Q93 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
The differential equation representing the family of curves $x \sin x + y^3 = 4ax$ is
- A$\dfrac{dy}{dx} = \dfrac{y^3 + x^2 \cos x}{3xy^2}$
- Correct answerB$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$
- C$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy}$
- D$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3x^2 y}$
Solution
The given equation of the family of curves is $x \sin x + y^3 = 4ax$.
Dividing the entire equation by $x$, we get:
$\sin x + \dfrac{y^3}{x} = 4a$
Differentiating both sides with respect to $x$, we obtain:
$\cos x + \dfrac{3y^2 \dfrac{dy}{dx} \cdot x - y^3 \cdot 1}{x^2} = 0$
Multiplying the equation by $x^2$ gives:
$x^2 \cos x + 3xy^2 \dfrac{dy}{dx} - y^3 = 0$
Rearranging the terms to solve for $\dfrac{dy}{dx}$:
$3xy^2 \dfrac{dy}{dx} = y^3 - x^2 \cos x$
$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$
Answer: $\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$
Dividing the entire equation by $x$, we get:
$\sin x + \dfrac{y^3}{x} = 4a$
Differentiating both sides with respect to $x$, we obtain:
$\cos x + \dfrac{3y^2 \dfrac{dy}{dx} \cdot x - y^3 \cdot 1}{x^2} = 0$
Multiplying the equation by $x^2$ gives:
$x^2 \cos x + 3xy^2 \dfrac{dy}{dx} - y^3 = 0$
Rearranging the terms to solve for $\dfrac{dy}{dx}$:
$3xy^2 \dfrac{dy}{dx} = y^3 - x^2 \cos x$
$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$
Answer: $\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$
Hence, the correct answer is option: 'B'.