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Found in: Page 77

### 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

# $\mathbf{x}\equiv \mathbf{0}$In problems 33-40, Solve the equation given in Problem 2.${\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{{\mathbf{2}}}{\mathbf{dx}}{\mathbf{-}}{\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

$\frac{\mathbf{1}}{\mathbf{4}}\mathbf{log}\left|\frac{\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}}{\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{1}}\right|\mathbf{-}\mathbf{x}\mathbf{=}C$and

See the step by step solution

## Solving the given equation

The given equation is,

${\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{dx}\mathbf{-}\mathbf{dy}\mathbf{=}\mathbf{0}$

Simplify the above equation,

$\begin{array}{l}{\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{dx}\mathbf{=}\mathbf{dy}\\ \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}······\left(\mathbf{1}\right)\end{array}$

Let,

$\begin{array}{l}\mathbf{t}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\\ \frac{\mathbf{dt}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{4}\\ \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{dt}}{\mathbf{dx}}\mathbf{+}\mathbf{4}\end{array}$

Substitute the $\mathbf{t}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}$ and $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{dt}}{\mathbf{dx}}\mathbf{+}\mathbf{4}$ in the equation (1),

$\frac{\mathbf{dt}}{\mathbf{dx}}\mathbf{+}\mathbf{4}\mathbf{=}{\left(\mathbf{t}\right)}^{\mathbf{2}}$

Simplify the above equation,

$\frac{\mathbf{dt}}{\mathbf{dx}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{4}$

Cross multiplication on both sides in the above equation,

$\frac{\mathbf{dt}}{{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{4}}\mathbf{=}\mathbf{dx}$

## Integrating

Taking integration of both sides in the above equation

$\int \frac{\mathbf{dt}}{{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{2}}^{\mathbf{2}}}\mathbf{=}\int \mathbf{dx}$

One knows that,

$\int \frac{\mathbf{dx}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{a}}\mathbf{log}\left|\frac{\mathbf{x}\mathbf{-}\mathbf{a}}{\mathbf{x}\mathbf{+}\mathbf{a}}\right|\mathbf{+}\mathbf{c}$

Solve the above equation,

$\begin{array}{l}\frac{\mathbf{1}}{\mathbf{2}\left(\mathbf{2}\right)}\mathbf{log}\left|\frac{\mathbf{t}\mathbf{-}\mathbf{2}}{\mathbf{t}\mathbf{+}\mathbf{2}}\right|\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{c}\\ \frac{\mathbf{1}}{\mathbf{4}}\mathbf{log}\left|\frac{\mathbf{t}\mathbf{-}\mathbf{2}}{\mathbf{t}\mathbf{+}\mathbf{2}}\right|\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{c}\end{array}$

## Finding the solution

Substitute the $\mathbf{t}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}$ in the above equation,

$\frac{\mathbf{1}}{\mathbf{4}}\mathbf{log}\left|\frac{\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)\mathbf{-}\mathbf{2}}{\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)\mathbf{+}\mathbf{2}}\right|\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{c}$

Simplify the above equation,

$\frac{\mathbf{1}}{\mathbf{4}}\mathbf{log}\left|\frac{\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}}{\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{1}}\right|\mathbf{-}\mathbf{x}\mathbf{=}C$

Also, note that $\mathbf{x}\equiv \mathbf{0}$ is a solution.

Hence the solution is $\frac{\mathbf{1}}{\mathbf{4}}\mathbf{log}\left|\frac{\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}}{\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{1}}\right|\mathbf{-}\mathbf{x}\mathbf{=}C$ and $\mathbf{x}\equiv \mathbf{0}$