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 the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.
$(2 x + {yx}^{- 1}) dx + (xy - 1) dy = 0$

The given equation is having an integrating factor that is a function of x alone.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

General form of separable, linear, exact or integrating factors

Separable equation: If the right-hand side of the equation $\frac{dy}{dx} = f (x, y)$ can be expressed as a function g(x) that depends only on x times a function p(y) that depends only on y, then the differential equation is called separable.

Linear equation: Standard form of linear equation is $\frac{dy}{dx} + P (x) y = Q (x)$ .

Exact differential form: The differential form $M (x, y) dx + N (x, y) dy$ is said to be exact in a rectangle R if there is a function such that

$\frac{\partial F}{\partial x} (x, y) = M (x, y) and \frac{\partial F}{\partial y} (x, y) = N (x, y)$

Special integrating factors: If $\frac{(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})}{N}$ is continuous and depends only on x. If $\frac{(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})}{M}$ is continuous and depends only on y.

Evaluate the given equation

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

Evaluate it.

$\begin{array}{l} (2 x + {yx}^{- 1}) dx + (xy - 1) dy = 0 \\ \frac{dy}{dx} = \frac{2 x + {yx}^{- 1}}{xy - 1} \end{array}$

Compare the given equation with general form of separable and linear equation.

So, the given equation is neither separable nor linear.

Testing for exactness

Given, $(2 x + {yx}^{- 1}) dx + (xy - 1) dy = 0$

Let $M = 2 x + {yx}^{- 1}, N = xy - 1$

Then,

$\begin{matrix} \frac{\partial M}{\partial y} = x^{- 1} \\ \frac{\partial N}{\partial x} = y \end{matrix}$

So, $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$

Therefore, the given equation is not exact.

Computing integrating factor

If $\frac{(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})}{N}$ then the given function is x alone.

If $\frac{(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})}{M}$ then the given function is y alone.

Then, substitute the values to prove it.

$\begin{matrix} \frac{(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})}{N} = \frac{x^{- 1} - y}{xy - 1} \\ = \frac{\frac{1 - xy}{x}}{xy - 1} \\ = - \frac{(xy - 1)}{x (xy - 1)} \\ = - \frac{1}{x} \end{matrix}$ $\begin{matrix} \frac{(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})}{N} = \frac{x^{- 1} - y}{xy - 1} \\ = \frac{\frac{1 - xy}{x}}{xy - 1} \\ = - \frac{(xy - 1)}{x (xy - 1)} \\ = - \frac{1}{x} \end{matrix}$

So, we obtain an integrating factor that is a function of x alone.

Hence, the given equation is having an integrating factor that is a function of x alone.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems , identify the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.
$(2 x + {yx}^{- 1}) dx + (xy - 1) dy = 0$

Step by Step Solution

General form of separable, linear, exact or integrating factors

Evaluate the given equation

Testing for exactness

Computing integrating factor

Most popular questions for Math Textbooks

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Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems , identify the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.2x+yx-1dx+xy-1dy=0

Step by Step Solution

General form of separable, linear, exact or integrating factors

Evaluate the given equation

Testing for exactness

Computing integrating factor

Most popular questions for Math Textbooks

Want to see more solutions like these?

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In Problems , identify the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.
$(2 x + {yx}^{- 1}) dx + (xy - 1) dy = 0$