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Q16E

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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. $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}{\mathbf{=}}\frac{\mathbf{y}\mathbf{\left(}\mathbf{lny}\mathbf{-}\mathbf{lnx}\mathbf{+}\mathbf{1}\mathbf{\right)}}{\mathbf{x}}$

Homogeneous equation for the given equation is ${\mathbf{y}}{\mathbf{=}}{{\mathbf{xe}}}^{{\mathbf{Cx}}}$.

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, $\frac{d\mathrm{y}}{d\mathrm{x}}=\frac{\mathrm{y}\left(\mathrm{lny}-\mathrm{lnx}+1\right)}{\mathrm{x}}$.

Evaluate it.

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

localid="1655201008520"

## Step 3: Substitution method

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

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

By Differentiating,

$\begin{array}{rcl}\frac{d\mathrm{y}}{d\mathrm{x}}& =& \mathrm{v}+\mathrm{x}\frac{d\mathrm{v}}{d\mathrm{x}}\\ \mathrm{vln}\mathrm{v}+\mathrm{v}& =& \mathrm{v}+\mathrm{x}\frac{d\mathrm{v}}{d\mathrm{x}}\\ \mathrm{vln}\mathrm{v}& =& \mathrm{x}\frac{d\mathrm{v}}{d\mathrm{x}}\\ \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}& =& \frac{1}{\mathrm{x}}\mathrm{dx}\end{array}$

## Step 4: Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl}\int \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}& =& \int \frac{1}{\mathrm{x}}\mathrm{dx}\\ \int \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}& =& \mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{1}\end{array}$

Integrate $\begin{array}{rcl}& & \int \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}\end{array}$ separately.

Let us take $\mathrm{w}=\mathrm{ln}\mathrm{v}$. Then, $\mathrm{dv}=\mathrm{v}\mathrm{dw}$

Now,

$\begin{array}{rcl}\int \frac{1}{\mathrm{vw}}\mathrm{vdw}& =& \int \frac{1}{\mathrm{w}}\mathrm{dw}\\ & =& \mathrm{ln}\left|\mathrm{w}\right|\end{array}$

Substitute $\mathrm{w}=\mathrm{ln}\mathrm{v}$.

$\begin{array}{rcl}\int \frac{1}{v\mathrm{ln}v}dv& =& \mathrm{ln}\left|w\right|\\ & =& \mathrm{ln}\left|\mathrm{ln}v\right|\end{array}$

Then,

$\begin{array}{rcl}\mathrm{ln}\left|\mathrm{ln}\mathrm{v}\right|& =& \mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{1}\\ \mathrm{ln}\mathrm{v}& =& {\mathrm{e}}^{\mathrm{ln}\left|\mathrm{x}\right|+{\mathrm{C}}_{1}}\\ \mathrm{ln}\mathrm{v}& =& {\mathrm{xe}}^{{\mathrm{C}}_{1}}\\ \mathrm{ln}\mathrm{v}& =& {\mathrm{xC}}_{2}\\ \mathrm{v}& =& {\mathrm{e}}^{{\mathrm{xC}}_{2}}\\ \mathrm{v}& =& {\mathrm{e}}^{\mathrm{xC}}\end{array}$

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

localid="1655200766733" $\begin{array}{rcl}\frac{\mathrm{y}}{\mathrm{x}}& =& {\mathrm{e}}^{\mathrm{Cx}}\\ \mathrm{y}& =& {\mathrm{xe}}^{\mathrm{Cx}}\end{array}$

Therefore, Homogeneous equation for the given equation is ${\mathbf{y}}{\mathbf{=}}{{\mathbf{xe}}}^{{\mathbf{Cx}}}$.