Question

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

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

Short 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-by-step solution

     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.

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

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 role="math" localid="1664190043647" $\mathbf{sin}\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{cos}\mathbf{x}\mathbf{=}\mathbf{ln}\left|\mathbf{y}\right|\mathbf{+}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{+}\pi \mathbf{-}\mathbf{1}$