Question

Solve the differential equation $$ \sin x \frac{d y}{d x}+2 y \cos x=1 $$ subject to the boundary condition \(y(\pi / 2)=1\).

Short answer

The solution is \[ y = - \csc x \cot x + \csc^2 x \]

Step-by-step solution

- Rewrite the Differential Equation

First, observe the given differential equation: \[ \sin x \frac{dy}{dx} + 2y \cos x = 1 \]Rewrite the differential equation in the standard linear form: \[ \frac{dy}{dx} + P(x)y = Q(x) \]Divide both sides of the equation by \sin x: \[ \frac{dy}{dx} + \frac{2 \cos x}{\sin x} y = \frac{1}{\sin x} \]Since \frac{2 \cos x}{\sin x} = 2 \cot x, the equation becomes: \[ \frac{dy}{dx} + 2 \cot x y = \csc x \]

- Identify Integrating Factor

The standard linear differential equation form is \[ \frac{dy}{dx} + P(x)y = Q(x) \]where \( P(x) = 2 \cot x \).The integrating factor (IF) is determined by: \[ IF = e^{\int P(x) dx} \]Substitute \( P(x) = 2 \cot x \) to find the integrating factor: \[ IF = e^{\int 2 \cot x dx} \]Since \int \cot x \, dx = \log(|\sin x|), we get: \[ IF = e^{2 \log(|\sin x|)} = |\sin x|^2 \]

- Multiply Through by the Integrating Factor

Multiply every term of the differential equation \[ \frac{dy}{dx} + 2 \cot x y = \csc x \]by the integrating factor \( |\sin x|^2 \): \[ |\sin x|^2 \frac{dy}{dx} + 2 |\sin x|^2 \cot x y = |\sin x|^2 \csc x \]This simplifies to: \[ \sin^2 x \frac{dy}{dx} + 2 y \sin x \cos x = \sin x \]

- Simplify and Integrate

Notice the left side can be rewritten using the product rule as: \[ \frac{d}{dx}(y \sin^2 x) = \sin x \]Thus, on integrating both sides with respect to \(x\): \[ y \sin^2 x = \int \sin x \, dx \]Integrate the right side: \[ y \sin^2 x = - \cos x + C \]

- Solve for y

Isolate \( y \) by dividing both sides by \( \sin^2 x \): \[ y = - \frac{\cos x}{\sin^2 x} + \frac{C}{\sin^2 x} \]Thus: \[ y = - \csc x \cot x + C \csc^2 x \]

- Apply Boundary Condition

Use the boundary condition \( y(\pi / 2) = 1 \): \[ 1 = - \csc(\pi/2) \cot(\pi/2) + C \csc^2(\pi/2) \]Recall values at \( x = \pi/2 \): \csc(\pi/2) = 1, \cot(\pi/2) = 0. Then: \[ 1 = 0 + C \]So, \( C = 1 \) and substitute back: \[ y = - \csc x \cot x + \csc^2 x \]

Key concepts

Linear Differential Equations

To solve differential equations, it's helpful to first recognize the type of equation you're dealing with. A linear differential equation has the standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \). This form is useful because it can be solved using specific techniques like the integrating factor method. When you encounter an equation like the given problem, you need to rearrange it to fit this standard linear form. Here, we divide the entire equation by \( \sin x \) to isolate the derivative \( \frac{dy}{dx} \). We thus transform our original equation:

Integrating Factor

An important method for solving linear differential equations is using an integrating factor (IF). The integrating factor makes the equation integrable. For an equation in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is found by: \( IF = e^{\int P(x) dx} \). In our case, we had \( P(x) = 2 \cot x \). Performing the integration: \( IF = e^{\int 2 \cot x \, dx} \), which simplifies to: \( IF = |\sin x|^2 \). This is then multiplied through the entire differential equation to make it easily integrable.

Initial Value Problems

Initial value problems require you to find a specific solution that passes through a given point - in other words, you must satisfy a given condition, \( y(x_0) = y_0 \). In our example, the boundary condition is \( y(\frac{\pi}{2}) = 1 \). After solving the general equation, you substitute \( \frac{\pi}{2} \) for \( x \) in your solution to find specific constants. This step ensures that the solution fits the initial condition exactly, giving you a unique solution to the problem.

Trigonometric Functions

Trigonometric functions often appear in differential equations, requiring you to use their specific properties and integrals. Knowing integrals, such as \( \int \cot x \, dx = \log(|\sin x|) \), is essential. Additionally, understanding the relationships and identities between these functions, like \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \), is crucial for simplifying and solving your equations. In our example, we used these properties to find the integrating factor and to eventually isolate and solve for \( y \). Knowledge of trigonometric values, especially at specific points like \( \pi/2 \), is also very helpful.