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 solve the differential equation \( y \frac{dx}{dy} = x (\log_e x - \log_e y + 1) \), we first rearrange the equation by separating variables.

The given equation is:

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

Rearranging, we have:

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

Bringing all terms involving \( x \) and \( y \) on one side:

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

This can be rewritten as:

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

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

Substituting this into our differential equation, we integrate along:

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

This leads us 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 given by the point \((e, 1)\).

From the initial condition \((x, y) = (e, 1)\), substituting in, we have:

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

This gives the solution satisfying \(\left| \log_e \frac{x}{y} \right| = y\), matching the given point \((e, 1)\).

Therefore, the correct solution curve is:

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

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 \( \ln t = p \):

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