Q12E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

If it is, then solve it.

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

Step-by-Step Solution

Answer

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

Step 1: 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.

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

Here

 

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

Step 3: 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}$