Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Multiply both sides of the equation \(\frac{d y}{d x}+P(x) y=Q(x)\) by the factor \(e^{\int P(x) d x+C}\). (a) Show that \(e^{\int P(x) d x+C}\) is an integrating factor for every value of \(C\). (b) Solve the resulting equation for \(y\), and show that it agrees with the general solution given before Example 1 .

Step-by-Step Solution

Answer

The integrating factor is valid. Solving yields the known general solution for the equation.

Step 1: Identify the Integrating Factor

Recognize that the equation \( \frac{dy}{dx} + P(x)y = Q(x) \) is a first-order linear differential equation. An integrating factor \( \mu(x) \) is of the form \( e^{\int P(x)\,dx} \), where \( \int P(x) \) is an indefinite integral.

Step 2: Multiply the Equation

To make the left side of the differential equation a perfect derivative, multiply the entire equation by the integrating factor \[ e^{\int P(x)\,dx + C} \],giving\[ e^{\int P(x)\,dx + C} \left( \frac{dy}{dx} + P(x)y \right) = e^{\int P(x)\,dx + C} Q(x). \]

Step 3: Simplify Using a Constant Property

Note that multiplying by \( e^C \) is equivalent to multiplying by a constant, thereby not affecting the integrating factor's properties. Hence, \( e^{\int P(x)\,dx + C} = e^C \cdot e^{\int P(x)\,dx} \) is still an appropriate integrating factor for any \( C \). This proves part (a).

Step 4: Recognize the Derivative

The left side of the equation, \( e^{\int P(x)\,dx} \frac{dy}{dx} + e^{\int P(x)\,dx} P(x) y \), can be rewritten as the derivative of the product:\[ \frac{d}{dx} \left( e^{\int P(x)\,dx} y \right). \]

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \),\[ \int \frac{d}{dx} \left( e^{\int P(x)\,dx} y \right) \,dx = \int e^{\int P(x)\,dx} Q(x) \,dx. \]The left side simplifies to\[ e^{\int P(x)\,dx} y. \]

Step 6: Solve for \( y \)

Solving for \( y \), we get:\[ y = \frac{1}{e^{\int P(x)\,dx}} \left( \int e^{\int P(x)\,dx} Q(x) \,dx + C_1 \right), \]where \( C_1 \) is a constant of integration.

Step 7: Compare with General Solution

The solution corresponds to previously known general solutions of first-order linear equations, where \( y \) is obtained in terms of the integrating factor. This satisfies the condition and agrees with the general solution given in standard examples.

Key Concepts

Integrating Factor

In differential equations, an integrating factor is a function that transforms a non-exact differential equation into an exact one. This allows us to integrate and solve it more easily.
For a first-order linear differential equation such as \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is typically of the form \( e^{\int P(x) \,dx} \). This factor helps simplify the appearance of the equation by ensuring that the left-hand side becomes the derivative of a product.
For example:

• The integrating factor, when used, transforms the left side into the derivative \( \frac{d}{dx} \left( e^{\int P(x) \,dx} y \right) \).
• By multiplying through by \( e^{\int P(x) \,dx+C} \), the equation becomes easier to integrate, regardless of the constant \( C \), due to the properties of exponents.

This technique greatly simplifies solving differential equations by reducing them to a form that is quick to integrate.

First-Order Linear Differential Equation

A first-order linear differential equation is an equation that involves the first derivative of a function. In its standard form, it appears as \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are known functions of \( x \).
This kind of equation is linear in both the function \( y \) and its first derivative \( \frac{dy}{dx} \). The primary goal in solving these equations is to find the function \( y \) that satisfies this relationship for all \( x \) within a given domain.
Why is this important?

• Linear differential equations often model real-world phenomena, making their solutions crucial in fields like physics, engineering, and economics.
• Once the equation is expressed in its standard form, techniques like spotting the integrating factor can be employed to find the solution.

The process typically involves transforming and simplifying the equation into one that is straightforward to integrate.

General Solution

The general solution of a differential equation is a solution that contains every particular solution of the equation. It generally includes a constant of integration, representing an entire family of solutions.
For the given first-order linear differential equation: \[ y = \frac{1}{e^{\int P(x)\,dx}} \left( \int e^{\int P(x)\,dx} Q(x) \,dx + C_1 \right) \]
This formula represents the general solution, as it includes the arbitrary constant \( C_1 \). This constant reflects the initial conditions or any particular solution.
Why is a general solution useful?

• It represents a broad solution set, allowing us to choose specific solutions by setting the constant based on boundary or initial conditions.
• This adaptability makes general solutions highly valuable in various scientific computations and modeling scenarios.

Therefore, deriving the general solution is a critical step in the process of problem-solving within differential equations.

Indefinite Integral

An indefinite integral is the antiderivative of a function, indicating the general form of an area under a curve. It represents a family of functions, of which the derivatives are the given function.
Expressed as \( \int f(x) \, dx \), an indefinite integral includes a constant of integration, \( C \), reflecting the infinite number of antiderivatives of a function.
How does it fit into solving differential equations?

• In first-order linear differential equations, an indefinite integral is used to find the integrating factor, \( e^{\int P(x) \,dx} \).
• Also employed in finding the general solution, it provides the necessary flexibility to accommodate the constant of integration.

The use of indefinite integrals in these equations opens the path to integrate the resulting expressions simply, thus solving the equation systematically.