Problem 11 - Calculus: Early Transcendentals

Question

Solve the differential equation. \( xy' - 2y = x^2, x > 0 \)

Step-by-Step Solution

Answer

The solution is \( y = x^2(\ln x + C) \).

Step 1: Rewrite the Differential Equation

The given differential equation is \( xy' - 2y = x^2 \). We can start by rewriting it in standard linear form. The standard form for a first-order linear differential equation is \( y' + P(x)y = Q(x) \). Divide through by \( x \) to achieve this form: \( y' - \frac{2}{x}y = x \).

Step 2: Identify the Integrating Factor

For our rewritten equation, \( P(x) = -\frac{2}{x} \). The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \). Calculating gives us: \( \mu(x) = e^{-\int \frac{2}{x} \ dx} = e^{-2 \ln |x|} = x^{-2} \) (since \( e^{\ln a} = a \)).

Step 3: Multiply the Differential Equation by the Integrating Factor

Next, multiply the entire differential equation by the integrating factor \( x^{-2} \) to make the left side an exact derivative: \( x^{-2} y' - \frac{2}{x} \cdot x^{-2} y = x \cdot x^{-2} \). Simplifies to: \( \frac{d}{dx}(x^{-2}y) = x^{-1} \).

Step 4: Integrate Both Sides

Integrate both sides with respect to \( x \), \( \int \frac{d}{dx}(x^{-2}y) \, dx = \int x^{-1} \, dx \) This simplifies to: \( x^{-2}y = \ln |x| + C \), where \( C \) is the constant of integration.

Step 5: Solve for y

Finally, solve for \( y \) by multiplying both sides by \( x^2 \): \( y = x^2(\ln |x| + C) \). Since \( x > 0 \), we can simplify to: \( y = x^2(\ln x + C) \).

Key Concepts

Integrating Factor

In solving linear differential equations, the integrating factor is a crucial auxiliar tool. It simplifies the equation into a form that can be integrated directly.
To find the integrating factor, we identify function \( P(x) \) from the standard linear equation format, \( y' + P(x)y = Q(x) \). In our example, the function is \( P(x) = -\frac{2}{x} \). To determine the integrating factor, we use the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

This simplifies the equation to a point where integration becomes straightforward.
For example:

• Calculate \( \int -\frac{2}{x} \, dx \) which equals \(-2 \ln|x|\).
• This simplifies to the integrating factor: \( \mu(x) = e^{-2 \ln |x|} = x^{-2} \).

By multiplying the original differential equation by \( x^{-2} \), we transform it into an exact balance of derivatives on one side, paving the way for straightforward integration.

Linear Differential Equation

Linear differential equations are foundational in calculus and differential equations. They involve derivatives of a function and are usually represented in the form \( y' + P(x)y = Q(x) \).
This structure is not just linear in \( y' \) but also in \( y \) itself, meaning there are no nonlinear terms like \( y^2 \) or \( (y')^2 \). In the problem we address, we initially have:

• \( xy' - 2y = x^2 \)

To convert it into standard form, divide everything by \( x \):

• Resulting in: \( y' - \frac{2}{x}y = x \).

This transformation might seem minor, but it is significant because it sets us up perfectly to apply an integrating factor.
Understanding this structure is vital as it ensures we correctly apply the appropriate method for solving the equation.
Once in this form, solving the equation becomes a systematic process.

Exact Derivative

The exact derivative simplifies the integration process. By converting the differential equation using the integrating factor, the left-hand side can be condensed into the expression of a derivative.
In the solved equation:

• We have: \( x^{-2}y'-\frac{2}{x}x^{-2}y = x^{-1} \).

Multiplying all terms by the integrating factor \( x^{-2} \) converts the left side to an exact derivative:

• \( \frac{d}{dx}(x^{-2}y) \).

This transformation is critical.
Once you express the equation as an exact derivative, integration is straightforward since the derivative of a function, when integrated, returns the original function.

• It transforms the equation to: \( x^{-2}y = \int x^{-1} \, dx \).
The process underscores why finding the exact expression of a derivative is a simple yet powerful tool in solving differential equations.