Question

If the solution of \[ \left( 1 + 2e^\frac{x}{y} \right) dx + 2e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0 \] is \[ x + \lambda y e^\frac{x}{y} = c \quad \text{(where \(c\) is an arbitrary constant), then \( \lambda \) is:} \]

• A. 1
• B. 2
• C. 3
• D. None of these

Hint:

When solving differential equations, identifying patterns in the given equation can help choose a useful substitution that simplifies the problem.

Answer 1

The Correct Option is B

The appearance of \( \frac{x}{y} \) in the given equation suggests the substitution: \[ \frac{x}{y} = u \] Differentiating: \[ dx = u dy + y du \] Rewriting the given equation using this substitution: \[ (1 + 2e^u) (u dy + y du) + 2e^u (1 - u) dy = 0 \] Expanding: \[ (1 + 2e^u) y du + (1 + 2e^u) u dy + 2e^u (1 - u) dy = 0 \] Factorizing the terms: \[ (1 + 2e^u) du + \frac{dy}{y} = 0 \] This is now a variable separable form: \[ \frac{(1 + 2e^u)}{1 + 2e^u} du + \frac{dy}{y} = 0 \] Integrating both sides: \[ \log |1 + 2e^u| + \log |y| = \log |c| \] \[ (1 + 2e^u) y = c \] Substituting \( u = \frac{x}{y} \): \[ (x + 2 y e^\frac{x}{y}) = c \] Comparing with the given general solution, we get: \[ \lambda = 2 \]