Problem 1 - Advanced Engineering Mathematics

Question

In Problems \(1-4\), 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. $$ y=c_{1} e^{x}+c_{2} e^{-x},(-\infty, \infty) ; y^{\prime \prime}-y=0, y(0)=0, y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution to the initial-value problem is \(y = \frac{1}{2}e^x - \frac{1}{2}e^{-x}\).

Step 1: Understand the Given Family of Functions

The general solution of the differential equation is \(y = c_1 e^x + c_2 e^{-x}\). This equation is valid for the entire real number line \((-\infty, \infty)\). The differential equation it solves is \(y'' - y = 0\).

Step 2: Apply Initial Conditions

We are given two initial conditions: \(y(0) = 0\) and \(y'(0) = 1\). We'll use these to find the particular values of \(c_1\) and \(c_2\) that make this a solution to the initial-value problem.

Step 3: Compute the Derivative

First, find the derivative of \(y\): \[y' = c_1 e^x - c_2 e^{-x}\]

Step 4: Apply the First Initial Condition

Substitute \(x = 0\) into \(y\) to apply the initial condition \(y(0) = 0\):\[y(0) = c_1 e^0 + c_2 e^0 = c_1 + c_2 = 0\]From this, we have \(c_1 = -c_2\).

Step 5: Apply the Second Initial Condition

Substitute \(x = 0\) into \(y'\) to apply the condition \(y'(0) = 1\):\[y'(0) = c_1 e^0 - c_2 e^0 = c_1 - c_2 = 1\]Since \(c_1 = -c_2\) from Step 4, we have:\[c_1 + c_1 = 1\]\[2c_1 = 1\]\[c_1 = \frac{1}{2}\]By substituting \(c_1\) back into \(c_1 = -c_2\), we find \(c_2 = -\frac{1}{2}\).

Step 6: Write the Particular Solution

Substitute \(c_1\) and \(c_2\) into the general solution to find the particular solution:\[y = \frac{1}{2}e^x - \frac{1}{2}e^{-x}\]

Key Concepts

Initial Value Problem

An initial value problem is a type of mathematical problem where a differential equation is solved along with specific starting conditions. These starting conditions, known as "initial conditions," generally specify the value of the unknown function and possibly its derivatives at a certain point. In our exercise, the differential equation given is \(y'' - y = 0\), and the initial conditions are \(y(0) = 0\) and \(y'(0) = 1\).

• These initial conditions help in determining the exact solution within the family of solutions.
• The goal is to find a unique solution that follows both the differential equation and satisfies these initial conditions.

By incorporating these conditions into the general solution, we isolate the values for the constants in the solution formula, making it specific to our problem.

General Solution

The general solution of a differential equation is a broad solution that includes an arbitrary number of constants, typically denoted as \(c_1, c_2\), etc., which represent an infinite family of potential solutions. In our exercise, the general solution is \(y = c_1 e^x + c_2 e^{-x}\).

• This expression solves the differential equation \(y'' - y = 0\) for a range of \(c_1\) and \(c_2\).
• It represents all possible solutions of the differential equation before specific initial conditions are applied.
• The constants are adjusted later to find the particular solution that fits the initial conditions provided in the problem.

The general solution thus prepares the groundwork for deriving a particular solution.

Particular Solution

A particular solution is derived from a general solution by applying specific initial conditions which result in fixed values for the arbitrary constants. Once these constants are determined, the solution becomes particular to the initial value problem.In our exercise, after applying the initial conditions \(y(0) = 0\) and \(y'(0) = 1\), we found the constants \(c_1 = \frac{1}{2}\) and \(c_2 = -\frac{1}{2}\). When substituted back into the general solution \(y = c_1 e^x + c_2 e^{-x}\), we get the particular solution:\[ y = \frac{1}{2} e^x - \frac{1}{2} e^{-x} \]

• This particular solution satisfies both the differential equation and the initial conditions.
• It represents a single, unique solution that fits the context of the problem.

Thus, the process of finding a particular solution is crucial in solving initial value problems.

Exponential Functions

Exponential functions play a crucial role in the solutions of differential equations, particularly those involving growth and decay, as seen here with \(e^x\) and \(e^{-x}\).

• These functions are characterized by their rates of change, which are proportional to their current values.
• In our differential equation, the exponential functions \(e^x\) and \(e^{-x}\) form the basis of the general solution.
• Their unique property of being their own derivatives makes them particularly useful in solving linear differential equations.

In the context of this exercise, they facilitate the construction of solutions that meet the differential equation's requirements and adapt to the provided initial conditions.