Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
If $\frac{dy}{dx} + y\frac{dx}{dy} = x.y(-2) = 1$, then :
$x + 3y^2 = 1$
$2x + y + 3 = 0$
$x + y + 1 = 0$
$x^2 - 4y = 0$
Step-by-Step Solution
Key Concept: Clairaut's equation yields both a family of linear general solutions and a singular (envelope) solution found via the discriminant method.
This is Clairaut's equation $y = xp - p^2$ where $p = rac{dy}{dx}$. The general solution is $y = cx - c^2$. Setting the discriminant condition for singular solution: the envelope is found from $ rac{\partial}{\partial c}(y - cx + c^2) = 0 \Rightarrow -x + 2c = 0 \Rightarrow c = rac{x}{2}$. Substituting back: $y = x \cdot rac{x}{2} - ( rac{x}{2})^2 = rac{x^2}{4}$.
Correct Answer: 3,4