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Q-2-2.6-1E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 76

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $\frac{dy}{dx} = f (x, y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

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

When the right-hand side of the equation $\frac{dy}{dx} = f (x, y)$ can be expressed as a function of the combination ax + by, where a and b are constants, that is, $\frac{dy}{dx} = G (ax + by)$ 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 $\frac{dy}{dx} + P (x) y = Q (x) y^{n}$ , 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 $(a_{1} x + b_{1} y + c_{1}) dx + (a_{2} x + b_{2} y + c_{2}) dy = 0$ .

Step 2: Evaluate the given equation

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

By Evaluating,

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

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

$\frac{du}{dx} = \frac{dy}{dx} - 4 \frac{dy}{dx} = \frac{du}{dx} + 4$

Now,

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

Step 3: Substitution method

Integrating Equation (1),

$\int \frac{du}{u^{2} - 2 u - 3} = \int dx \ln [{(\frac{3 - u}{u + 1})}^{\frac{1}{4}}] = x + C$

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

Substitute u = y - 4x.

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

It seems that the given equation is linear coefficient.

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

Most popular questions for Math Textbooks

In problems, 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ .

$\cos (x + y) dy = \sin (x + y) dx$

Use the method discussed under “Bernoulli Equations” to solve problems .

$\frac{d y}{d x} = \frac{2 y}{x} - x^{2} y^{2}$

Question: (a) Show that the equation $\frac{dy}{dx} = f (x, y)$ is homogeneous if and only if $f (tx, ty) = f (x, y)$ .

(b) A function $H (x, y)$ is called homogeneous of order n if . $H (tx, ty) = t^{n} H (x, y)$

Show that the equation $M (x, y) dx + N (x, y) dy = 0$ is homogeneous if $M (x, y)$ and $N (x, y)$ are both homogeneous of the same order.

Question: In Problems 1-30, solve the equation.

$\frac{dy}{dx} + y \tan x + \sin x = 0$

In problem 7-16, solve the equation. $x \frac{dy}{dx} = \frac{1}{y^{3}}$

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