\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{+}}{\mathbf{ln}}\left|\frac{{\mathbf{x}}^{\mathbf{6}}}{{\mathbf{y}}^{\mathbf{2}}}\right|{\mathbf{=}}{\mathbf{C}}$

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, $\left(\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\right) \mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Evaluate it.

$\begin{array}{rcl}\left(\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\right) \mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dy}& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dy}& \mathbf{=}& \mathbf{-}\left(\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{dx}\\ \frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{-}\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}}\\ & \mathbf{=}& \frac{{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}}\\ & & \end{array}$

Now take LCM.

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \frac{{\mathbf{y}}^{\mathbf{3}}\mathbf{-}\mathbf{3}{\mathbf{yx}}^{\mathbf{2}}}{{\mathbf{xy}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}\\ & \mathbf{=}& \frac{\frac{{\mathbf{y}}^{\mathbf{3}}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-}\frac{3\mathbf{y}}{\mathbf{x}}}{\frac{{\mathbf{y}}^{2}}{{\mathbf{x}}^{2}}\mathbf{-}\mathbf{1}}\\ & \mathbf{=}& \frac{{\left(\frac{\mathbf{y}}{\mathbf{x}}\right)}^{\mathbf{3}}\mathbf{-}\mathbf{3}\left(\frac{\mathbf{y}}{\mathbf{x}}\right)}{{\left(\frac{\mathbf{y}}{\mathbf{x}}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\\ & & \end{array}$

## Substitution method

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

Then $y=vx$

By Differentiating,

role="math" localid="1655179163129" $\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{{\mathbf{v}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathbf{v}}{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{{\mathbf{v}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathbf{v}}{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{v}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \frac{{\mathbf{v}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}{\mathbf{v}}^{\mathbf{3}}\mathbf{+}\mathbf{v}}{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ & & \end{array}$

$\begin{array}{rcl}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}& \mathbf{=}& \mathbf{-}\mathbf{2}\frac{\mathbf{v}}{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\\ \frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{\mathbf{v}}\mathbf{dv}& \mathbf{=}& \mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{dx}\\ \left(\mathbf{v}\mathbf{-}\frac{\mathbf{1}}{\mathbf{v}}\right)\mathbf{dv}& \mathbf{=}& \mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{dx}\\ & & \end{array}$

Now, integrate on both sides.

$\int \left(\mathbf{v}\mathbf{-}\frac{\mathbf{1}}{\mathbf{v}}\right)\mathbf{dv}\mathbf{=}\mathbf{-}\int \frac{\mathbf{2}}{\mathbf{x}}\mathbf{dx}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{-}\mathbf{ln}\left|\mathbf{v}\right|\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}\mathbf{C}\phantom{\rule{0ex}{0ex}}{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{ln}\left|\mathbf{v}\right|\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}\mathbf{2}\mathbf{C}\phantom{\rule{0ex}{0ex}}{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{ln}\left|{\mathbf{v}}^{\mathbf{2}}\right|\mathbf{+}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{4}}\right|\mathbf{=}\mathbf{C}\phantom{\rule{0ex}{0ex}}$

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

$\begin{array}{rcl}{\left(y}{x}\right)}^{\mathbf{2}}\mathbf{-}\mathbit{I}\mathbit{n}|{\left(y}{x}\right)}^{2}|\mathbf{+}\mathbit{I}\mathbit{n}|{x}^{4}|& \mathbf{=}& \mathbit{C}\\ \frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\mathbf{ln}\left|\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\right|\mathbf{+}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{4}}\right|& \mathbf{=}& \mathbf{C}\\ \frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\mathbf{ln}\left|\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{4}}}\right|& \mathbf{=}& \mathbf{C}\\ \frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathbf{ln}\left|\frac{{\mathbf{x}}^{\mathbf{6}}}{{\mathbf{y}}^{\mathbf{2}}}\right|& \mathbf{=}& \mathbf{C}\end{array}$

Therefore, Homogeneous equation for the given equation is $\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{+}}{\mathbf{ln}}\left|\frac{{\mathbf{x}}^{\mathbf{6}}}{{\mathbf{y}}^{\mathbf{2}}}\right|{\mathbf{=}}{\mathbf{C}}$ ### Want to see more solutions like these? 