Problem 57 - Thomas Calculus

Question

Solve the initial value problems. \begin{equation}\frac{d^{2} y}{d x^{2}}=2 e^{-x}, \quad y(0)=1 \quad \text { and } \quad y^{\prime}(0)=0\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Integrate to Find the First Derivative

Start by integrating the given second-order differential equation with respect to \( x \) to find the first derivative \( \frac{dy}{dx} \). This requires integrating \( \int 2e^{-x} \, dx \), which gives \( \frac{dy}{dx} = -2e^{-x} + C_1 \), where \( C_1 \) is the constant of integration.

Step 2: Apply Initial Condition for First Derivative

Use the initial condition \( y'(0) = 0 \) to find the constant \( C_1 \). Substitute \( x=0 \) into \( \frac{dy}{dx} = -2e^{-x} + C_1 \) to get \( 0 = -2e^{0} + C_1 \), leading to \( C_1 = 2 \). Now, \( \frac{dy}{dx} = -2e^{-x} + 2 \).

Step 3: Integrate to Find the General Solution

Integrate the expression for \( \frac{dy}{dx} \) to find \( y(x) \). Integrate \( \int (-2e^{-x} + 2) \, dx \). This gives \( y = 2e^{-x} + 2x + C_2 \).

Step 4: Apply Initial Condition for Function Value

Use the initial condition \( y(0) = 1 \) to find the constant \( C_2 \). Substitute \( x = 0 \) into the expression: \( y = 2e^{0} + 2(0) + C_2 = 1 \). This simplifies to \( 2 + C_2 = 1 \), resulting in \( C_2 = -1 \). Thus, the particular solution is \( y = 2e^{-x} + 2x - 1 \).

Step 5: Verify the Solution

Ensure the solution satisfies the initial conditions. Differentiate \( y = 2e^{-x} + 2x - 1 \) and verify \( y'(0) = 0 \) and \( y(0) = 1 \). Calculate \( \, \frac{dy}{dx} = -2e^{-x} + 2 \). At \( x = 0 \), \( y' = -2e^{0} + 2 = 0 \) and \( y(0) = 2e^{0} + 0 - 1 = 1 \), confirming both initial conditions are satisfied.

Key Concepts

Initial Value Problems

An initial value problem is a type of differential equation that comes with specific starting conditions. These conditions determine the precise solution from all possible solutions of the differential equation.
Initial value problems are crucial in mathematical modeling as they give definite answers based on the initial conditions of the system.
In our exercise, we're given two initial conditions:

• The function value at zero, \( y(0) = 1 \)
• The derivative at zero, \( y'(0) = 0 \)

These conditions specify where our solution must begin, making it unique and eliminating any ambiguity about the trajectory of the function."},{

Second-order Differential Equation

A second-order differential equation involves the second derivative of a function. It is used to describe systems where the rate of change of a rate of change needs to be monitored, such as in physics for motion equations.
In our problem, the second-order differential equation is \( \frac{d^2 y}{d x^2} = 2 e^{-x} \). The task is to integrate twice to revert from the second derivative back to the function \( y(x) \).
This often involves finding constants of integration at each step, which are determined using the initial conditions of the problem. Understanding how to solve these equations is crucial for various fields of science and engineering.

Integration with Respect to x

Integration is the reverse process of differentiation, permitting us to find original functions from their rates of change. When dealing with a second-order differential equation, we first integrate to find the first derivative, and then integrate again to find the function itself.
In this exercise, we begin by integrating the given second-order differential equation \( \int 2e^{-x} \, dx \).

• First integral gives: \( \frac{dy}{dx} = -2e^{-x} + C_1 \)
• Second integral of \( \frac{dy}{dx} \) results in the function \( y(x) = 2e^{-x} + 2x + C_2 \)

Each integration step introduces a new constant, adjusted later to match the initial conditions. The process of integrating differential equations is a fundamental technique in calculus and is essential for solving initial value problems.

General Solution of Differential Equation

The general solution of a differential equation includes all possible solutions of the equation, expressed with arbitrary constants that can take various values.
When initial conditions are applied, a particular solution is found, defining the function uniquely by fixing the constants. In this case, using the second-order differential equation, our general solution was:

• \( y(x) = 2e^{-x} + 2x + C_2 \)

After applying the initial conditions \( y(0) = 1 \) and \( y'(0)=0 \), we determined:

• \( C_1 = 2 \)
• \( C_2 = -1 \)

Thus, the particular solution becomes \( y = 2e^{-x} + 2x - 1 \). This unique equation satisfies both initial conditions provided, showcasing the importance of initial conditions in deriving a specific solution from the general form.