Q9E - 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.

$\mathbf{(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

 

Step-by-Step Solution

Answer

The solution is $y=\left(C-3x\right)/\left(x^{2-1}\right)$.

Step 1: Evaluate whether the equation is exact

Here$\mathbf{(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

 

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

 

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3} \\ \mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1} \\ \frac{\mathbf{\partial }\mathbf{M}}{\mathbf{\partial }\mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{=}\frac{\mathbf{\partial }\mathbf{N}}{\mathbf{\partial }\mathbf{x}} \end{gathered}$$

 This equation is exact.

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

Here

 $$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3} \\ \mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}\int \mathbf{(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \end{gathered}$$

Step 3: Determine the value of g(y)

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

 Now$\mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}$

$$\begin{gathered} {\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{C} \\ \left(x^2-1\right)\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{=}\mathbf{C} \\ \left(x^2-1\right)\mathbf{y}\mathbf{=}\mathbf{C}\mathbf{-}\mathbf{3}\mathbf{x} \\ \mathbf{y}\mathbf{=}\left(C-3x\right)\mathbf{/}\left(x^2-1\right) \end{gathered}$$ 

 

 

Therefore, the solution is$\mathit{y}\mathbf{=}\left(C-3x\right)\mathbf{/}\left(x^2-1\right)$ $\mathit{y}\mathbf{=}\left(C-3x\right)\mathbf{/}\left(x^2-1\right)$