Problem 54 - College Algebra and Calculus: An Applied Approach

Question

Find the particular solution \(y=f(x)\) that satisfies the differential equation and initial condition. $$ f^{\prime}(x)=\frac{x^{2}-5}{x^{2}}, x>0 ; \quad f(1)=2 $$

Step-by-Step Solution

Answer

The particular solution to the differential equation that meets the initial condition \(f(1)=2\) will therefore be \(f(x)=x + \frac{5}{x} - 4\).

Step 1: Integrating the function

To solve the ordinary differential equation, integrate the given function: \(f'(x)=\frac{x^{2}-5}{x^{2}}\). The integral of the function is as follows: \(\int f'(x) dx = \int \frac{x^{2}-5}{x^{2}} dx\). By splitting the fraction, this integral can be rewritten as: \(\int (1 - \frac{5}{x^{2}}) dx\).

Step 2: Calculate the integral

Calculate the integral of the function we simplified in the previous step: \(\int (1 - \frac{5}{x^{2}}) dx\). The antiderivative is calculated to be \(x + \frac{5}{x} + C\), where \(C\) is the constant of integration.

Step 3: Apply the initial condition

Now it's time to apply the initial condition \(f(1) = 2\) to find the constant \(C\). Substitute \(x = 1\) and \(f(x) = 2\) into the antiderivative expression: \(2 = 1 + 5 + C\). Solving for \(C\) will yield \(C = -4\).

Key Concepts

Initial Conditions

Initial conditions are crucial in solving differential equations because they help us find the particular solution out of infinitely many potential solutions. In this exercise, the initial condition is given as \( f(1) = 2 \). This means when \( x = 1 \), the value of the function \( f(x) \) should be 2.

Here's why initial conditions matter:

• An ordinary differential equation typically has a family of solutions, determined up to a constant \( C \).
• Initial conditions specify a unique solution by providing enough information to solve for \( C \).

In our example, after finding the antiderivative \( x + \frac{5}{x} + C \), the initial condition \( f(1) = 2 \) allows us to determine that \( C = -4 \). This value of \( C \) plugs back into our solution to make it unique.

Antiderivatives

Antiderivatives are fundamental in solving differential equations, especially the most common type, ordinary differential equations (ODEs). They are the reverse operation of differentiation.

When solving for antiderivatives:

• Identify the function you need to integrate.
• Perform the integration to find the family of antiderivatives, which includes a constant of integration \( C \).

In this exercise, given \( f'(x) = \frac{x^2 - 5}{x^2} \), we split the function and find the antiderivative:\[ \int \left(1 - \frac{5}{x^2}\right) dx = x + \frac{5}{x} + C \]The constant \( C \) represents the infinite set of vertical translations of the function. The correct value of \( C \) is determined later using initial conditions.

Ordinary Differential Equations

Ordinary differential equations, or ODEs, contain functions and their derivatives and are central to modeling many real-world phenomena in fields like physics, engineering, and biology.

Key points about ODEs:

• They involve ordinary derivatives of a function, not partial derivatives.
• The complexity of solving ODEs can vary greatly depending on the order and type of the equation.

In our problem, the ODE is represented by \( f'(x) = \frac{x^2 - 5}{x^2} \), making it a first-order ODE because only the first derivative \( f'(x) \) appears. Solving this involves finding the function \( f(x) \) whose derivative matches the given expression, utilizing the connection between differentiation and integration.