Question

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

$\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)}\mathbf{dx}\mathbf{-}\mathbf{(}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The solution is $\text{sin}\text{x}\text{cos}{\text{y +x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{= C}$

Step-by-step solution

Evaluate whether the equation is exact

Here$\mathbf{(}\mathbf{cos}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)}\mathbf{dx}\mathbf{-}\mathbf{(}\mathbf{sin}\mathbf{x}\mathbf{sin}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

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

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{cos}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)} \\ \mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{sin}\mathbf{x}\mathbf{sin}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{)} \end{gathered}$$

$\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{cos}\mathbf{x}\mathbf{sin}\mathbf{y}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$

This equation is exact.

Find the value of F (x, y)

Here

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{cos}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)} \\ \mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}\int \mathbf{(}\mathbf{cos}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}\mathbf{sin}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \end{gathered}$$

determine the value of g(y)

$$\begin{gathered} \frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)} \\ \mathbf{-}\mathbf{sin}\mathbf{x}\mathbf{sin}\mathbf{y}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{sin}\mathbf{x}\mathbf{sin}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{)} \\ \mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{y} \\ \mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}} \end{gathered}$$

Now $\text{F(x,y) = sin}\text{x}\text{cos}{\text{y +x}}^{\text{2}}{\text{-y}}^{\text{2}}{\text{=C}}_1$

The solution of the differential equation is$\text{sin}\text{x}\text{cos}{\text{y +x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{= C}$

Hence the solution is$\mathbf{\text{sin}}\mathbf{}\mathbf{\text{x}}\mathbf{}\mathbf{\text{cos}}\mathbf{}{\mathbf{\text{y +x}}}^{\mathbf{\text{2}}}{\mathbf{\text{-y}}}^{\mathbf{\text{2}}}\mathbf{\text{= C}}$$\text{sin}\text{x}\text{cos}{\text{y +x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{= C}$