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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

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# Use the method discussed under “Homogeneous Equations” to solve problems 9-16. $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{3}\mathbf{xy}}$

Homogeneous equation for the given equation is ${\mathbf{\left(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{\right)}}^{{\mathbf{3}}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{C}}$.

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, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^{2}-{\mathrm{y}}^{2}}{3\mathrm{xy}}$.

Evaluate it.

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

## Step 3: Substitution method

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

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

By Differentiating,

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1}{3}\left(\frac{1}{\mathrm{v}}-\mathrm{v}\right)& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1}{3\mathrm{v}}-\frac{\mathrm{v}}{3}-\mathrm{v}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1-{\mathrm{v}}^{2}-3{\mathrm{v}}^{2}}{3\mathrm{v}}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{1-4{\mathrm{v}}^{2}}{3\mathrm{v}}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{\mathrm{v}}{1-4{\mathrm{v}}^{2}}\mathrm{dv}& =& \frac{1}{3\mathrm{x}}\mathrm{dx}\end{array}$

## Step 4: Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl}\int \frac{\mathrm{v}}{1-4{\mathrm{v}}^{2}}\mathrm{dv}& =& \int \frac{1}{3\mathrm{x}}\mathrm{dx}\\ \int \frac{\mathrm{v}}{1-4{\mathrm{v}}^{2}}\mathrm{dv}& =& \mathrm{ln}\left|\mathrm{x}\right|+\mathrm{C}\end{array}$

Find the value of $\begin{array}{rcl}& & \int \frac{\mathrm{v}}{1-4{\mathrm{v}}^{2}}\mathrm{dv}\end{array}$ separately.

Let $\begin{array}{rcl}\mathrm{w}& =& 1-4{\mathrm{v}}^{2}\\ \frac{d\mathrm{w}}{d\mathrm{v}}& =& -8\mathrm{v}\\ \mathrm{dv}& =& -\frac{1}{8\mathrm{v}}\mathrm{dw}\end{array}$

$\begin{array}{rcl}\int \frac{\mathrm{v}}{1-4{\mathrm{v}}^{2}}\mathrm{dv}& =& -\int \frac{\mathrm{v}}{\mathrm{w}}\frac{1}{8\mathrm{v}}\mathrm{dw}\\ & =& -\frac{1}{8}\int \frac{1}{\mathrm{w}}\mathrm{dw}\\ & =& -\frac{1}{8}\mathrm{ln}\left|\mathrm{w}\right|\end{array}$

Then, $\begin{array}{rcl}\int \frac{\mathrm{v}}{1-4{\mathrm{v}}^{2}}\mathrm{dv}& =& -\frac{1}{8}\mathrm{ln}\left|1-4{\mathrm{v}}^{2}\right|\end{array}$.

And role="math" localid="1655139371679" $\begin{array}{rcl}-\frac{1}{8}\mathrm{ln}\left|1-4{\mathrm{v}}^{2}\right|& =& \frac{1}{3}\mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{1}\end{array}$

## Step 5: Substitution method

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

$\begin{array}{rcl}-\frac{1}{8}\mathrm{ln}\left|1-4{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{2}\right|& =& \frac{1}{3}\mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{1}\\ -3\mathrm{ln}\left|1-4{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{2}\right|& =& 8\mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{2}\\ -3\mathrm{ln}\left|\frac{{\mathrm{x}}^{2}-4{\mathrm{y}}^{2}}{{\mathrm{x}}^{2}}\right|& =& 8\mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{2}\\ & & \end{array}$

Where $\mathrm{ln}\left(\frac{\mathrm{M}}{\mathrm{N}}\right)=\mathrm{ln}\mathrm{M}-\mathrm{ln}\mathrm{N}$

Then,

$\begin{array}{rcl}-3\mathrm{ln}\left|{\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right|+3\mathrm{ln}\left|{\mathrm{x}}^{2}\right|& =& 8\mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{2}\\ -3\mathrm{ln}\left|{\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right|+3\mathrm{ln}\left|{\mathrm{x}}^{2}\right|-8\mathrm{ln}\left|\mathrm{x}\right|& =& {\mathrm{C}}_{2}\\ \mathrm{ln}\left|{\mathrm{x}}^{6}\right|-\mathrm{ln}\left|{\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}\right|-\mathrm{ln}\left|{\mathrm{x}}^{8}\right|& =& {\mathrm{C}}_{2}\\ \mathrm{ln}\left|\frac{{\mathrm{x}}^{6}}{{\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}}\right|-\mathrm{ln}\left|{\mathrm{x}}^{8}\right|& =& {\mathrm{C}}_{2}\end{array}$

$\begin{array}{rcl}\mathrm{ln}\left|\frac{1}{{\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}{\mathrm{x}}^{2}}\right|& =& {\mathrm{C}}_{2}\\ \frac{1}{{\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}{\mathrm{x}}^{2}}& =& {\mathrm{e}}^{{\mathrm{C}}_{2}}\\ \frac{1}{{\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}{\mathrm{x}}^{2}}& =& {\mathrm{C}}_{3}\\ {\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}{\mathrm{x}}^{2}& =& \frac{1}{{\mathrm{C}}_{3}}\\ {\left({\mathrm{x}}^{2}-4{\mathrm{y}}^{2}\right)}^{3}{\mathrm{x}}^{2}& =& \mathrm{C}\end{array}$

Therefore, Homogeneous equation for the given equation is ${\mathbf{\left(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{\right)}}^{{\mathbf{3}}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{C}}$

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