Problem 65 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(63-66\). $$ \frac{d^{2} y}{d x^{2}}=2 e^{-x}, \quad y(0)=1 \quad \text { and } \quad y^{\prime}(0)=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Problem Type

We have a second-order differential equation with given initial conditions: \( \frac{d^{2} y}{d x^{2}}=2 e^{-x} \), \( y(0)=1 \), and \( y'(0)=0 \). The task is to find the function \( y(x) \).

Step 2: Integrate to Find First Derivative

Integrate the given second-order differential equation, \( \frac{d^{2} y}{d x^{2}} = 2 e^{-x} \), with respect to \( x \) to find the first derivative, \( \frac{dy}{dx} \). \[ \frac{dy}{dx} = \int 2 e^{-x} \, dx = -2 e^{-x} + C_1 \] Here, \( C_1 \) is the constant of integration.

Step 3: Apply Initial Condition to First Derivative

Use the initial condition \( y'(0) = 0 \) to solve for \( C_1 \).\[ y'(0) = -2 e^{0} + C_1 = 0 \] \[ -2 + C_1 = 0 \] \[ C_1 = 2 \] Now, the first derivative becomes \( \frac{dy}{dx} = -2 e^{-x} + 2 \).

Step 4: Integrate to Find the Function y(x)

Integrate the expression for \( \frac{dy}{dx} \) to find \( y(x) \). \[ y(x) = \int (-2 e^{-x} + 2) \, dx = 2 e^{-x} + 2x + C_2 \] \( C_2 \) is a new constant of integration.

Step 5: Apply Initial Condition to Function y(x)

Now use the initial condition \( y(0) = 1 \) to solve for \( C_2 \).\[ y(0) = 2 e^{0} + 2 \times 0 + C_2 = 1 \] \[ 2 + C_2 = 1 \] \[ C_2 = -1 \] Thus, the function becomes \( y(x) = 2 e^{-x} + 2x - 1 \).

Key Concepts

Initial Value Problems

Initial value problems are a specific type of differential equation where we are given certain specified values, known as initial conditions, at the start of the process. In mathematical terms, we're solving for a function with certain values already defined at the beginning, such as its value and derivative at a particular point.
In our exercise, the problem is a second-order differential equation. This means it involves the second derivative of the unknown function. Initial conditions like these:

• \( y(0)=1 \)
• \( y'(0)=0 \)

help specify the values of the function, \( y(x) \), and its first derivative at \( x=0 \).
By specifying these conditions, we give essential information to find a unique solution to the differential equation. Without initial conditions, there can be infinitely many solutions. So, initial value problems help us pin down the exact path or behavior of the function, making our solution specific and applicable to the given situation.

Integration

Integration is one of the foundational techniques in calculus. It's essentially the reverse process of differentiation. When dealing with differential equations, integration helps us find the antiderivative or the original function from its derivative.
In our example, we started with the second derivative, \( \frac{d^2 y}{d x^2} = 2 e^{-x} \), and integrated with respect to \( x \) to find the first derivative, \( \frac{dy}{dx} \).

• The integration resulted in: \( \frac{dy}{dx} = -2 e^{-x} + C_1 \)

This process repeated once more, integrating the first derivative to find the function \( y(x) \).

• Integration here led us to \( y(x) = 2 e^{-x} + 2x + C_2 \)

Integration smoothly translates the rate of change back into the original quantity, providing us the tools to solve complex equations when we know some of the function's behavior.

Constant of Integration

The constant of integration arises naturally during the process of integration. When we integrate a function, we obtain a family of functions differing by a constant. This represents how the slope or rate could be adjusted but still remain valid solutions.
For instance, when we integrated to find \( \frac{dy}{dx} \), we added \( C_1 \):

• \( \frac{dy}{dx} = -2 e^{-x} + C_1 \)

After solving with the initial condition \( y'(0)=0 \), we found \( C_1 = 2 \). Such a constant is crucial because it allows us to adjust our function to fit initial conditions, maintaining consistency with any additional data.
Similarly, when integrating the first derivative to find \( y(x) \), another constant, \( C_2 \), emerged:

• \( y(x) = 2 e^{-x} + 2x + C_2 \)

Using \( y(0)=1 \), we determined \( C_2 = -1 \).
Constants of integration are key to tailoring indefinite integration to specific conditions, and thus finding the one precise solution from a family of solutions.

Functions and Derivatives

Functions are mathematical objects that define how one quantity changes in relation to another. In calculus, the derivative of a function measures how the function's value shifts as its input changes. It’s like the mathematical lens through which we observe the changing speed or rate of occurrence.
Our task involved a function, \( y(x) \), whose second derivative was given as \( \frac{d^2 y}{d x^2} = 2 e^{-x} \). The challenge was to recover the original function from its derivatives using integration and initial conditions.
The process is straightforward:

• The derivative of a function gives us the slope or rate of change.
• The second derivative, its further rate of change or curvature.
• With each integration, we step backwards from derivative to the function level.

By matching with initial conditions, the exact form of the function is established. Understanding functions and their derivatives is crucial, as they provide insights into the behavior and pattern of changes occurring within any mathematical model or real-world application.