Question

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y^{\prime} \sin x=y \ln y\) \(y=e\) when \(x=\pi / 3\)

Short answer

General solution: \( y = e^{C} (\csc x + \cot x) \). Particular solution: \( y = \frac{e}{\sqrt{3}} (\csc x + \cot x) \).

Step-by-step solution

- Rewrite the Differential Equation

The given differential equation is: \[ y' \ \frac{dy}{dx} \sin x = y \ln y \]. First, rewrite it in a separable form.

- Separate the Variables

Separate the variables by dividing both sides by \(y \ln y \sin x\) : \[ \frac{1}{y \ln y} \frac{dy}{dx} = \frac{1}{\sin x} \]. Next, rearrange to: \[ \frac{1}{y \ln y} dy = \frac{dx}{\sin x} \].

- Integrate Both Sides

Integrate both sides of the equation: \[\int \frac{1}{y \ln y} dy = \int \frac{dx}{\sin x} \].

- Solve Each Integral

Complete the integrations: \[ \int \frac{1}{y \ln y} dy = \int \frac{dx}{\sin x} \]. Use substitution \( u = \ln y \) -> \( du = \frac{1}{y} dy \): \[ \int \frac{1}{u} du = \int \frac{dx}{\sin x} \]. The integration results in: \[ \ln |\ln y| = \ln |\csc x + \cot x| + C \].

- Find the General Solution

Solve for \(y\): \[ \ln y = e^{C'} (\csc x + \cot x) \]. Let \(C' = e^C\): \[ y = e^{C} (\csc x + \cot x) \].

- Apply the Boundary Condition

Use the boundary condition \( y = e \) when \( x = \frac{\pi}{3} \): \[ e = e^{C} (\csc \frac{\pi}{3} + \cot \frac{\pi}{3}) \].

- Solve for C

Evaluate the terms: \[ \csc \frac{\pi}{3} = 2/\sqrt{3}, \cot \frac{\pi}{3} = 1/\sqrt{3} \]. So: \[ e = e^{C} (\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}) = e^{C} \sqrt{3} \]. Solve for C: \[ e^{C} = \frac{e}{\sqrt{3}} \] -> \[ C = 1 - \ln \sqrt{3} \].

- Write the Particular Solution

Substitute the value of C back into the general solution: \[ y = e^{1 - \ln \sqrt{3}} (\csc x + \cot x) \]. Simplify for particular solution: \[ y = \frac{e}{\sqrt{3}} (\csc x + \cot x) \].

Key concepts

differential equations

Differential equations involve unknown functions and their derivatives. They describe various phenomena in engineering, physics, economics, and more.
They are powerful tools for modeling and understanding the behavior of systems.
In this exercise, the differential equation is represented as:
\[ y' \sin(x) = y \ln(y). \]
This equation relates the derivative of the unknown function y with respect to x to y itself.

integration techniques

Integration techniques are essential for solving differential equations. They involve finding the antiderivative or integral of a function.
In separable differential equations, we often use substitution and basic integration rules.
In this exercise, we have:

• Rewriting the equation to make it separable: \[ \frac{1}{y \ln y} \frac{dy}{dx} = \frac{1}{\sin x}. \]
• Separating the variables: \[ \frac{1}{y \ln y} dy = \frac{dx}{\sin x}. \]
• Integrating both sides: \[ \int \frac{1}{y \ln y} dy = \int \frac{dx}{\sin x}. \]
• Using substitution to simplify: \[ u = \ln y, \ du = \frac{1}{y} dy \] on the left side.

This simplifies integration and helps us find the general solution.

boundary conditions

Boundary conditions specify the values of the solution at certain points. These are crucial for finding particular solutions.
In this exercise, the condition is:
\(y = e\) when \(x = \frac{\pi}{3}\).
This means that at \(x = \frac{\pi}{3}\), the solution y must equal the constant e.

• We use this information to find the constant C in the general solution.
  
• Substituting these values helps us tailor the general solution to the particular needs of the problem.
  

Boundary conditions turn the general solution into a specific answer linked to initial or border values.

general solution

The general solution contains an arbitrary constant and represents a family of solutions. For this differential equation, we first find:
\( \ln y = e^{C'} (\csc x + \cot x) \)
where \(C'\) is the arbitrary constant.
Properties of the general solution:

• Represents all possible solutions.
• Involves integration constants.
• Before applying boundary conditions, it encompasses every scenario that fits the differential equation.

In simplified form, integrating both sides gives us:
\( y = e^C (\csc x + \cot x) \).

particular solution

A particular solution meets the initial or boundary conditions provided in the problem. We use the information from the conditions:
\( y = e \) when \( x = \frac{\pi}{3} \).
Given the general solution:
\( y = e^C (\csc x + \cot x) \)
and boundary conditions, we find C:
\( e = e^C (\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}) = e^C \sqrt{3} \)
Simplifying, we get:

• \(C = 1 - \ln \sqrt{3} \)
• Particular solution: \( y = \frac{e}{\sqrt{3}} (\csc x + \cot x) \)

This matches the solution to the initial conditions.

separable differential equations

Separable differential equations can be rearranged to isolate variables on different sides of the equation. This makes them easier to solve:
The first step is rewriting the given equation:

• \( y' \sin x = y \ln y \)
• Separate the variables: \( \frac{1}{y \ln y} \frac{dy}{dx} = \frac{1}{\sin x} \)
• Rearrange: \( \frac{1}{y \ln y} dy = \frac{dx}{\sin x} \)

Simplifies integration:
Use substitution and integration rules:

• Integrate each part separately: \( \int \frac{1}{y \ln y} dy = \int \frac{dx}{\sin x} \)
• Find the general solution and then apply boundary conditions.

This method is efficient and essential for solving simple differential equations.