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

Answer: (C)

$\quad \left| \log_e \frac{x}{y} \right| = y$

Answer 2

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 $\ln t = p$:

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