Question

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

If it is, then solve it.

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

Short answer

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

Step-by-step solution

Evaluate whether the equation is exact

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

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

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y} \\ \mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y} \\ \frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}1\mathbf{=}1\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}} \end{gathered}$$

This equation is exact.

Find the value of F (x, y)

Here

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

Determine the value of g(y)

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

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

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

Therefore, the solution isrole="math" localid="1664170796558" ${\mathbf{\text{x}}}^{\mathbf{\text{2}}}{\mathbf{\text{+ yx -y}}}^{\mathbf{\text{2}}}\mathbf{}\mathbf{\text{=}}\mathbf{}\mathbf{\text{C}}$