Q97 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
The differential equation of the family of all parabolas whose axis is the $y$-axis is $\ldots$
- A$\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} = 0$
- Correct answerB$x\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 0$
- C$\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = 0$
- D$x\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} = 0$
Solution
The equation of the family of all parabolas whose axis is the $y$-axis is given by:
$y = ax^2 + c$
where $a$ and $c$ are arbitrary constants.
Differentiating both sides with respect to $x$, we get:
$\dfrac{dy}{dx} = 2ax$
$\Rightarrow 2a = \dfrac{1}{x} \dfrac{dy}{dx}$
Differentiating again with respect to $x$, we get:
$\dfrac{d^2y}{dx^2} = 2a$
Substituting the value of $2a$ from the first derivative into the second derivative, we get:
$\dfrac{d^2y}{dx^2} = \dfrac{1}{x} \dfrac{dy}{dx}$
$\Rightarrow x\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 0$
Answer: $x\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 0$
$y = ax^2 + c$
where $a$ and $c$ are arbitrary constants.
Differentiating both sides with respect to $x$, we get:
$\dfrac{dy}{dx} = 2ax$
$\Rightarrow 2a = \dfrac{1}{x} \dfrac{dy}{dx}$
Differentiating again with respect to $x$, we get:
$\dfrac{d^2y}{dx^2} = 2a$
Substituting the value of $2a$ from the first derivative into the second derivative, we get:
$\dfrac{d^2y}{dx^2} = \dfrac{1}{x} \dfrac{dy}{dx}$
$\Rightarrow x\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 0$
Answer: $x\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 0$
Hence, the correct answer is option: 'B'.