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

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

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

Evaluate it.

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

Now take LCM.

## Step 3: Substitution method

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

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

By Differentiating,

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

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

Now, integrate on both sides.

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

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

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

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