Problem 9 - Advanced Engineering Mathematics

Question

In Problems \(7-10, x=c_{1} \cos t+c_{2} \sin t\) is a two-parameter family of solutions of the second-order DE \(x^{\prime \prime}+x=0\). Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. $$ x(\pi / 6)=\frac{1}{2}, \quad x^{\prime}(\pi / 6)=0 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{2\sqrt{3}}{5} \cos t - \frac{1}{5} \sin t \).

Step 1: Identify the General Solution

The given differential equation is \( x'' + x = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients. The general solution is given as \( x = c_1 \cos t + c_2 \sin t \). This is the form of the two-parameter family of solutions.

Step 2: Compute the First Derivative

To apply the initial conditions, we need to compute the first derivative of the solution. Differentiate \( x = c_1 \cos t + c_2 \sin t \) with respect to \( t \):\[ x' = -c_1 \sin t + c_2 \cos t \].

Step 3: Apply the First Initial Condition

The first initial condition is \( x(\pi / 6) = \frac{1}{2} \). Substitute \( t = \pi / 6 \) into the general solution:\[ c_1 \cos(\pi / 6) + c_2 \sin(\pi / 6) = \frac{1}{2} \].Since \( \cos(\pi / 6) = \frac{\sqrt{3}}{2} \) and \( \sin(\pi / 6) = \frac{1}{2} \), the equation becomes:\[ c_1 \frac{\sqrt{3}}{2} + c_2 \frac{1}{2} = \frac{1}{2} \].

Step 4: Apply the Second Initial Condition

The second initial condition is \( x'(\pi / 6) = 0 \). Substitute \( t = \pi / 6 \) into the derivative:\[ -c_1 \sin(\pi / 6) + c_2 \cos(\pi / 6) = 0 \].With \( \cos(\pi / 6) = \frac{\sqrt{3}}{2} \) and \( \sin(\pi / 6) = \frac{1}{2} \), the equation becomes:\[ -c_1 \frac{1}{2} + c_2 \frac{\sqrt{3}}{2} = 0 \].

Step 5: Solve the System of Equations

You now have two equations:1. \( c_1 \frac{\sqrt{3}}{2} + c_2 \frac{1}{2} = \frac{1}{2} \)2. \( -c_1 \frac{1}{2} + c_2 \frac{\sqrt{3}}{2} = 0 \)Solve the first equation for \( c_2 \):\[ c_2 = 1 - \frac{\sqrt{3}}{2}c_1 \] (from the first equation)Substitute \( c_2 \) into the second equation:\[ -c_1 \frac{1}{2} + (1 - \frac{\sqrt{3}}{2}c_1)\frac{\sqrt{3}}{2} = 0 \]Solve this equation to find \( c_1 \):\[ -c_1 \frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{3}{4}c_1 = 0 \]\[ \frac{2\sqrt{3}}{4} = c_1 ( \frac{1}{2} + \frac{3}{4} ) \]\[ c_1 \frac{5}{4} = \frac{2\sqrt{3}}{4} \]\[ c_1 = \frac{2\sqrt{3}}{5} \]Now, use this to find \( c_2 \):\[ c_2 = 1 - \frac{\sqrt{3}}{2} \cdot \frac{2\sqrt{3}}{5} \]\[ = 1 - \frac{6}{5} \]\[ = -\frac{1}{5} \].

Step 6: Solution for the IVP

Substitute \( c_1 \) and \( c_2 \) back into the general solution:\[ x = \frac{2\sqrt{3}}{5} \cos t - \frac{1}{5} \sin t \].This is the particular solution that satisfies the initial value problem.

Key Concepts

Initial Value Problem

An "Initial Value Problem" is a type of differential equation problem where the solution is not only required to satisfy the differential equation but also one or more initial conditions. These initial conditions specify the value of the solution and possibly its derivatives at a particular point in time. Initial value problems are common in real-world applications where a specific initial state is known, and the task is to determine how this state evolves over time.
For example, in our exercise, we are given the differential equation \( x'' + x = 0 \) along with initial conditions \( x(\pi / 6) = \frac{1}{2} \) and \( x'(\pi / 6) = 0 \). The goal here is to find a unique function \( x(t) \) that satisfies both the differential equation and these initial conditions.
This problem-solving approach involves using the given initial conditions to find the specific values of the constants in the general solution of the differential equation. These initial values "anchor" the solution, providing a unique path that the solution follows as time progresses.

Linear Homogeneous Differential Equation

A "Linear Homogeneous Differential Equation" is a differential equation in which the dependent variable and all its derivatives appear linearly. Additionally, there is no constant term or function of the independent variable added on the right side of the equation, meaning the equation is set to zero, hence "homogeneous."
In simpler terms, an equation like \( x'' + x = 0 \) is linear because the terms involve either the function \( x \) or its derivatives but do not involve any products or nonlinear operations such as squares of \( x \) or \( x' \). It is homogeneous because the equation equals zero.
Linear homogeneous differential equations often appear in physics and engineering because their simplicity allows for straightforward solutions. The general solution to these equations often involves exponential functions, sines, or cosines, reflecting oscillatory or exponential growth/decay behavior.

Constant Coefficients

In the realm of differential equations, "Constant Coefficients" refer to the fact that the coefficients multiplying the terms of the equation do not change; they remain constant. This is starkly different from differential equations with variable coefficients, where the coefficients are functions of the independent variable.
The equation \( x'' + x = 0 \) is an example of a linear homogeneous differential equation with constant coefficients because the coefficients (in this case, the implicit coefficient of 1 for both \( x'' \) and \( x \)) are constants.
Such constant coefficient equations can be systematically solved to find their general solutions. These solutions typically use strategies involving characteristic equations or the method of undetermined coefficients. The constant nature of the coefficients simplifies the problem, making this class of equations a fundamental subject in differential equation studies for beginners.