Problem 4 - Advanced Engineering Mathematics

Question

The given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. $$ \begin{aligned} &y=c_{1}+c_{2} \cos x+c_{3} \sin x,(-\infty, \infty) ; y^{\prime \prime \prime}+y^{\prime}=0 \\ &y(\pi)=0, y^{\prime}(\pi)=2, y^{\prime \prime}(\pi)=-1 \end{aligned} $$

Step-by-Step Solution

Answer

The specific solution is \(y = -1 - \cos x - 2 \sin x\).

Step 1: Assess the General Solution

The given family of functions is \(y = c_1 + c_2 \cos x + c_3 \sin x\). First, recognize that this is the general solution for the differential equation \(y''' + y' = 0\). Our task is to determine constants \(c_1\), \(c_2\), and \(c_3\) using the initial conditions given.

Step 2: Establish Initial Conditions Equations

Substitute \(x = \pi\) into the general solution and its derivatives:- From \(y(\pi) = 0\), we have: \[ c_1 + c_2 \cos \pi + c_3 \sin \pi = 0 \] Simplifying, \(c_1 - c_2 = 0\).- Differentiating, \(y' = -c_2 \sin x + c_3 \cos x\), and from \(y'(\pi) = 2\):\[ -c_2 \sin \pi + c_3 \cos \pi = 2 \] Simplifying, \(-c_3 = 2\) therefore, \(c_3 = -2\).- Second derivative, \(y'' = -c_2 \cos x - c_3 \sin x\), and from \(y''(\pi) = -1\):\[ -c_2 \cos \pi - c_3 \sin \pi = -1 \] Simplifying, \(c_2 = -1\).

Step 3: Solve for Remaining Constant

Using the equation from the first initial condition, \(c_1 - c_2 = 0\):- Substitute \(c_2 = -1\) into \(c_1 - c_2 = 0\): \[ c_1 + 1 = 0 \]- Solve for \(c_1\), giving \(c_1 = -1\).

Step 4: Write the Member of the Family

With the constants found (\(c_1 = -1\), \(c_2 = -1\), \(c_3 = -2\)), the specific solution to the differential equation that satisfies the initial value problem is given by:\[ y = -1 - \cos x - 2 \sin x \]

Key Concepts

General Solution

In mathematics, a general solution to a differential equation represents an entire family of solutions that span all possible behaviors of the system described by the equation. For our problem, the general solution is given as \( y = c_1 + c_2 \cos x + c_3 \sin x \), where \( c_1, c_2, \) and \( c_3 \) are constants. These constants are not immediately known and must be determined based on additional information or conditions. The beauty of the general solution is that it encompasses an infinite number of specific solutions, each corresponding to particular choices of constants.

• \( c_1, c_2, c_3 \) are arbitrary constants.
• Each choice of constants gives a different, specific function from the family of solutions.
• The general solution is defined over a range, in this case from \(-\infty \) to \( \infty \).

Differential Equation

A differential equation involves derivatives of a function and is a key tool for modeling phenomena in various fields such as physics, biology, and engineering. In this problem, the differential equation is \( y''' + y' = 0 \). This equation relates the third derivative and the first derivative of the function \( y \). The goal is to find all functions \( y \) that satisfy this equation across the specified interval.Each order of derivative tells us different things about the function:

• The first derivative, \( y' \), represents the rate of change or the slope of the function.
• The third derivative, \( y''' \), can relate to the function’s jerk or how its rate of acceleration is changing.

Initial Conditions

Initial conditions are specific values given for a function and its derivatives at a particular point, used to determine the exact solution from the general solution. For example, in this exercise, we have initial conditions given by \( y(\pi) = 0 \), \( y'(\pi) = 2 \), and \( y''(\pi) = -1 \).These conditions help us determine the values of the constants \( c_1, c_2, \) and \( c_3 \) in our general solution. By substituting the initial conditions into the general solution and its derivatives, we can form a system of equations. Solving this system allows us to pinpoint the exact member of the family of solutions that fulfills these conditions.

• Each initial condition corresponds to a specific known value of the function or its derivative.
• The number of initial conditions required generally equals the order of the highest derivative in the differential equation.

Solution of Differential Equations

Finding a solution to a differential equation involves determining a function that satisfies both the equation itself and any given initial or boundary conditions. After determining the constants \( c_1, c_2, \) and \( c_3 \) using the initial conditions, we can write the specific solution.For this problem, the solution to the initial value problem was found to be \( y = -1 - \cos x - 2 \sin x \). This function satisfies:

• The differential equation \( y''' + y' = 0 \)
• All given initial conditions: \( y(\pi) = 0 \), \( y'(\pi) = 2 \), and \( y''(\pi) = -1 \)

Thus, solving a differential equation often involves two key steps: identifying the general solution and then applying the initial conditions to find a particular solution, ensuring the function properly models the situation described by those conditions.