Problem 32 - Essential Calculus

Question

Solve the differential equation. Use a graphing utility to graph three solutions, one of which passes through the given point. Determine the function \(f\) if \(f^{\prime \prime}(x)=-\frac{4}{(x-1)^{2}}-2, f(2)=3\) and \(f^{\prime}(2)=0, x>1\).

Step-by-Step Solution

Answer

The solution will be the function \( f \) that satisfies both the differential equation and the given initial conditions.

Step 1: Solve homogeneous differential equation

First, we solve for the homogenous solution by setting the second derivative of \( f \), \( f^{\prime \prime}(x) = 0 \). This is a simple differential equation and the solution is \( f(x) = Ax + B \), where \( A \) and \( B \) are constants.

Step 2: Solve non-homogeneous differential equation

Solve the non-homogeneous differential equation by substituting the general solution into the given second order differential equation \( f^{\prime \prime}(x) = -\frac{4}{(x-1)^2} - 2 \). This will lead to the particular solution.

Step 3: Find General Solution

The general solution will be the sum of the homogeneous and particular solutions.

Step 4: Find initial conditions

Use the initial conditions \( f(2)=3 \) and \( f^{\prime}(2)=0 \) to find the constants.

Step 5: Find the specific function \( f \)

Once the constants are found, substitute them back into the general solution to find the specific function \( f \).

Key Concepts

Homogeneous Solution

When tackling differential equations, a homogeneous solution is derived by initially ignoring any non-homogeneous parts of the equation. This means you first solve the differential equation assuming the independent variables do not have an external force or term applied. In mathematical terms, you set the non-zero functions on the right side of the equation to zero.

Here's a step-by-step breakdown:

• Consider the differential equation of the form \( f''(x) = 0 \). This indicates no additional forces or inputs are acting on the system.
• The solution to this homogeneous equation is generally a linear expression, \( f(x) = Ax + B \), where both \( A \) and \( B \) are unknown constants. These constants will be determined based on other conditions later on.

This part of the solution represents the "natural" behavior of the system without any interference. It lays the groundwork on which other types of solutions are built.

Particular Solution

The particular solution arises when the differential equation includes non-homogeneous terms, which are functions of the independent variable such as \(-\frac{4}{(x-1)^2} - 2\). The goal is to find a function that accounts for these terms.

Here's how you can approach finding a particular solution:

• We substitute a guessed form of the solution into the differential equation. This educated guess should ideally reflect the type of non-homogeneous term present. In this situation, given the term \(-\frac{4}{(x-1)^2} - 2\), seek a form that could adequately incorporate these specific forms.
• By plugging this guessed form into the original differential equation, solve for any unknown coefficients or terms in your guessed solution.

Once determined, the particular solution effectively describes the component of the system's behavior explicitly due to the external terms.

Finally, the complete or general solution of the differential equation blends both the homogeneous and particular solutions, providing a full representation of the function's behavior.

Initial Conditions

Initial conditions in differential equations serve as the critical keys that unlock the specific solutions to the problems. When you solve a differential equation, you generally arrive at a general solution that contains arbitrary constants.

In this exercise, our initial conditions are \( f(2) = 3 \) and \( f'(2) = 0 \). Here's how to use them:

• Substitute these initial conditions into your general solution (which includes both the homogeneous and particular solutions) to determine the arbitrary constants.
• The first condition confirms the value of the function at a particular point, \( x = 2 \).
• The second condition gives you information about the slope or derivative at the same point, hence understanding more the nature of the curve.

By applying these conditions, you pinpoint the exact solution that not only satisfies the differential equation but also exactly fits the specified behavior at given points. This process turns a general solution into the specific function we seek to describe.