Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve each differential equation. $$ \sin x \frac{d y}{d x}+2 y \cos x=\sin 2 x ; y=2 \text { when } x=\frac{\pi}{6} $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{1}{3} \sin x + \frac{11}{24} \csc^2 x \).

Step 1: Identify the Differential Equation Type

The given equation \( \sin x \frac{dy}{dx} + 2y \cos x = \sin 2x \) is a first-order linear differential equation. We recognize it by its structure \( a(x) \frac{dy}{dx} + b(x) y = g(x) \).

Step 2: Rearrange into Standard Form

Divide the entire equation by \( \sin x \) to express it in the standard linear form: \( \frac{dy}{dx} + \frac{2 \cos x}{\sin x} y = \frac{\sin 2x}{\sin x} \). This simplifies to \( \frac{dy}{dx} + 2 \cot x \, y = 2 \cos x \).

Step 3: Identify the Integrating Factor

For a linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = 2 \cot x \), so \( \mu(x) = e^{\int 2 \cot x \, dx} = e^{2 \ln |\sin x|} = \sin^2 x \).

Step 4: Multiply through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \sin^2 x \) to obtain \( \sin^2 x \frac{dy}{dx} + 2y \sin x \cos x = 2 \sin^2 x \cos x \).

Step 5: Rewrite as a Derivative

Observe that the left side of the equation is now the derivative of the product of \( y \) and the integrating factor: \( \frac{d}{dx} (y \sin^2 x) = 2 \sin^2 x \cos x \).

Step 6: Integrate Both Sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx}(y \sin^2 x) dx = \int 2 \sin^2 x \cos x \, dx \). The left side simplifies to \( y \sin^2 x \), and upon substituting \( \sin^2 x \cdot \cos x = \frac{1}{3} \sin^3 x + C \) on the right, we have \( y \sin^2 x = \frac{1}{3} \sin^3 x + C \).

Step 7: Solve for the Constant of Integration

Use the initial condition \( y=2 \) when \( x = \frac{\pi}{6} \). Substitute to find \( C \): \( 2 \left(\frac{1}{2}\right)^2 = \frac{1}{3} \left(\frac{1}{2}\right)^3 + C \). This simplifies to \( \frac{1}{2} = \frac{1}{24} + C \), leading to \( C = \frac{12}{24} - \frac{1}{24} = \frac{11}{24} \).

Step 8: Express the General Solution

Substitute \( C \) back into the integrated solution to find \( y \): \( y \sin^2 x = \frac{1}{3} \sin^3 x + \frac{11}{24} \). Therefore, \( y = \frac{1}{3} \sin x + \frac{11}{24} \csc^2 x \).

Key Concepts

Integrating Factor

When dealing with first-order linear differential equations, the integrating factor is a key tool. It transforms the equation into a form that is easier to solve. Think of the integrating factor as a clever function we multiply through the entire equation to help us spot patterns and solutions.
For a differential equation in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is calculated as \( e^{\int P(x) \, dx} \).

• In our exercise, we identified the function \( P(x) = 2\cot x \).
• Calculating the integrating factor, we find \( \mu(x) = e^{\int 2 \cot x \, dx} = e^{2 \ln |\sin x|} = \sin^2 x \).

This integrating factor \( \sin^2 x \) essentially groups the terms into a product rule derivative, making the equation solvable through integration. The integrating factor is all about making your differential equation behave nicely!

Initial Condition

An initial condition is a piece of information that allows us to find a specific solution to a differential equation, rather than just a family of solutions. It provides a starting point, helping us pin down the constant of integration that appears after we solve the equation by integration.
In the given problem, the initial condition is \( y = 2 \) when \( x = \frac{\pi}{6} \). Here's how it helps:

• Once we've solved the differential equation and arranged the solutions in terms of \( y \) and \( x \), the initial condition helps us find the constant \( C \) in our general solution.
• Using the initial condition, we substitute \( x \) and \( y \) into our equation and solve for \( C \).

This very specific solution, tailored by the initial condition, fits our initial criteria and gives a practical answer rather than a theoretical possibility. It's a roadmap, guiding us to the precise path we need to take.

Integration

Integration is the process we use to reverse differentiation, bridging the gap between derivatives and the original functions. It's a crucial step in solving differential equations, allowing us to find a family of functions from their derivatives.
In first-order linear differential equations, after applying the integrating factor, you'll notice that the equation can often be rewritten as the derivative of a product. This is where integration steps in:

• We integrate both sides of the equation, so on the left, you capture an expression involving \( y \), and on the right, you tackle a function of \( x \).
• In our exercise, we needed to integrate \( \int \frac{d}{dx}(y \sin^2 x) \, dx \) and \( \int 2 \sin^2 x \cos x \, dx \).

Integration gives the solution a shape, providing the canvas to identify our solution's constants and fit them with initial conditions. Without integration, we would be stuck with derivatives that tell us how functions change, but not what the functions actually are.

Standard Linear Form

In the world of differential equations, the standard linear form is like an organizational blueprint. It arranges the equation into a neat and manageable form, making it easier to apply techniques like integrating factors. The format is \( \frac{dy}{dx} + P(x) y = Q(x) \).
Here's why it's important:

• It helps us immediately identify relevant components like \( P(x) \) and \( Q(x) \), crucial for calculating the integrating factor.
• Transforming a given equation into this form simplifies analysis and problem-solving.

In our exercise, the transformation process involved dividing through by \( \sin x \) to align the given differential equation with the standard form. This allowed the use of the integrating factor method effectively, shedding light on the path to the solution. Understanding and converting to standard linear form is like setting the stage for the solving performance that follows.