Question

Solve: \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \).

Hint:

When finding the IF for \( \int \frac{1}{x \log x} dx \), remember that \( \log(\log x) \) is the integral, which allows the exponential to cancel perfectly.

Answer 1

Step 1: Understanding the Concept:
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \). We use the Integrating Factor (IF).
Step 2: Detailed Explanation:
Divide the equation by \( x \log x \):
\[ \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2} \]
Here, \( P = \frac{1}{x \log x} \) and \( Q = \frac{2}{x^2} \).
Integrating Factor (\( IF \)):
\[ IF = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} \]
Let \( t = \log x \implies dt = \frac{1}{x} dx \).
\[ IF = e^{\int \frac{1}{t} dt} = e^{\log t} = t = \log x \]
The general solution is:
\[ y \cdot (IF) = \int Q \cdot (IF) dx + C \]
\[ y \log x = \int \frac{2}{x^2} \log x dx + C \]
Using integration by parts on \( \int 2x^{-2} \log x dx \):
Let \( u = \log x \), \( dv = 2x^{-2} dx \). Then \( du = \frac{1}{x} dx \), \( v = -\frac{2}{x} \).
\[ y \log x = -\frac{2}{x} \log x - \int -\frac{2}{x^2} dx \]
\[ y \log x = -\frac{2}{x} \log x - \frac{2}{x} + C = -\frac{2}{x}(\log x + 1) + C \]
Step 3: Final Answer:
The solution is \( y \log x = -\frac{2}{x}(1 + \log x) + C \).