Problem 15 - Calculus: Early Transcendentals

Question

Solve the initial-value problem. \( x^2y' + 2xy = \ln x, y(1) = 2 \)

Step-by-Step Solution

Answer

The solution is \( y = \frac{x \ln x - x + 3}{x^2}. \)

Step 1: Identify the Type of Differential Equation

The given differential equation is \( x^2y' + 2xy = \ln x \). Since the equation can be rewritten as \( y' + \frac{2}{x}y = \frac{\ln x}{x^2} \), it is a linear first-order differential equation of the form \( y' + P(x)y = Q(x) \).

Step 2: Calculate the Integrating Factor

To solve the equation, find the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{2}{x} \), so: \[ \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = x^2. \]

Step 3: Multiply the Differential Equation by the Integrating Factor

Multiply every term of the differential equation by the integrating factor \( x^2 \). The equation becomes: \[ x^2 \/ (x^2 y' + 2xy) = x^2 \ln x/x^2 \], which simplifies to: \[ x^2y' + 2xy = \ln x. \]

Step 4: Express the Left Side as a Derivative

The left-hand side of the equation \( x^2y' + 2xy \) can be rewritten as a derivative:\[ \frac{d}{dx}(x^2y) = \ln x. \]

Step 5: Integrate Both Sides

Integrate both sides with respect to \(x\). The left side becomes:\[ \int \frac{d}{dx}(x^2y) \, dx = x^2y, \]and the right side becomes:\[ \int \ln x \, dx. \]Using integration by parts, let \( u = \ln x \) and \( dv = dx \), then \( du = \frac{1}{x} dx \) and \( v = x \). The integral becomes:\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C. \]

Step 6: Apply the Initial Condition

Using the initial condition \( y(1) = 2 \), substitute into the equation:\[ 1^2 \, y = 1 \ln 1 - 1 + C. \]This simplifies to: \[ 2 = 0 - 1 + C \Rightarrow C = 3. \]

Step 7: Find the Solution Function

Using all the steps, substitute \( C = 3 \) back into our expression:\[ x^2y = x \ln x - x + 3, \]and solve for \( y(x) \):\[ y = \frac{x \ln x - x + 3}{x^2}. \]

Key Concepts

Linear First-Order Differential Equation

A linear first-order differential equation is a type of equation that involves the first derivative of a function and can be expressed in the form \( y' + P(x)y = Q(x) \). In this context, 'linear' refers to how the function and its derivative appear in the equation—specifically, they are not raised to any power other than one. Linear first-order equations are straightforward to work with because they follow a predictable pattern.

In our original problem, we have an equation \( x^2y' + 2xy = \ln x \). When simplified, this becomes \( y' + \frac{2}{x}y = \frac{\ln x}{x^2} \). This confirms our equation fits the form of a linear first-order differential equation. Such equations are solvable using methods like an integrating factor, as we'll discuss next.

Integrating Factor

The integrating factor is a crucial tool for solving linear first-order differential equations. It is a function, usually denoted \( \mu(x) \), that simplifies the process of solving these equations by allowing us to rewrite them in a more manageable form.

• To find the integrating factor, we use the formula \( \mu(x) = e^{\int P(x) \, dx} \).
• For our equation, \( P(x) = \frac{2}{x} \), which leads to \( \mu(x) = e^{2 \ln |x|} = x^2 \).

Multiplying the entire differential equation by this integrating factor turns it into an equation where the left-hand side is the derivative of a product. This transformation is eyes open into easier steps and highlights why integrating factors are indispensable in solving such equations.

Integration by Parts

Integration by parts is a technique used to integrate products of functions, described by the formula \( \int u \, dv = uv - \int v \, du \). It's a key method when dealing with expressions that are difficult to integrate directly.

• The formula is derived from the product rule of differentiation, and is useful specifically for integrals involving products of different types of functions.
• In solving our differential equation, we need to integrate \( \ln x \). By choosing \( u = \ln x \) and \( dv = dx \), we find \( du = \frac{1}{x} dx \) and \( v = x \).
• This leads to \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C \).

The calculation illustrates how integration by parts can untangle even complex logarithmic functions when solving differential equations.

Initial Condition

The initial condition is an additional piece of information provided with a differential equation, specifying the value of the solution at a particular point. It's typically used to find the constant of integration once we have the general solution to the differential equation.

• For the problem at hand, the initial condition is \( y(1) = 2 \).
• After integrating both sides of the equation, an arbitrary constant \( C \) appears. We apply the initial condition to determine \( C \).
• Substituting \( x = 1 \) and \( y = 2 \) into the integrated expression, we solve for \( C \), finding that \( C = 3 \).

The initial condition ensures uniqueness of the solution, tailoring the general formula we obtain to the specific situation described in the problem. This step completes our journey from a differential equation to a concrete solution that fits the criteria exactly.