Problem 20 - Calculus: Early Transcendentals

Question

Find the function \( f \) such that \( f'(x) = xf(x) - x \) and \( f(0) = 2. \)

Step-by-Step Solution

Answer

The function is \( f(x) = 1 + e^{(x^2/2)} \).

Step 1: Understand the Problem

The given problem is to find the function \( f(x) \) such that its derivative \( f'(x) \) matches the equation \( f'(x) = xf(x) - x \) and has the initial condition \( f(0) = 2 \). This involves solving a first-order differential equation with an initial condition.

Step 2: Rearrange the Differential Equation

Rearrange the given differential equation to form a standard form by making \( y = f(x) \), then the equation becomes \( y' - xy = -x \). The equation is now in the form of \( y' - xy = -x \).

Step 3: Identify the Integrating Factor

The integrating factor \( \mu(x) \) for a first-order linear differential equation of the form \( y' + p(x)y = q(x) \) is given by \( \mu(x) = e^{\int p(x) \, dx} \). Here, \( p(x) = -x \), so \( \mu(x) = e^{-\int x \, dx} = e^{-(x^2/2)} \).

Step 4: Multiply Through by the Integrating Factor

Multiply the entire differential equation \( y' - xy = -x \) by the integrating factor \( e^{-(x^2/2)} \). This gives \( e^{-(x^2/2)}y' - xe^{-(x^2/2)}y = -xe^{-(x^2/2)} \).

Step 5: Simplify Using the Integrating Factor

Recognize that the left-hand side is equivalent to the derivative of \( e^{-(x^2/2)}y \), implying \( \frac{d}{dx}[e^{-(x^2/2)}y] = -xe^{-(x^2/2)} \).

Step 6: Integrate Both Sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx}[e^{-(x^2/2)}y] \, dx = \int -xe^{-(x^2/2)} \, dx \). This simplifies to \( e^{-(x^2/2)}y = -\int xe^{-(x^2/2)} \ dx + C \), where \( C \) is a constant.

Step 7: Solve the Integral

The integral \( \int xe^{-(x^2/2)} \ dx \) can be solved using substitution methods or recognizing the Gaussian integral form, which results in \( -e^{-(x^2/2)} + C \). Thus, \( e^{-(x^2/2)}y = e^{-(x^2/2)} - C \) simplifies to \( y = 1 - Ce^{(x^2/2)} \).

Step 8: Apply the Initial Condition

Use the initial condition \( f(0) = 2 \) to find \( C \). Substituting \( x = 0 \) gives \( 2 = 1 - C \). Solving for \( C \) gives \( C = -1 \).

Step 9: Determine the Solution Function

Substitute \( C = -1 \) back into \( y = 1 - Ce^{(x^2/2)} \), resulting in \( f(x) = 1 + e^{(x^2/2)} \). Thus, the function \( f(x) \) that satisfies the differential equation and initial condition is \( f(x) = 1 + e^{(x^2/2)} \).

Key Concepts

First-order Differential Equations

A differential equation relates a function with its derivatives. A first-order differential equation involves the first derivative of the unknown function. It can be expressed in the standard form as:
\[ y' + p(x)y = q(x) \]
where \( y' \) denotes the derivative of \( y \) with respect to \( x \), \( p(x) \) and \( q(x) \) are functions of \( x \).
First-order differential equations are fundamental in modeling real-world phenomena like population growth or radioactive decay. They describe how a quantity changes at each moment, given its current state.
It's the simplest type of differential equation, relying only on the first derivative, offering a stepping stone in understanding more complex equations.

Initial Value Problems

An initial value problem is a type of differential equation paired with a condition that specifies the value of the unknown function at a particular point. For example, the problem \( f'(x) = xf(x) - x \) with \( f(0) = 2 \) is an initial value problem.
Here's the breakdown of its components:

• The differential equation itself specifies how the function changes.
• The initial condition \( f(0) = 2 \) provides a specific "starting point" for the solution.

Initial conditions are crucial because they ensure the uniqueness of the solution. Without these, a differential equation may have infinite possible solutions.
By providing a start or initial condition, we gain a specific function that uniquely solves the differential equation in its given context.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve linear first-order differential equations. The goal is to convert the differential equation into an exact differential, which can then be integrated directly.
To apply this method:

• Put the differential equation in standard form \( y' + p(x)y = q(x) \).
• Calculate the integrating factor \( \mu(x) \), using \( \mu(x) = e^{\int p(x) \, dx} \).
• Multiply every term by \( \mu(x) \), transforming the left-hand side into the derivative of \( \mu(x)y \).
• Integrate both sides to find the solution.

This approach efficiently solves many differential equations where direct integration isn't feasible. The integrating factor modifies the equation, making it easier to solve by matching the derivative form directly.
This transformation is the key step where the method unravels the solution path.