Q25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 21–26, solve the initial value problem.

 $\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{x}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{)}\mathbf{dy}\mathbf{,}\mathbf{y}\mathbf{(}\mathit{\pi }\mathbf{)}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The solution is $\mathbf{sin} \mathbf{x}\mathbf{-}\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{=}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\pi \mathbf{-}\mathbf{1}$ .

Step 1: Evaluate the equation is exact

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

 

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

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

 

 

 This equation is not exact.

Step 2: Find the solution

But this equation can be separable.

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

  

 Integrating on both sides, the result is 

 $\mathbf{-}\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{x}\mathbf{=}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\mathbf{C}$

 

Therefore, the solution of the differential equation is$\mathbf{-}\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{x}\mathbf{=}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\mathbf{C}$  

Step 3: Apply the initial conditions

Apply the initial conditions $\mathbf{y}\mathbf{(}\pi \mathbf{)}\mathbf{=}\mathbf{1}$.

 $$\begin{gathered} \mathbf{-}\pi \mathbf{cos} \pi \mathbf{+} \mathbf{sin} \pi \mathbf{=}\mathbf{ln} \left|\mathbf{1}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{+}\mathbf{C} \\ \mathbf{C}\mathbf{=}\pi \mathbf{-}\mathbf{1} \end{gathered}$$

 The solutions is 

 .

 $$\begin{gathered} \mathbf{-}\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{x}\mathbf{=}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\pi \mathbf{-}\mathbf{1} \\ \mathbf{sin} \mathbf{x}\mathbf{-}\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{=}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\pi \mathbf{-}\mathbf{1} \end{gathered}$$

Hence the solution is  $\mathbf{sin} \mathbf{x}\mathbf{-}\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{=}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\pi \mathbf{-}\mathbf{1}$