Question

The solution of the differential equation \( \frac{d}{dt} \log \frac{x}{t} = x \) is _______.

• A. \( x = e^{ct} \)
• B. \( x + e^{ct} = 0 \)
• C. \( x = e^t + t \)
• D. \( xe^{ct} = 0 \)

Hint:

For differential equations, use separation of variables and verify solutions by substitution.

Answer 1

The Correct Option is A

\[ \frac{d}{dt} \log \frac{x}{t} = \frac{d}{dt} (\log x - \log t) = \frac{1}{x} \cdot \frac{dx}{dt} - \frac{1}{t} = x. \] \[ \frac{dx}{dt} = x (x + \frac{1}{t}). \] Separate variables: \( \frac{dx}{x (x + \frac{1}{t})} = dt \).
Partial fractions: \( \frac{1}{x (x + \frac{1}{t})} = \frac{A}{x} + \frac{B}{x + \frac{1}{t}} \). Solve: \[ 1 = A (x + \frac{1}{t}) + B x \quad \Rightarrow \quad A = t, \quad B = -t. \] \[ \int \left( \frac{t}{x} - \frac{t}{x + \frac{1}{t}} \right) dx = \int dt. \] \[ t \ln x - t \ln \left( x + \frac{1}{t} \right) = t + c \quad \Rightarrow \quad \ln \frac{x}{x + \frac{1}{t}} = 1 + \frac{c}{t}. \] \[ \frac{x}{x + \frac{1}{t}} = e^{1 + \frac{c}{t}} \quad \Rightarrow \quad \text{No simple form matches options.} \] Options seem incorrect; solution is complex.