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Q-2-2.6-1E

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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

# In problems identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{G}}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$.${\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{{\mathbf{2}}}{ }{\mathbf{dx}}{\mathbf{-}}{\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

The given equation is the form of both homogeneous and Bernoulli.

See the step by step solution

## Step 1: General form of homogeneous, Bernoulli, linear coefficients of the form of y'=G(ax+by).

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 $\frac{\mathbf{y}}{\mathbf{x}}$ alone, then we say the equation is homogeneous.

Equations of the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$

When 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 combination ax + by, where a and b are constants, that is, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$then the substitution z = ax + by transforms the equation into a separable one.

Bernoulli’s equation

A first-order equation that can be written in the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{y}}^{\mathbf{n}}$ , where P(x) and Q(x) are continuous on an interval (a,b) and n is a real number, is called a Bernoulli equation.

Equation of Linear coefficients

We have used various substitutions for y to transform the original equation into a new equation that we could solve. In some cases, we must transform both x and y into new variables, say u and v.

This is the situation for equations with linear coefficients-that is, equations of the form $\left({\mathbf{a}}_{\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{y}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{x}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$.

## Step 2: Evaluate the given equation

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

By Evaluating,

$\begin{array}{rcl}{\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}} \mathbf{dx}\mathbf{-}\mathbf{dy}& \mathbf{=}& \mathbf{0}\\ \frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& {\left(\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\\ & & \end{array}$

Let u = y - 4x. Then, differentiate it to find the value of $\frac{\mathbf{du}}{\mathbf{dx}}$.

$\frac{\mathbf{du}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{4}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{du}}{\mathbf{dx}}\mathbf{+}\mathbf{4}$

Now,

$\begin{array}{rcl}\frac{\mathbf{du}}{\mathbf{dx}}\mathbf{+}\mathbf{4}& \mathbf{=}& {\left(\mathbf{u}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\\ \frac{\mathbf{du}}{\mathbf{dx}}& \mathbf{=}& \left({\mathbf{u}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{u}\mathbf{+}\mathbf{1}\right)\mathbf{-}\mathbf{4}\\ & \mathbf{=}& {\mathbf{u}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}\\ \frac{\mathbf{du}}{{\mathbf{u}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}}& \mathbf{=}& \mathbf{dx}······\left(1\right)\\ & & \end{array}$

## Step 3: Substitution method

Integrating Equation (1),

$\int \frac{\mathbf{du}}{{\mathbf{u}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}}\mathbf{=}\int \mathbf{dx}\phantom{\rule{0ex}{0ex}}\mathbf{ln}\left[{\left(\frac{\mathbf{3}\mathbf{-}\mathbf{u}}{\mathbf{u}\mathbf{+}\mathbf{1}}\right)}^{\frac{\mathbf{1}}{\mathbf{4}}}\right]\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}$

$\begin{array}{rcl}\frac{\mathbf{3}\mathbf{-}\mathbf{u}}{\mathbf{u}\mathbf{+}\mathbf{1}}& \mathbf{=}& {\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\\ \mathbf{-}\mathbf{1}\mathbf{+}\frac{\mathbf{4}}{\mathbf{u}\mathbf{+}\mathbf{1}}& \mathbf{=}& {\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\\ \frac{\mathbf{4}}{\mathbf{u}\mathbf{+}\mathbf{1}}& \mathbf{=}& {\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\mathbf{+}\mathbf{1}\\ \mathbf{u}\mathbf{+}\mathbf{1}& \mathbf{=}& \frac{\mathbf{4}}{{\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\mathbf{+}\mathbf{1}}\\ \mathbf{u}& \mathbf{=}& \frac{\mathbf{4}}{{\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\mathbf{+}\mathbf{1}}\mathbf{-}\mathbf{1}\\ & & \end{array}$

Substitute u = y - 4x.

$\begin{array}{rcl}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}& \mathbf{=}& \frac{\mathbf{4}}{{\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\mathbf{+}\mathbf{1}}\mathbf{-}\mathbf{1}\\ \mathbf{y}& \mathbf{=}& \frac{\mathbf{4}}{{\mathbf{Ce}}^{\mathbf{4}\mathbf{x}}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\\ & & \end{array}$

It seems that the given equation is linear coefficient.

Therefore, the given equation is the form of linear coefficient.