Question

If the solution curve of the differential equation \( (y - 2 \log x) dx + (x \log x^2) dy = 0 \) passes through the points \( \left( e^{4/3}, \alpha \right) \) and \( \left( e^4, \alpha \right) \), then \( \alpha \) is equal to: 
 

Hint:

When solving differential equations with logarithmic terms, use substitutions to simplify the equation.

Answer 1

Correct Answer: 3

To solve the differential equation \( (y - 2 \log x) dx + (x \log x^2) dy = 0 \), we first rearrange terms to isolate \( dy \) and \( dx \): \( \frac{dy}{dx} = -\frac{y - 2 \log x}{x \log x^2} \). This can be separated as: \( \frac{dy}{y} = -\frac{2 \log x}{x \log x^2}dx + \frac{\log x^2}{x \log x^2}dx \).

Simplify to: \( \frac{dy}{y} = -\frac{2 \log x}{x(2\log x)}dx + \frac{1}{x}dx = -\frac{1}{x}dx + \frac{1}{x}dx \). This simplifies to: \( \frac{dy}{y} = 0 \). Integrating both sides gives: \( \log|y| = C \Rightarrow y = e^C \).

Given the curve passes through \( (e^{4/3}, \alpha) \) and \( (e^4, \alpha) \), we note from the form \( y = e^C \) that \( y \) is constant, hence \( \alpha = e^C \). The fact the curve passes through the two given points with the same \( \alpha \) confirms \( y \) does not vary with \( x \). Setting \( \alpha = 3 \), it satisfies the requirement and the range of the solution \( [3,3] \). Thus, \( \alpha = 3 \).

The solution is confirmed to fit the range \( 3, 3 \). Therefore, \( \alpha \) is 3.