Problem 44 - Advanced Engineering Mathematics

Question

\(y=c_{1} \cos 3 x+c_{2} \sin 3 x\) is a two-parameter family of solutions of the second-order DE \(y^{\prime \prime}+9 y=0\). If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. $$ y^{\prime}(\pi / 3)=1, y^{\prime}(\pi)=0 $$

Step-by-Step Solution

Answer

Side conditions lead to unique realization: \( c_2 = -\frac{1}{3} \).

Step 1: Differentiate the General Solution

Start by differentiating the given solution \( y = c_{1} \cos 3x + c_{2} \sin 3x \). Use the derivatives of sine and cosine: \( \frac{d}{dx}[\cos 3x] = -3 \sin 3x \) and \( \frac{d}{dx}[\sin 3x] = 3 \cos 3x \). Thus, the first derivative will be \( y' = -3c_{1} \sin 3x + 3c_{2} \cos 3x \).

Step 2: Apply the First Boundary Condition

Apply the first boundary condition \( y'(\pi/3) = 1 \). Substitute \( \pi/3 \) into the derivative equation: \(-3c_{1} \sin (\pi) + 3c_{2} \cos (\pi) = 1\). Since \( \sin(\pi) = 0\) and \(\cos(\pi) = -1\), you obtain: \( 3c_{2}(-1) = 1\).- Simplify to find \( c_{2} = -\frac{1}{3} \).

Step 3: Apply the Second Boundary Condition

Apply the second boundary condition \( y'(\pi) = 0 \). Substitute \( \pi \) into the derivative equation: \(-3c_{1} \sin (3\pi) + 3c_{2} \cos (3\pi) = 0\). Since \( \sin(3\pi) = 0\) and \( \cos(3\pi) = -1\), you obtain: \( 3c_{2}(-1) = 0\).- Plug in \( c_{2} = -\frac{1}{3} \) from Step 2 and simplify to find \(-\frac{1}{3}(3)(-1) + c_{1}(0) = 0\), confirming that \( c_{2} \) is consistent.

Step 4: Solve for Missing Constant

Use the constistiency to solve for \( c_{1} \). Since \( c_{1} \) cancels out, it suggests that there might be a reinvestigation needed in a previous step if \( c_1 \) wasn't used in obtaining the knowns and a revisitation to clarify missing dependencies is viable. However, the simplification showed no other assignment.

Key Concepts

Boundary Conditions

Boundary conditions are crucial for solving differential equations, especially when looking for specific solutions that satisfy given constraints. These conditions specify the values or behavior of a solution at certain points. In the exercise, we are given two boundary conditions:

• \( y'\left(\frac{\pi}{3}\right) = 1 \)
• \( y'(\pi) = 0 \)

These conditions allow us to determine the solution that fits the exact requirements of the problem. By utilizing these conditions, we can find specific values for our constants, like \( c_1 \) and \( c_2 \) in the differential equation solution, that make the equation satisfy the boundary constraints. This process highlights the importance of boundary conditions, as different conditions would lead to different specific solutions.

Second-Order Differential Equations

Second-order differential equations involve the second derivative of a function. They are used to describe a wide range of phenomena, from mechanical vibrations to electrical circuits. The given exercise involves a second-order differential equation: \( y'' + 9y = 0 \). This is a homogeneous equation, meaning there are no terms independent of the function or its derivatives.
A general solution to this type of equation can usually be expressed as a linear combination of trigonometric or exponential functions. Here, it's expressed using trigonometric functions: \( y = c_1 \cos 3x + c_2 \sin 3x \). The constants \( c_1 \) and \( c_2 \) are determined by applying boundary conditions, which constrain the general solution to specific cases.
Working through such equations helps in understanding how to manipulate and interpret differential equations in real-world scenarios, linking mathematical theory with practical applications.

Trigonometric Functions

Trigonometric functions like sine and cosine play a vital role in solving second-order differential equations. In the context of our equation \( y'' + 9y = 0 \), trigonometric functions represent oscillatory solutions, which often appear in problems involving waves or harmonic oscillators.
The general solution \( y = c_1 \cos 3x + c_2 \sin 3x \) uses these functions to reflect periodic behavior. Differentiating these functions, as seen in the solution steps, requires applying the chain rule, leading to derivatives like \(-3 \sin 3x\) and \(3 \cos 3x \).

• \( \frac{d}{dx}[\cos 3x] = -3 \sin 3x \)
• \( \frac{d}{dx}[\sin 3x] = 3 \cos 3x \)

Understanding these derivatives is essential not only for solving the given problem but also for any quizzes or tests that involve trigonometric functions in differential equations. These functions are tools to model situations where things cyclically repeat over time, such as sound waves or pendulums.