Problem 30 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$\frac{d^{2} y}{d \theta^{2}}+y=0, \quad y(\pi / 3)=0, y^{\prime}(\pi / 3)=2$$

Step-by-Step Solution

Answer

The solution is \( y(\theta) = -\sqrt{3} \cos \theta + \sin \theta \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \). This is a linear homogeneous second-order differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

For the homogeneous differential equation \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \), we use the characteristic equation \( r^2 + 1 = 0 \).

Step 3: Solve the Characteristic Equation

Solve \( r^2 + 1 = 0 \). This gives \( r = \pm i \) since \( r = \sqrt{-1} \).

Step 4: Write the General Solution

Using the solutions \( r = \pm i \), the general solution is \( y(\theta) = c_1 \cos \theta + c_2 \sin \theta \).

Step 5: Apply Initial Condition \( y(\pi/3) = 0 \)

Substitute \( \theta = \pi/3 \) into the general solution: \( 0 = c_1 \cos(\pi/3) + c_2 \sin(\pi/3) \). This simplifies to \( 0 = \frac{c_1}{2} + \frac{c_2\sqrt{3}}{2} \).

Step 6: Apply Initial Condition \( y'(\pi/3) = 2 \)

Differentiating the general solution gives \( y'(\theta) = -c_1 \sin \theta + c_2 \cos \theta \). Substitute \( \theta = \pi/3 \), which gives \( 2 = -c_1 \sin(\pi/3) + c_2 \cos(\pi/3) \). This simplifies to \( 2 = -\frac{c_1\sqrt{3}}{2} + \frac{c_2}{2} \).

Step 7: Solve the System of Equations

Solve the system of equations from steps 5 and 6: 1. \( \frac{c_1}{2} + \frac{c_2\sqrt{3}}{2} = 0 \)2. \( -\frac{c_1\sqrt{3}}{2} + \frac{c_2}{2} = 2 \).Multiplying equation 1 by 2 gives \( c_1 + c_2\sqrt{3} = 0 \) so \( c_1 = -c_2\sqrt{3} \). Substitute into equation 2 and solve for \( c_2 \) and \( c_1 \).

Step 8: Calculate Constants \( c_1 \) and \( c_2 \)

Substitute \( c_1 = -c_2\sqrt{3} \) into equation 2: \( -\frac{(-c_2\sqrt{3})\sqrt{3}}{2} + \frac{c_2}{2} = 2 \).This becomes \( \frac{3c_2}{2} + \frac{c_2}{2} = 2 \) or \( 2c_2 = 2 \), yielding \( c_2 = 1 \).Substitute \( c_2 = 1 \) back to find \( c_1 = -\sqrt{3} \).

Step 9: Write the Solution with the Constants

Substitute \( c_1 = -\sqrt{3} \) and \( c_2 = 1 \) into the general solution: \( y(\theta) = -\sqrt{3} \cos \theta + \sin \theta \).

Key Concepts

Initial-Value Problem

An initial-value problem in the realm of differential equations is where you need to find a specific solution to a differential equation that satisfies given conditions called initial conditions. These initial conditions are typically specified values of the unknown function and its derivatives at a certain point.
This exercise presents an initial-value problem where the second-order differential equation is given alongside initial conditions: \( y(\pi/3) = 0 \) and \( y'(\pi/3) = 2 \).

• The first part of solving an initial-value problem involves finding the general solution to the differential equation without accounting for any conditions.
• Subsequently, you apply the initial conditions to this general solution to find specific constants, thus deriving a particular solution.

By satisfying these conditions, the solution is tailored to the problem's specific scenario.

Characteristic Equation

To solve a differential equation like the one in our problem, we often convert the differential equation into a simpler algebraic form known as the characteristic equation. This step is crucial when solving linear differential equations with constant coefficients.

• The given equation \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \) translates into the characteristic equation by substituting derivatives with powers of a variable, usually \( r \).
• In this example, the characteristic equation becomes \( r^2 + 1 = 0 \).

Characteristics equations help identify the nature of the solutions. Here, solving \( r^2 + 1 = 0 \) gives complex roots \( r = \pm i \), indicating solutions that involve trigonometric functions.

Homogeneous Differential Equation

A homogeneous differential equation is one where every term is dependent on the unknown function or its derivatives. There are no standalone terms (constant or explicitly function-free).
In our exercise, \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \) is defined as homogeneous because each term either involves \( y \) or its derivatives. This kind of equation typically has solutions in the form of trigonometric or exponential functions based on the characteristic equation's roots:

• Real roots lead to exponential functions in the solution.
• Complex roots, as we encounter here, give solutions involving sine and cosine functions.

For our problem, the trigonometric nature help us derive a general solution in terms of sine and cosine: \( y(\theta) = c_1 \cos\theta + c_2 \sin\theta \).

Second-Order Differential Equation

Second-order differential equations are equations involving the second derivative of a function. They are central in modeling systems with acceleration, such as physical and mechanical systems.
Our problem features the second-order differential equation \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \). Solving second-order equations usually follows a structured method:

• Identify if the equation is homogeneous or non-homogeneous.
• Derive the characteristic equation and solve it to find its roots.
• Formulate the general solution based on those roots.
• Apply specific initial or boundary conditions to refine the solution to a particular one.

These steps ensure that the obtained solution is robust and correctly aligned with any real-world phenomena the equation represents.