Question

The solution curve of the differential equation \( y \frac{dx}{dy} = x (\log_e x - \log_e y + 1) \), \( x > 0 \), \( y > 0 \) passing through the point \( (e, 1) \) is

• A. \(\quad \left| \log_e \frac{y}{x} \right| = x\)
• B. \(\quad \left| \log_e \frac{y}{x} \right| = y^2\)
• C. \(\quad \left| \log_e \frac{x}{y} \right| = y\)
• D. \(\quad 2 \left| \log_e \frac{x}{y} \right| = y + 1\)

Answer 1

The Correct Option is C

Given:

\[ \frac{dx}{dy} = \frac{x}{y} \left( \ln \left( \frac{x}{y} \right) + 1 \right) \]

Let:

\[ \frac{x}{y} = t \quad \implies \quad x = ty \]

Differentiating:

\[ \frac{dx}{dy} = t + y \frac{dt}{dy} \]

Substitute:

\[ t + y \frac{dt}{dy} = t \left( \ln(t) + 1 \right) \]

Rearranging:

\[ \frac{dt}{dy} = \frac{t \ln(t)}{y} \]

Integrating both sides:

\[ \int \frac{1}{t} dt = \int \frac{dy}{y} \]

Let \( \ln t = p \):

\[ dp = \frac{1}{t} dt \quad \implies \quad \ln \left( \frac{x}{y} \right) = y \]