Problem 24 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(\frac{d^{2} y}{d x^{2}}=2 ; y=5\) and \(\frac{d y}{d x}=3\) when \(x=0\)

Step-by-Step Solution

Answer

The function is \( y = x^2 + 3x + 5 \).

Step 1: Identify Given Information

We are given:1. The second derivative, \(\frac{d^{2} y}{d x^{2}} = 2\).2. The value of \( y = 5 \) when \( x = 0 \).3. The value of \( \frac{d y}{d x} = 3 \) when \( x = 0 \).

Step 2: Integrate the Second Derivative

To find the first derivative, integrate the second derivative: \[ \frac{d^2 y}{d x^2} = 2 \] Integrate with respect to \( x \): \[ \frac{d y}{d x} = 2x + C_1 \]

Step 3: Apply Initial Condition for First Derivative

We know \( \frac{d y}{d x} = 3 \) when \( x = 0 \). Substitute these values into the integrated equation: \[ 3 = 2(0) + C_1 \] \[ C_1 = 3 \] Thus, \( \frac{d y}{d x} = 2x + 3 \).

Step 4: Integrate the First Derivative

To find \( y \), integrate the first derivative: \[ \frac{d y}{d x} = 2x + 3 \] Integrate with respect to \( x \): \[ y = x^2 + 3x + C_2 \]

Step 5: Apply Initial Condition for y

We know \( y = 5 \) when \( x = 0 \). Substitute these values into the integrated equation: \[ 5 = (0)^2 + 3(0) + C_2 \] \[ C_2 = 5 \] Thus, \( y = x^2 + 3x + 5 \).

Key Concepts

Initial Conditions

Initial conditions are specific values given for the function and its derivatives at a certain point, usually to determine the constants after integration. For this problem, we have two initial conditions:

• When \( x = 0 \), \( y = 5 \)
• When \( x = 0 \), \( \frac{d y}{d x} = 3 \)

These initial conditions help us find the constants \( C_1 \) and \( C_2 \) after integrating the given second-order differential equation.

Integration

Integration is a fundamental technique in calculus used to find the antiderivative of a function. Here, we integrate twice to solve the second-order differential equation.

• First, integrate the second derivative \( \frac{d^2 y}{d x^2} = 2 \):
  \( \frac{d y}{d x} = 2x + C_1 \)
• Then, integrate \( \frac{d y}{d x} = 2x + 3 \) to find \( y \):
  \( y = x^2 + 3x + C_2 \)

Each time we integrate, we add an arbitrary constant because the antiderivative of a function is not unique without initial conditions.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. For this exercise, we start with a second-order derivative \( \frac{d^2 y}{d x^2} = 2 \).

• The first derivative \( \frac{d y}{d x} \) tells us how \( y \) changes with respect to \( x \).
• The second derivative \( \frac{d^2 y}{d x^2} \) tells us how the first derivative changes with respect to \( x \).

By integrating the second derivative, we find the first derivative and then integrate again to find the original function \( y \). This step-by-step process is crucial in solving differential equations.