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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 9–20, determine whether the equation is exact.If it is, then solve it. $\left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right){\mathbf{dx}}{\mathbf{+}}\left(\frac{\mathbf{x}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\mathbf{-}\mathbf{2}\mathbf{y}\right){\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

The solution is ${\text{x}}^{\text{2}}{\text{- y}}^{\text{2}}\text{+ arctan}\left(\text{xy}\right)\text{= C}$.

See the step by step solution

## Step 1: Evaluate whether the equation is exact

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

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

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

This equation is exact.

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

Here

$\mathbf{M}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}\left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right)\phantom{\rule{0ex}{0ex}}\mathbf{F}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}\int \mathbf{M}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}\int \left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right)\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{xy}\mathbf{\right)}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}$

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

Now $\mathbf{F}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{xy}\mathbf{\right)}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}$

Therefore, the solution of the differential equation is

${\text{x}}^{\text{2}}{\text{+ tan}}^{\text{- 1}}{\text{(xy) - y}}^{\text{2}}\text{= C}\phantom{\rule{0ex}{0ex}}{\text{x}}^{\text{2}}{\text{- y}}^{\text{2}}\text{+ arctan}\left(\text{xy}\right)\text{= C}\phantom{\rule{0ex}{0ex}}$

Hence the solution is ${{\mathbf{\text{x}}}}^{{\mathbf{\text{2}}}}{{\mathbf{\text{- y}}}}^{{\mathbf{\text{2}}}}{\mathbf{\text{+ arctan}}}\mathbf{\left(}\mathbf{\text{xy}}\mathbf{\right)}{\mathbf{\text{= C}}}$

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