Problem 17 - Thomas Calculus

Question

Solve the initial value problems. \(\theta \frac{d y}{d \theta}+y=\sin \theta, \quad \theta>0, \quad y(\pi / 2)=1\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{-\cos \theta + \frac{\pi}{2}}{\theta} \).

Step 1: Recognize the Equation Form

The given differential equation \( \theta \frac{d y}{d \theta} + y = \sin \theta \) is a first-order linear ordinary differential equation (ODE). We notice it has the standard form of \( p(\theta) \frac{dy}{d\theta} + q(\theta)y = g(\theta) \), where \( p(\theta) = \theta \), \( q(\theta) = 1 \), and \( g(\theta) = \sin \theta \).

Step 2: Simplify the Equation

To simplify, divide every term by \( \theta \) (since \( \theta > 0 \)). This gives \( \frac{d y}{d \theta} + \frac{1}{\theta} y = \frac{\sin \theta}{\theta} \). Now, the equation is in the standard linear form \( \frac{dy}{d\theta} + P(\theta)y = Q(\theta) \) with \( P(\theta) = \frac{1}{\theta} \) and \( Q(\theta) = \frac{\sin \theta}{\theta} \).

Step 3: Find the Integrating Factor

The integrating factor \( \mu(\theta) \) is given by \( e^{\int P(\theta) d\theta} = e^{\int \frac{1}{\theta} d\theta} = e^{\ln |\theta|} = \theta \) for \( \theta > 0 \).

Step 4: Form the Integral

Multiply the entire equation by the integrating factor \( \theta \) to transform it into a perfect derivative: \( \theta \frac{d y}{d \theta} + y = \sin \theta \). Notice this becomes \( \frac{d}{d\theta}(\theta y) = \sin \theta \).

Step 5: Integrate Both Sides

Integrate the equation \( \frac{d}{d\theta}(\theta y) = \sin \theta \) with respect to \( \theta \):\[ \int \frac{d}{d\theta}(\theta y) \, d\theta = \int \sin \theta \, d\theta \]This simplifies to \[ \theta y = -\cos \theta + C \], where \( C \) is the integration constant.

Step 6: Solve for the Unknown Constant

Use the initial condition \( y(\frac{\pi}{2}) = 1 \) to determine \( C \):\[ \left(\frac{\pi}{2}\right)y \left(\frac{\pi}{2}\right) = -\cos \left(\frac{\pi}{2}\right) + C \rightarrow \frac{\pi}{2}(1) = 0 + C \]So \( C = \frac{\pi}{2} \).

Step 7: Write the General Solution

Substitute \( C = \frac{\pi}{2} \) back into \( \theta y = -\cos \theta + C \):\[ \theta y = -\cos \theta + \frac{\pi}{2} \].

Step 8: Solve for \( y \)

Divide through by \( \theta \) to solve for \( y \):\[ y = \frac{-\cos \theta + \frac{\pi}{2}}{\theta} \].This is the solution to the differential equation.

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation is a foundational concept in calculus and differential equations. Such equations take the form \( a(x) \frac{dy}{dx} + b(x)y = c(x) \), where \( a(x), b(x), \) and \( c(x) \) are functions of \( x \), and \( y \) is the unknown function. The equation involves only the first derivative of \( y \).

This form makes them particularly manageable because they can often be solved using straightforward methods, such as separation of variables or an integrating factor. They are crucial for modeling various natural phenomena and processes in engineering, physics, and economics.

Integrating Factor

An integrating factor is a tool used to solve linear differential equations. It is a function that, when multiplied with the differential equation, transforms it into a form that can be easily integrated. For a first-order linear equation in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \).

This factor simplifies the equation into \( \frac{d}{dx} (\mu(x)y) = \mu(x)Q(x) \), allowing both sides to be integrated directly. This method is widely used because it provides a systematic way to bypass complex algebraic manipulations.

Ordinary Differential Equations

Ordinary differential equations (ODEs) differ from partial differential equations in that they involve functions of a single variable and their derivatives. They form the backbone of mathematical modeling of dynamic systems. ODEs can be categorized by order, linearity, and the presence of coefficients.

First-order ODEs, involving the first derivative, often describe the rates of change and are common in real-world scenarios such as population growth and decay. Solving an ODE usually means finding a function or set of functions that satisfy the given equation.

Initial Conditions

Initial conditions are essential in the context of differential equations, especially when solving initial value problems. These conditions specify the value of the unknown function at a particular point, which is crucial because differential equations have many possible solutions.

In the problem \( \theta \frac{d y}{d \theta} + y = \sin \theta \), with \( y(\pi/2) = 1 \), the initial condition \( y(\pi/2) = 1 \) pinpoints the exact solution by providing a specific starting point. This allows us to determine any constants that appear in the solution, ensuring it meets the problem's requirements.