Problem 39 - 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. $$ y(0)=0, y(\pi / 6)=-1 $$

Step-by-Step Solution

Answer

The solution is \( y = -\sin 3x \).

Step 1: Differentiate the General Solution

First, we differentiate the function \( y = c_1 \cos 3x + c_2 \sin 3x \) with respect to \( x \).The first derivative is: \[ y' = -3c_1 \sin 3x + 3c_2 \cos 3x \]The second derivative is:\[ y'' = -9c_1 \cos 3x - 9c_2 \sin 3x \]

Step 2: Verify the Differential Equation

Substitute \( y \), \( y'' \), and the equation body into the original differential equation \( y'' + 9y = 0 \) to verify it is satisfied:\[ -9c_1 \cos 3x - 9c_2 \sin 3x + 9(c_1 \cos 3x + c_2 \sin 3x) = 0 \]This simplifies to:\[ 0 = 0 \]The verification confirms that the general solution \( y = c_1 \cos 3x + c_2 \sin 3x \) satisfies the differential equation.

Step 3: Apply the Boundary Condition \( y(0) = 0 \)

Substitute \( x = 0 \) into the general solution:\[ y(0) = c_1 \cos 0 + c_2 \sin 0 = c_1 \cdot 1 + c_2 \cdot 0 = c_1 \]Given \( y(0) = 0 \), we obtain:\[ c_1 = 0 \]

Step 4: Apply the Boundary Condition \( y(\pi/6) = -1 \)

With \( c_1 = 0 \), substitute \( x = \pi/6 \) into the solution:\[ y(\pi/6) = c_1 \cos \left(3 \cdot \frac{\pi}{6}\right) + c_2 \sin \left(3 \cdot \frac{\pi}{6}\right) = 0 + c_2 \sin \left(\frac{\pi}{2}\right) \]Since \( \sin(\frac{\pi}{2}) = 1 \), this becomes:\[ y(\pi/6) = c_2 \\]Given \( y(\pi/6) = -1 \), we find:\[ c_2 = -1 \]

Step 5: Write the Particular Solution

Substitute the values of \( c_1 \) and \( c_2 \) into the general solution:\[ y = 0 \cdot \cos 3x - 1 \cdot \sin 3x \]Thus, the particular solution satisfying the boundary conditions is:\[ y = -\sin 3x \]

Key Concepts

Second-Order Differential Equations

Second-order differential equations are crucial in both mathematics and science as they involve derivatives of an unknown function up to the second order. These equations often describe physical systems, such as oscillations, waves, and other dynamic phenomena.

A second-order differential equation has the general form \( a(x) y'' + b(x) y' + c(x) y = f(x) \). In this context, the given example \( y'' + 9y = 0 \) is a homogeneous linear second-order differential equation where the solution revolves around solving for \( y \) that depends on the variable \( x \).

Homogeneous indicates that the function \( f(x) \) is zero, which often simplifies the process of finding a solution as we look for complementary functions that satisfy the equation. The importance lies in understanding how to manipulate and solve for the constant coefficients that usually define the behavior of the system modeled by the equation.

Trigonometric Solutions

In solving second-order differential equations like \( y'' + 9y = 0 \), trigonometric functions often emerge as part of the solution, especially in systems modeling oscillations. The general solution provided here \( y = c_1 \cos 3x + c_2 \sin 3x \) is pivotal because trigonometric functions naturally fit problems involving periodicity and repetitive motion.

• The constants \( c_1 \) and \( c_2 \) are arbitrary, reflecting the two degrees of freedom typical in second-order equations.
• Using identities like \( \cos ax \) and \( \sin ax \) can lead to solutions useful for boundary conditions and initial values.

Understanding the trigonometric components is essential since they provide insight into wave-like behaviors seen in mechanical and electromagnetic systems. The coefficients in the cosine and sine terms (3 in this case) reflect the frequency of oscillations. By mastering these trigonometric solutions, you will be well-equipped to handle a range of real-world applications.

General and Particular Solutions

The concept of general and particular solutions is a foundational piece in understanding differential equations. The general solution of a differential equation encompasses all possible solutions, incorporating arbitrary constants. In our specific case, the general form is \( y = c_1 \cos 3x + c_2 \sin 3x \), which includes arbitrary constants \( c_1 \) and \( c_2 \).

• General solutions are useful for understanding the full breadth of possible behaviors described by the DE.
• Particular solutions are derived by applying specific boundary conditions or initial values.

Boundary conditions like \( y(0) = 0 \) and \( y(\pi/6) = -1 \) help narrow the scope to find a particular solution that satisfies these specific constraints. This process removes ambiguity and provides a precise function from the general solution that meets the given conditions. In this example, the particular solution was found as \( y = - \sin 3x \), representing a single, clearly defined behavior of the system with specified limits at two points.