Problem 31 - Essential Calculus

Question

Determine the function \(f\) if \(f^{\prime \prime}(x)=\frac{2}{x^{2}}, f(1)=1,\) and \(f^{\prime}(1)=1, x>0\)

Step-by-Step Solution

Answer

The function \(f(x)\) that satisfies the given conditions is given by: \(f(x)=-2 \ln|x| + 3x -2\).

Step 1: Integrate the second derivative to obtain the first derivative

Integrating \(f^{\prime \prime}(x)=\frac{2}{x^{2}}\) with respect to \(x\) gives \(f^{\prime}(x)=-\frac{2}{x} + c_{1}\) where \(c_{1}\) is the constant of integration.

Step 2: Apply the condition \(f^{\prime}(1)=1\)

Substitute \(x=1\) into the expression for \(f^{\prime}(x)\) to give \(f^{\prime}(1)=-\frac{2}{1} + c_{1}=1 => c_{1}=3\).

Step 3: Integrate the first derivative to obtain the original function

Integrating \(f^{\prime}(x)=-\frac{2}{x} + 3\) with respect to \(x\) gives \(f(x) = -2 \ln|x| + 3x +c_{2}\), where \(c_{2}\) is another constant of integration.

Step 4: Apply the condition \(f(1)=1\)

Substitute \(x=1\) into the expression for \(f(x)\) to give \(f(1)=-2\ln|1|+3(1)+c_{2}=1\). Thus, \(c_{2}=-2\).

Key Concepts

Integration

Integration is a fundamental technique in calculus, used to find functions when their derivatives are known. In this case, we began with the second derivative of a function, expressed as \( f''(x) = \frac{2}{x^2} \). The goal is to reverse the process of differentiation, essentially "undoing" it to find the original function step by step.

When we integrate a derivative, we add a constant called the constant of integration. This constant appears because when differentiating, any constant would disappear. Thus, its value is not readily apparent from just the derivative.

The method involves:

• Identifying the given derivative.
• Performing the integration step by step, reducing the order of the derivative each time.
• Applying initial conditions to solve for unknown constants.

In our example, integrating \( f''(x) = \frac{2}{x^2} \) with respect to \( x \) provides \( f'(x) = -\frac{2}{x} + c_1 \), enabling further work to find the original function.

Differential Equations

Differential equations are equations that involve the derivatives of a function. They can describe various phenomena such as growth rates, physical laws, and more. Solving them typically involves finding a function that satisfies the equation.

In this problem, we were given a second-order differential equation, \( f''(x) = \frac{2}{x^2} \). Here's a friendly breakdown of how to tackle such problems:

• Recognize the type of differential equation and its order (second-order here).
• Use integration to "step back" through the derivatives. Each integration reduces the order of the derivative.
• Find the complementary function, which introduces the constant of integration.

Finally, applying known initial conditions is crucial as it adjusts these constants to fit the specific scenario. Understanding differential equations unlocks solving many real-world problems and equations seen in physics, engineering, and beyond.

Initial Conditions

Initial conditions provide the specific values needed to determine the constants of integration unique to a scenario. These conditions typically come in form of values such as function evaluations at certain points, like \( f(1)=1 \) and \( f'(1)=1 \) in our exercise.

Here's how we effectively use initial conditions:

• After integrating, substitute the given points into the integrated equation.
• Solve the resulting equations for the unknown constants.
• Apply these constants to the integral to customize the solution to adhere to initial conditions.

This process ensures the solution is specific to the given parameters. In our exercise, substituting \( f'(1)=1 \) allowed us to solve for \( c_1 \), and \( f(1)=1 \) helped determine \( c_2 \). Initial conditions are the key components that personalize mathematical solutions to their specific scenarios.