Problem 7 - Calculus Volume 3

Question

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of \(c_{1}\) and \(c_{2}\). What do the solutions have in common? $$ [\mathbf{T}] y^{\prime \prime}+2 y^{\prime}-3 y=0 ; \quad y(x)=c_{1} e^{x}+c_{2} e^{-3 x} $$

Step-by-Step Solution

Answer

The function \( y(x) = c_1 e^x + c_2 e^{-3x} \) solves the differential equation, with solutions having exponential growth/decay.

Step 1: Identify the Differential Equation and the Function

The differential equation given is \( y'' + 2y' - 3y = 0 \). The function given is \( y(x) = c_1 e^x + c_2 e^{-3x} \). We need to verify that this function is a solution to the given differential equation.

Step 2: Differentiate the Function for First Derivative

First, find the first derivative of \( y(x) = c_1 e^x + c_2 e^{-3x} \). This is \( y'(x) = c_1 e^x - 3c_2 e^{-3x} \).

Step 3: Differentiate the Function for Second Derivative

Next, find the second derivative of \( y(x) \). This is \( y''(x) = c_1 e^x + 9c_2 e^{-3x} \).

Step 4: Substitute into the Differential Equation

Substitute \( y(x) \), \( y'(x) \), and \( y''(x) \) into the differential equation: \( y'' + 2y' - 3y = 0 \).Calculating terms:- \( y'' = c_1 e^x + 9c_2 e^{-3x} \)- \( 2y' = 2(c_1 e^x - 3c_2 e^{-3x}) = 2c_1 e^x - 6c_2 e^{-3x} \)- \( -3y = -3(c_1 e^x + c_2 e^{-3x}) = -3c_1 e^x - 3c_2 e^{-3x} \)

Step 5: Verify the Equation

Now, sum the terms: \( (c_1 e^x + 9c_2 e^{-3x}) + (2c_1 e^x - 6c_2 e^{-3x}) - (3c_1 e^x + 3c_2 e^{-3x}) \).Simplifying: - Combine like terms: \( (c_1+2c_1-3c_1)e^x + (9c_2-6c_2-3c_2)e^{-3x} \)- This simplifies to \( 0 e^x + 0 e^{-3x} = 0 \).Thus, the function satisfies the differential equation.

Step 6: Graphical Analysis of Solutions

For various values of \( c_1 \) and \( c_2 \), graph the function \( y(x) = c_1 e^x + c_2 e^{-3x} \) using a graphing utility. Observe the shape and characteristics of the solutions.

Step 7: Common Features of Solutions

All solutions have the same general form but differ in their particular initial conditions or shapes based on \( c_1 \) and \( c_2 \). The solutions decay or grow exponentially.

Key Concepts

Second Derivative

The second derivative, often denoted as \( y'' \), represents the rate of change of the first derivative. Essentially, it's the derivative of the derivative. It helps us understand the curvature or concavity of a function. To find the second derivative of a function like \( y(x) = c_1 e^x + c_2 e^{-3x} \), we first need the first derivative, which is \( y'(x) = c_1 e^x - 3c_2 e^{-3x} \).
- The second derivative is then calculated as \( y''(x) = c_1 e^x + 9c_2 e^{-3x} \).
This derivative is vital when verifying solutions to differential equations, as seen in the problem where we plugged it back into the equation \( y'' + 2y' - 3y = 0 \) to check its validity. By examining \( y'' \), we gain insights into the behavior of solutions, such as whether they curve upwards or downwards as they evolve.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They are a key feature in the problem, appearing as \( e^x \) and \( e^{-3x} \) in the function \( y(x) = c_1 e^x + c_2 e^{-3x} \).

• \( e^x \) represents exponential growth, leading to increased values as \( x \) becomes larger.
• \( e^{-3x} \) represents exponential decay, as the value decreases with increasing \( x \). Each term's behavior is crucial for understanding the overall function's dynamics.

Exponential functions are often used in differential equations because they exhibit consistent growth or decay rates, making them essential for modeling natural phenomena. In our context, varying coefficients \( c_1 \) and \( c_2 \) act as parameters that modify the solution's specific characteristics, determining whether it grows, shrinks, or exhibits a mix of both patterns.

Initial Conditions

Initial conditions in a differential equation are the values at the beginning of a solution or when \( x = 0 \). They play a crucial role in determining the particular solution to a differential equation from a family of solutions described by a general form. In our exercise, \( y(x) = c_1 e^x + c_2 e^{-3x} \) forms a general solution with parameters \( c_1 \) and \( c_2 \).

• By changing \( c_1 \) and \( c_2 \), we get different particular solutions. Each set of \( c_1 \) and \( c_2 \) values alters the curve's shape and properties.
• Graphical representation helps to visualize these changes, showing us how initial conditions impact the behavior of solutions over time.

Initial conditions help ensure that the solution not only satisfies the differential equation but also matches any physical or real-world scenarios being modeled. The problem invites us to use a graphing utility to observe how different initial conditions result in different solution curves, emphasizing this concept's importance.