Problem 42 - 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. $$ y(0)=1, y^{\prime}(\pi)=5 $$

Step-by-Step Solution

Answer

The specific solution is \( y = \cos 3x - \frac{5}{3} \sin 3x \).

Step 1: Understanding the Problem

We are asked to find a specific solution for the differential equation \( y'' + 9y = 0 \) given the family of solutions \( y = c_1 \cos 3x + c_2 \sin 3x \). The specific solution must satisfy the boundary conditions \( y(0) = 1 \) and \( y'(\pi) = 5 \).

Step 2: Boundary Condition at \( x = 0 \)

Substitute \( x = 0 \) into the family of solutions: \[ y(0) = c_1 \cos(3 \times 0) + c_2 \sin(3 \times 0) = c_1 \cdot 1 + c_2 \cdot 0 = c_1 \]Given that \( y(0) = 1 \), we have \[ c_1 = 1 \]

Step 3: Compute the Derivative

Find \( y' = \frac{d}{dx}(c_1 \cos 3x + c_2 \sin 3x) \): \[ y' = -3c_1 \sin 3x + 3c_2 \cos 3x \]

Step 4: Boundary Condition at \( x = \pi \)

Substitute \( x = \pi \) into the derivative solution for the second condition:\[ y'(\pi) = -3c_1 \sin(3 \pi) + 3c_2 \cos(3 \pi) \]Since \( \sin(3\pi) = 0 \) and \( \cos(3\pi) = -1 \): \[ y'(\pi) = 3c_2(-1) = -3c_2 \]With \( y'(\pi) = 5 \), solve for \( c_2 \):\[ -3c_2 = 5 \rightarrow c_2 = -\frac{5}{3} \]

Step 5: Formulating the Specific Solution

Substitute \( c_1 = 1 \) and \( c_2 = -\frac{5}{3} \) back into the family of solutions: \[ y = \cos 3x - \frac{5}{3} \sin 3x \]

Key Concepts

Boundary Conditions

Boundary conditions are critical in solving differential equations, especially second-order ones like the one in our exercise. They help to determine unique solutions from a set of possible solutions. By providing values that the solution must satisfy at specific points, boundary conditions effectively narrow down the choices.
In our problem, two boundary conditions were given:

• At \( x = 0 \), the solution \( y(0) \) must equal 1.
• The derivative \( y'(\pi) \) must equal 5 at \( x = \pi \).

The process began by applying the first boundary condition to find the value of \( c_1 \) and the second one to determine \( c_2 \). Using the given boundary conditions, we can find a unique, specific solution that satisfies the differential equation and meets these criteria.

Trigonometric Solutions

In solving second-order differential equations, trigonometric solutions often appear, especially when dealing with equations that have constant coefficients like \( y'' + 9y = 0 \). These equations commonly result in solutions involving sine and cosine functions due to their periodic nature and their properties when differentiated.
The general soution for our differential equation is given as:

• \( y = c_1 \cos(3x) + c_2 \sin(3x) \)

This form arises because solutions to the characteristic equation of the differential equation are imaginary, resulting in terms involving sine and cosine. These trigonometric functions simplify the process of applying boundary conditions and finding specific solutions.

Specific Solution

Finding a specific solution requires applying the boundary conditions to the general solution obtained from the differential equation. Once we have the general solution, like \( y = c_1 \cos 3x + c_2 \sin 3x \), which contains arbitrary constants \( c_1 \) and \( c_2 \), the task is to determine these constants.
For this exercise, using the boundary condition \( y(0) = 1 \), we set \( c_1 = 1 \) because \( \cos(0) = 1 \) and \( \sin(0) = 0 \). Then, substituting into the derivative for the boundary \( y'(\pi) = 5 \) results in the calculation of \( c_2 = -\frac{5}{3} \).
Hence, the specific solution satisfying both conditions is \( y = \cos(3x) - \frac{5}{3} \sin(3x) \). This solution is unique as it adheres to both the initial differential equation and the given boundary conditions.