Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, solve each differential equation. $$ \frac{d y}{d x}+2 y=x $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{1}{2}x - \frac{1}{4} + Ce^{-2x} \).

Step 1: Identify the Type of Differential Equation

The equation is in the standard form of a first-order linear differential equation, given by \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( P(x) = 2 \) and \( Q(x) = x \).

Step 2: Determine the Integrating Factor

For a linear differential equation, the integrating factor \( \mu(x) \) is given by \( \mu(x) = e^{\int P(x) \, dx} \).Since \( P(x) = 2 \), we have:\[ \mu(x) = e^{\int 2 \, dx} = e^{2x} \]

Step 3: Multiply Through by the Integrating Factor

Multiply each term of the differential equation by the integrating factor \( e^{2x} \):\[ e^{2x} \frac{dy}{dx} + 2 e^{2x} y = x e^{2x} \]

Step 4: Recognize the Left Side as a Derivative

The left-hand side of the equation, after multiplying by the integrating factor, becomes the derivative of the product \( e^{2x} y \):\[ \frac{d}{dx} (e^{2x} y) = x e^{2x} \]

Step 5: Integrate Both Sides

Integrate both sides of the equation with respect to \( x \):\[ \int \frac{d}{dx} (e^{2x} y) \, dx = \int x e^{2x} \, dx \]This gives:\[ e^{2x} y = \int x e^{2x} \, dx \]

Step 6: Solve the Right Side Integral

Use integration by parts to solve the integral \( \int x e^{2x} \, dx \).Let \( u = x \) and \( dv = e^{2x} \, dx \), then \( du = dx \) and \( v = \frac{1}{2} e^{2x} \).By integration by parts, \( \int u \, dv = uv - \int v \, du \):\[ \int x e^{2x} \, dx = \frac{1}{2} x e^{2x} - \int \frac{1}{2} e^{2x} \, dx \]\[ = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C \]

Step 7: Substitute the Integral Back In

Substitute the result of the integral back into the equation:\[ e^{2x} y = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C \]

Step 8: Solve for y

Divide each term by \( e^{2x} \) to isolate \( y \):\[ y = \frac{1}{2}x - \frac{1}{4} + Ce^{-2x} \]

Step 9: Provide the General Solution

The general solution to the differential equation is:\[ y(x) = \frac{1}{2}x - \frac{1}{4} + Ce^{-2x} \]

Key Concepts

First-order linear differential equations

The differential equation presented, \( \frac{dy}{dx} + 2y = x \), is a classic example of a first-order linear differential equation. Such equations can be recognized by their standard form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)
• Here, \( P(x) \) is a function of \( x \) that multiplies the dependent variable \( y \), and \( Q(x) \) is another function of \( x \).
• To solve these equations, the integrating factor method is typically used.

In our specific equation, we identify \( P(x) = 2 \) and \( Q(x) = x \). The strategy for solving begins by calculating an integrating factor, allowing us to treat the left side as a derivative. This transforms the equation into one that is easier to integrate and subsequently solve for \( y \).

Integrating factor

The integrating factor is a crucial component in solving first-order linear differential equations. It is a function, often denoted as \( \mu(x) \), that simplifies the differential equation by turning the left-hand side into a perfect derivative:

• The formula for the integrating factor is \( \mu(x) = e^{\int P(x) \, dx} \).
• In our case, with \( P(x) = 2 \), the integrating factor becomes \( \mu(x) = e^{2x} \).

When the entire differential equation is multiplied by the integrating factor, it allows the left-hand expression's transformation into \( \frac{d}{dx}(\mu(x)y) \). This simplification is key to manage the equation efficiently, reducing it to a form where straightforward integration is possible.

Integration by parts

Integration by parts is an essential technique used in calculus, especially when solving integrals that are products of a polynomial and an exponential function, like \( \int x e^{2x} \, dx \):

• This method is based on the product rule for differentiation and is given by: \( \int u \, dv = uv - \int v \, du \).
• In this scenario, choosing \( u = x \) and \( dv = e^{2x} \, dx \) allows us to calculate: \( du = dx \) and \( v = \frac{1}{2} e^{2x} \).
• Substituting into the integration by parts formula results in: \( \int x e^{2x} \, dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C \).

This approach helps resolve otherwise difficult integrals, playing a pivotal role in solving our differential equation.

General solution of differential equations

After applying all previous steps, we reach the general solution of the differential equation. This represents the family of curves meeting the criteria established by the differential equation.

• We derived \( y = \frac{1}{2} x - \frac{1}{4} + Ce^{-2x} \) as the solution.
• \( C \) is an arbitrary constant that modifies the curve depending on the initial conditions or specific constraints provided.
• This solution captures all possible behaviors of the original differential equation.

Understanding how these solutions arise helps in grasping not only the mechanics of solving differential equations but also the physical or real-world systems they model.