Problem 106 - University Calculus: Early Transcendentals

Question

Solve the initial value problems in Exercises. $$\frac{d^{2} y}{d x^{2}}=0 ; \quad y^{\prime}(0)=2, \quad y(0)=0$$

Step-by-Step Solution

Answer

The solution is \(y(x) = 2x\).

Step 1: Integrate the Second Derivative

The function given is \(\frac{d^{2} y}{d x^{2}}=0\). Integrate with respect to \(x\) to find the first derivative. Thus, \(\frac{d y}{d x} = C_1\), where \(C_1\) is a constant of integration.

Step 2: Integrate to Find the General Solution for y(x)

Integrate \(\frac{dy}{dx} = C_1\) with respect to \(x\) to find the function \(y(x)\). This gives \(y(x) = C_1 x + C_2\), where \(C_2\) is another constant of integration.

Step 3: Apply Initial Condition y'(0) = 2

Use the initial condition \(y'(0) = 2\). Substitute \(x = 0\) into \(y'(x) = C_1\) to get \(C_1 = 2\).

Step 4: Apply Initial Condition y(0) = 0

Use the second initial condition \(y(0) = 0\) by substituting \(x = 0\) into \(y(x) = C_1 x + C_2\), leading to \(0 = C_1 \cdot 0 + C_2\). This gives \(C_2 = 0\).

Step 5: Write the Particular Solution

Substitute the values of \(C_1\) and \(C_2\) back into the equation \(y(x) = C_1 x + C_2\). The particular solution is \(y(x) = 2x\).

Key Concepts

Initial Value Problems

Initial value problems are a type of differential equation where we know some specific values of the function and possibly its derivatives at certain points. These values are what we call initial conditions. They help us to find a specific solution to a differential equation rather than a general one. The problem usually involves finding a function which not only satisfies a given differential equation but also fits the initial conditions specified. For example, in the original exercise, we had the differential equation \( \frac{d^{2} y}{d x^{2}}=0 \) with initial conditions \( y'(0) = 2 \) and \( y(0) = 0 \). These conditions are crucial as they allow us to determine the unique "particular solution" out of the many possible solutions.

Constant of Integration

When solving differential equations, particularly during the integration step, we often encounter what is known as the "constant of integration." This constant represents any constant value that, when differentiated, results in zero. Thus, when integrating a function, we add this constant to show that there are infinitely many antiderivatives. For example:

• Upon integrating \( \frac{d^{2} y}{d x^{2}}=0 \), we get \( \frac{dy}{dx} = C_1 \).
• When integrating again to find \( y(x) \), we obtain \( y(x) = C_1 x + C_2 \).

These constants \( C_1 \) and \( C_2 \) are evaluated using the initial conditions to obtain particular values that fit the problem's requirement. In this exercise, applying the initial conditions helped us find that \( C_1 = 2 \) and \( C_2 = 0 \). This highlights the need for initial conditions to solve the equation effectively and derive the specific function out of the general solution.

Particular Solution

The idea of a "particular solution" arises from the fact that a differential equation can have an infinite number of solutions, each representing different scenarios. A particular solution is one such specific solution that satisfies both the differential equation and the given initial conditions. For instance, in our exercise, after deriving the general solution \( y(x) = C_1 x + C_2 \), we used the specific initial conditions \( y'(0) = 2 \) and \( y(0) = 0 \) to find exact values for the constants \( C_1 \) and \( C_2 \), thus obtaining the particular solution \( y(x) = 2x \).Therefore, when solving initial value problems, finding the particular solution is often the crucial step as it tells us the exact behavior of the system modeled by the differential equation at a specific context.