Problem 2 - 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. $$\begin{aligned}&y=c_{1} e^{4 x}+c_{2} e^{-x},(-\infty, \infty);\\\&y^{\prime \prime}-3 y^{\prime}-4 y=0, \quad y(0)=1, \quad y^{\prime}(0)=2\end{aligned}$$

Step-by-Step Solution

Answer

The specific solution is \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \).

Step 1: Understand the Problem

We are given a differential equation \( y'' - 3y' - 4y = 0 \) and a family of solutions \( y = c_{1} e^{4x} + c_{2} e^{-x} \). We need a solution that satisfies the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \).

Step 2: Compute the First Derivative

Differentiate the family of functions. The derivative is \( y' = 4c_{1} e^{4x} - c_{2} e^{-x} \).

Step 3: Apply Initial Conditions

Use the initial conditions to form equations: \( y(0) = c_{1} + c_{2} = 1 \) and \( y'(0) = 4c_{1} - c_{2} = 2 \).

Step 4: Solve for C1 and C2

Solve the system of equations. Adding \( c_{1} + c_{2} = 1 \) and \( 4c_{1} - c_{2} = 2 \) gives \( 5c_{1} = 3 \), so \( c_{1} = \frac{3}{5} \). Substituting \( c_{1} \) back, \( \frac{3}{5} + c_{2} = 1 \), so \( c_{2} = \frac{2}{5} \).

Step 5: Write Specific Solution

Substitute \( c_{1} \) and \( c_{2} \) into the general formula: \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \). This function is the specific solution that satisfies the initial conditions.

Key Concepts

Initial Value Problem

An initial value problem in the context of differential equations involves finding a specific solution that satisfies given conditions at a particular point. These conditions, known as initial conditions, allow us to determine the particular constants that make the solution unique. In our exercise, we are given the differential equation \( y'' - 3y' - 4y = 0 \) with the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \). To solve this initial value problem, we first need a general solution and then use these initial conditions to find the specific constants that will satisfy them. The goal is to determine the particular solution from the family of solutions such that the unique initial conditions hold true. By doing this, we find a curve that exactly passes through the given initial point with the correct slope.

General Solution

The general solution of a differential equation represents a family of functions that solve the equation, often involving arbitrary constants. These constants arise from the integration process when solving differential equations. In our context, the general solution is specified as \( y = c_{1} e^{4x} + c_{2} e^{-x} \), where \( c_{1} \) and \( c_{2} \) are undetermined constants. This general solution satisfies the differential equation across the entire specified interval \((-\infty, \infty)\).

• Each term, such as \( e^{4x} \) and \( e^{-x} \), is a mode of the solution, and the constants scale their importance in the solution.
• Depending on the initial conditions or boundary conditions, different values of \( c_{1} \) and \( c_{2} \) could be chosen, providing a richer set of solution possibilities.

To find the particular member of this family that fits specified initial conditions, these constants need to be determined by additional constraints, such as initial values.

First Derivative

The first derivative in the process of solving differential equations is crucial because it often helps to establish initial conditions that can pin down particular solutions. Differentiate the general solution \( y = c_{1} e^{4x} + c_{2} e^{-x} \) to find \( y' \). Here, the first derivative is calculated as \( y' = 4c_{1} e^{4x} - c_{2} e^{-x} \).

• The purpose of the first derivative in this problem is to use the initial condition \( y'(0) = 2 \). This provides a second equation when paired with the initial condition involving \( y \).
• Finding the first derivative allows us to capture the rate of change or slope of the function at specific points, such as when \( x = 0 \).

With both \( y(0) = c_{1} + c_{2} = 1 \) and \( y'(0) = 4c_{1} - c_{2} = 2 \), you now have a system of equations that can be solved to find the constants \( c_{1} \) and \( c_{2} \).

System of Equations

After establishing the general solution and its first derivative, we obtain a system of equations through the initial conditions provided. This system is composed of:

• \( c_{1} + c_{2} = 1 \)
• \( 4c_{1} - c_{2} = 2 \)

Solving this system of equations involves using methods such as substitution or elimination to find the values of the constants \( c_{1} \) and \( c_{2} \). Starting with the first equation, knowing \( c_{1} + c_{2} = 1 \), we attempt to isolate one variable. In our case, substituting the value found for \( c_{1} \) from the second equation into the first, we find:

• Adding the two equations eliminates \( c_{2} \): \( 5c_{1} = 3 \), giving \( c_{1} = \frac{3}{5} \)
• Using \( c_{1} \) in the equation \( c_{1} + c_{2} = 1 \), we find \( c_{2} = \frac{2}{5} \)

The solution to this system provides us with the specific constants that allow the differential equation's solution to fit the initial conditions exactly. By substituting \( c_{1} \) and \( c_{2} \) back into the general solution formula, we get the specific solution. This process exemplifies how initial conditions can transform a general solution into a precise function that matches particular criteria.