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

To resolve the differential equation \( y \frac{dx}{dy} = x (\log_e x - \log_e y + 1) \), we begin by rearranging the equation to separate variables.

The initial equation is:

\(y \frac{dx}{dy} = x (\log_e x - \log_e y + 1)\)

Upon rearrangement, we obtain:

\(\frac{dx}{dy} = \frac{x}{y} (\log_e x - \log_e y + 1)\)

Consolidating terms involving \( x \) and \( y \) to one side yields:

\(\frac{y}{x} \frac{dx}{dy} = \log_e x - \log_e y + 1\)

This expression can be further simplified to:

\(\frac{y}{x} \frac{dx}{dy} = \log_e \frac{x}{y} +1\)

Let \(u = \log_e \frac{x}{y}\). Then, the differential of \( u \) is \(du = \left(\frac{1}{x}\frac{dx}{dy} - \frac{1}{y} \right)\, dy\).

Substituting this into the differential equation, we proceed with integration:

\(\frac{dy}{dx} = \frac{\log_e \frac{x}{y} + 1}{\frac{y}{x}} = \left(\log_e \frac{x}{y} + 1\right) \frac{x}{y}\)

This transformation leads to:

\(y\, \frac{dx}{dy} = x \left(\log_e \frac{x}{y} + 1\right)\)

Integrating both sides with respect to \( y \), we apply the initial condition provided by the point \((e, 1)\).

Using the initial condition \((x, y) = (e, 1)\) for substitution, we find:

\(\log_e \frac{e}{1} = 1\)

This results in the solution satisfying \(\left| \log_e \frac{x}{y} \right| = y\), which is consistent with the given point \((e, 1)\).

Consequently, the determined solution curve is:

\(\left| \log_e \frac{x}{y} \right| = y\)