Q247 / 270 · NDA 2026 (Phase 1) · Mathematics
Passage: Consider the differential equation $e^{x+y}\dfrac{dy}{dx}=e^{x-y}$.
Question: What is $\dfrac{d^2y}{dx^2}\left(\dfrac{dx}{dy}\right)^2$ equal to ?
Question: What is $\dfrac{d^2y}{dx^2}\left(\dfrac{dx}{dy}\right)^2$ equal to ?
- Correct answerA$-2$
- B$-1$
- C$0$
- D$2$
Solution
Given the differential equation:
$e^{x+y}\dfrac{dy}{dx} = e^{x-y}$
$e^x e^y \dfrac{dy}{dx} = e^x e^{-y}$
Dividing both sides by $e^x$:
$\dfrac{dy}{dx} = \dfrac{e^{-y}}{e^y} = e^{-2y}$
Differentiating both sides with respect to $x$:
$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}(e^{-2y}) = -2e^{-2y} \dfrac{dy}{dx}$
Substituting $\dfrac{dy}{dx} = e^{-2y}$:
$\dfrac{d^2y}{dx^2} = -2e^{-2y}(e^{-2y}) = -2e^{-4y}$
From $\dfrac{dy}{dx} = e^{-2y}$, we get:
$\dfrac{dx}{dy} = \dfrac{1}{e^{-2y}} = e^{2y}$
Squaring both sides:
$\left(\dfrac{dx}{dy}\right)^2 = (e^{2y})^2 = e^{4y}$
Now, multiplying the two results:
$\dfrac{d^2y}{dx^2}\left(\dfrac{dx}{dy}\right)^2 = (-2e^{-4y})(e^{4y}) = -2e^0 = -2$
Answer: $-2$
$e^{x+y}\dfrac{dy}{dx} = e^{x-y}$
$e^x e^y \dfrac{dy}{dx} = e^x e^{-y}$
Dividing both sides by $e^x$:
$\dfrac{dy}{dx} = \dfrac{e^{-y}}{e^y} = e^{-2y}$
Differentiating both sides with respect to $x$:
$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}(e^{-2y}) = -2e^{-2y} \dfrac{dy}{dx}$
Substituting $\dfrac{dy}{dx} = e^{-2y}$:
$\dfrac{d^2y}{dx^2} = -2e^{-2y}(e^{-2y}) = -2e^{-4y}$
From $\dfrac{dy}{dx} = e^{-2y}$, we get:
$\dfrac{dx}{dy} = \dfrac{1}{e^{-2y}} = e^{2y}$
Squaring both sides:
$\left(\dfrac{dx}{dy}\right)^2 = (e^{2y})^2 = e^{4y}$
Now, multiplying the two results:
$\dfrac{d^2y}{dx^2}\left(\dfrac{dx}{dy}\right)^2 = (-2e^{-4y})(e^{4y}) = -2e^0 = -2$
Answer: $-2$
Hence, the correct answer is option: 'A'.