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

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

See the step by step solution

## General form of separable, linear, exact or integrating factors

• Separable 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 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 $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(\mathbf{x}\right)$.

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

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

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

## Evaluate the given equation

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

Evaluate it.

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

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, $\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

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

Then,

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

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

Therefore, the given equation is not exact.

## Computing integrating factor

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

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

Then, substitute the values to prove it.

$\begin{array}{c}\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{y}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\frac{\frac{\mathbf{1}\mathbf{-}\mathbf{xy}}{\mathbf{x}}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\mathbf{-}\frac{\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}{\mathbf{x}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\end{array}$$\begin{array}{c}\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{y}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\frac{\frac{\mathbf{1}\mathbf{-}\mathbf{xy}}{\mathbf{x}}}{\mathbf{xy}\mathbf{-}\mathbf{1}}\\ \mathbf{=}\mathbf{-}\frac{\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}{\mathbf{x}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\end{array}$

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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