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Expert-verified Found in: Page 1 ### 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 9–20, determine whether the equation is exact.If it is, then solve it.$\mathbf{\left(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{\right)}\mathbf{dx}\mathbf{+}\mathbf{\left(}{{\mathbf{x}}}^{{\mathbf{2}}}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{dy}\mathbf{=}\mathbf{0}$

The solution is $y=\left(C-3x\right)/\left({x}^{2-1}\right)$.

See the step by step solution

## Step 1: Evaluate whether the equation is exact

Here$\mathbf{\left(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{\right)}\mathbf{dx}\mathbf{+}\mathbf{\left(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{dy}\mathbf{=}\mathbf{0}$

The condition for exact is$\frac{\partial \mathbf{M}}{\partial \mathbf{y}}=\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$ .

$\mathbf{M}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\phantom{\rule{0ex}{0ex}}\mathbf{N}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{\partial }\mathbf{M}}{\mathbf{\partial }\mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{=}\frac{\mathbf{\partial }\mathbf{N}}{\mathbf{\partial }\mathbf{x}}\phantom{\rule{0ex}{0ex}}$

This equation is exact.

## Step 2: Find the value of F(x, y)

Here

$\mathbf{M}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\phantom{\rule{0ex}{0ex}}\mathbf{F}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}\int \mathbf{M}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}\int \mathbf{\left(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{\right)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}$

## Step 3: Determine the value of g(y)

$\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{N}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\phantom{\rule{0ex}{0ex}}\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\phantom{\rule{0ex}{0ex}}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{y}\phantom{\rule{0ex}{0ex}}$

Now$\mathbf{F}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}$

${\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{C}\phantom{\rule{0ex}{0ex}}\left({x}^{2}-1\right)\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{=}\mathbf{C}\phantom{\rule{0ex}{0ex}}\left({x}^{2}-1\right)\mathbf{y}\mathbf{=}\mathbf{C}\mathbf{-}\mathbf{3}\mathbf{x}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{=}\left(C-3x\right)\mathbf{/}\left({x}^{2}-1\right)$

Therefore, the solution is${\mathbit{y}}{\mathbf{=}}\left(C-3x\right){\mathbf{/}}\left({x}^{2}-1\right)$ ${\mathbit{y}}{\mathbf{=}}\left(C-3x\right){\mathbf{/}}\left({x}^{2}-1\right)$

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