Problem 1 - A First Course in Differential Equations with Modeling Applications

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. $$\begin{aligned}&y=c_{1} e^{x}+c_{2} e^{-x},(-\infty, \infty);\\\&y^{\prime \prime}-y=0, \quad y(0)=0, \quad y^{\prime}(0)=1\end{aligned}$$

Step-by-Step Solution

Answer

The particular solution is \(y = \frac{1}{2} e^x - \frac{1}{2} e^{-x}\).

Step 1: Identify the General Solution

The given general solution for the differential equation \(y'' - y = 0\) is \(y = c_1 e^x + c_2 e^{-x}\).

Step 2: Calculate the First Derivative

Differentiate the general solution \(y = c_1 e^x + c_2 e^{-x}\) with respect to \(x\) to find \(y'\). \[y' = \frac{d}{dx}(c_1 e^x + c_2 e^{-x}) = c_1 e^x - c_2 e^{-x}\]

Step 3: Apply Initial Condition \(y(0) = 0\)

Using the initial condition \(y(0) = 0\), substitute \(x = 0\) into \(y = c_1 e^x + c_2 e^{-x}\):\[y(0) = c_1 e^0 + c_2 e^0 = c_1 + c_2 = 0\]This gives us the equation \(c_1 + c_2 = 0\).

Step 4: Apply Initial Condition \(y'(0) = 1\)

Using the initial condition \(y'(0) = 1\), substitute \(x = 0\) into \(y' = c_1 e^x - c_2 e^{-x}\):\[y'(0) = c_1 e^0 - c_2 e^0 = c_1 - c_2 = 1\]This gives us the equation \(c_1 - c_2 = 1\).

Step 5: Solve the System of Equations

Solve the system of equations obtained from the initial conditions:1. \(c_1 + c_2 = 0\)2. \(c_1 - c_2 = 1\)Add the two equations:\[(c_1 + c_2) + (c_1 - c_2) = 0 + 1\]\[2c_1 = 1\]\[c_1 = \frac{1}{2}\]Substitute \(c_1 = \frac{1}{2}\) into \(c_1 + c_2 = 0\):\[\frac{1}{2} + c_2 = 0\]\[c_2 = -\frac{1}{2}\]

Step 6: Write the Particular Solution

With \(c_1 = \frac{1}{2}\) and \(c_2 = -\frac{1}{2}\), substitute these values back into the general solution:\[y = \frac{1}{2} e^x - \frac{1}{2} e^{-x}\]

Key Concepts

Initial Value Problem

An initial value problem (IVP) in the context of differential equations is a problem where we need to find a specific solution that satisfies both a differential equation and a set of initial conditions. Initial conditions specify the value(s) that the solution must take at particular points. For instance, in our original exercise, the initial conditions given are:

• \(y(0) = 0\) - This means that when \(x = 0\), \(y\) should equal zero.
• \(y'(0) = 1\) - This condition states that the derivative of \(y\) at \(x = 0\) should be one.

These conditions allow us to determine the particular solution of the differential equation from the general solution by solving for any arbitrary constants involved. Without such conditions, the differential equation might have infinitely many solutions.

General Solution

The general solution of a differential equation encompasses a family of functions that contains all possible solutions to the differential equation. This is achieved by including constants that represent degrees of freedom in the solution. In our specific problem, the general solution is:

• \(y = c_1 e^x + c_2 e^{-x}\)

Here, \(c_1\) and \(c_2\) are constants that can take any value, leading to a wide family of solutions. The task is to find the particular values of these constants that satisfy the initial conditions of the problem. Once these constants are determined using the initial values, we obtain the particular solution which is a specific member of this family.

System of Equations

A system of equations is a set of equations with multiple variables where the goal is to find common solutions that satisfy each equation in the set. In our exercise, the result of applying the initial conditions gave us two equations:

• \(c_1 + c_2 = 0\)
• \(c_1 - c_2 = 1\)

These form a system of linear equations in terms of \(c_1\) and \(c_2\). Solving this system involves finding values of \(c_1\) and \(c_2\) that work for both equations simultaneously. By adding these equations, we find \(2c_1 = 1\), leading to \(c_1 = \frac{1}{2}\). Substituting back to find \(c_2\) gives \(c_2 = -\frac{1}{2}\). This process is crucial in reducing the general solution to a single, specific solution.

First and Second Derivative

Differentiation, a fundamental concept in calculus, helps us understand the rate at which a function changes. In our case, the first and second derivatives relate to the differential equation of the problem. Given the general solution:

• First Derivative: \(y' = \frac{d}{dx}(c_1 e^x + c_2 e^{-x}) = c_1 e^x - c_2 e^{-x}\)
• Second Derivative: It's not calculated directly in the solution, but essential for the differential equation \(y'' - y = 0\).

The differential equation requires substituting the second derivative back into the equation. Using the first derivative helps in applying initial conditions and solving the IVP. Understanding both derivatives reveals the intricate behavior of the solution relative to changes in \(x\), allowing us to derive a deeper meaning from our solution.