Problem 53 - Thomas Calculus in SI Units

Question

Solve the initial value problems in Exercises \(51-54.\) $$\frac{d^{2} y}{d x^{2}}=2 e^{-x}, \quad y(0)=1 \quad \text { and } \quad y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

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

Step 1: Solve the Second Derivative

We begin by integrating the given second derivative \( \frac{d^2 y}{dx^2} = 2e^{-x} \) once to find the first derivative, \( \frac{dy}{dx} \). This gives us \( \frac{dy}{dx} = \int 2e^{-x} \, dx = -2e^{-x} + C_1 \), where \( C_1 \) is a constant of integration.

Step 2: Solve the First Derivative

Next, we integrate the expression for the first derivative to find \( y \). So we have \( y = \int (-2e^{-x} + C_1) \, dx = 2e^{-x} + C_1x + C_2 \), where \( C_2 \) is another constant of integration.

Step 3: Apply Initial Condition for y(0)

We use the initial condition \( y(0) = 1 \) to determine \( C_2 \). Substituting \( x = 0 \) into the equation for \( y \), we get \( 1 = 2e^{0} + 0 \cdot C_1 + C_2 \). This simplifies to \( 1 = 2 + C_2 \), giving us \( C_2 = -1 \).

Step 4: Apply Initial Condition for y'(0)

Now we use the initial condition \( y'(0) = 0 \) to find \( C_1 \). Substitute into the first derivative equation: \( 0 = -2e^{0} + C_1 \), which simplifies to \( 0 = -2 + C_1 \). Solving for \( C_1 \), we get \( C_1 = 2 \).

Step 5: Write the Final Solution for y(x)

Substitute the values of \( C_1 \) and \( C_2 \) back into the expression for \( y \). This gives the final solution: \( y = 2e^{-x} + 2x - 1 \).

Key Concepts

Differential Equations

Differential equations are equations that connect a function with its derivatives. In the context of our problem, we have a second-order differential equation: \[ \frac{d^2 y}{dx^2} = 2e^{-x} \]This means we are dealing with the second derivative of a function. Solving differential equations involves finding the unknown function that satisfies the given equation.A key purpose of solving differential equations is to describe physical phenomena. For instance, they can model the motion of celestial bodies or biological systems. Our task is to find the function \( y(x) \) which not only satisfies the equation but also aligns with initial conditions provided, like specific values for \( y \) or \( y' \) at certain points.

Integration

Integration is a mathematical technique used to find functions from their derivatives. Since our equation involves derivatives, integration is essential to reverse the differentiation process.
In our solution:

• We first integrate \( \frac{d^2 y}{dx^2} = 2e^{-x} \) to find the first derivative, \( \frac{dy}{dx} \). This yields:

\[ \frac{dy}{dx} = \int 2e^{-x} \, dx = -2e^{-x} + C_1 \]

• Next, we integrate \( \frac{dy}{dx} \) to find \( y \):

\[ y = \int (-2e^{-x} + C_1) \, dx = 2e^{-x} + C_1x + C_2 \]Through integration, we obtain the general expression for \( y(x) \), albeit with integration constants that need determination.

Constant of Integration

When integrating, we encounter terms known as constants of integration. These constants, denoted here as \( C_1 \) and \( C_2 \), reflect the indefinite nature of integration. They represent the functions' unknown parts which don't affect the derivative.For instance:

• In the integration of the second derivative to get the first derivative, we introduce \( C_1 \):

\[ \frac{dy}{dx} = -2e^{-x} + C_1 \]

• Upon further integration to find \( y \), \( C_2 \) appears:

\[ y = 2e^{-x} + C_1x + C_2 \]Determining these constants requires additional conditions, typically initial or boundary conditions, to provide particular solutions matching specific real-world problems.

Initial Conditions

Initial conditions are specific values provided to resolve the constants of integration within differential equations. They ensure that the derived function represents the particular solution rather than just a general family of solutions.In our problem, we use two initial conditions:

• \( y(0) = 1 \), allowing us to find \( C_2 \).
• \( y'(0) = 0 \) to determine \( C_1 \).

By plugging these conditions into our expressions, we concretize our solution:

• With \( y(0) = 1 \): \[ 1 = 2e^{0} + 0 \times C_1 + C_2 \Rightarrow C_2 = -1\]
• With \( y'(0) = 0 \): \[ 0 = -2e^{0} + C_1 \Rightarrow C_1 = 2\]

The application of these conditions refines the expression to a particular solution: \[ y = 2e^{-x} + 2x - 1 \] This specific solution fits both the differential equation and the initial conditions given at the start, providing one distinct path associated with the initial problem context.