Q19E - 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.

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

Step-by-Step Solution

Answer

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

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 $\frac{\partial \mathbf{M}}{\partial \mathbf{y}}=\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$ .

 

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

 This equation is exact.

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

Here

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right) \\ \mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}\int \left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right)\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{xy}\mathbf{\right)}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \end{gathered}$$

Step 3: Determine the value of g(y)

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

 

Therefore, the solution of the differential equation is 

  

 $$\begin{gathered} {\text{x}}^{\text{2}}{\text{ + tan}}^{\text{ - 1}}{\text{(xy) - y}}^{\text{2}}\text{ = C} \\ {\text{x}}^{\text{2}}{\text{ - y}}^{\text{2}}\text{ + arctan}\left(\text{xy}\right)\text{ = C} \end{gathered}$$

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