NDA Mathematics Question (2026) — Solution

Step-by-step solution

Given the differential equation: e^ x+y dy dx = e^ x-y e^x e^y dy dx = e^x e^ -y Dividing both sides by e^x: dy dx = e^ -y e^y = e^ -2y Differentiating both sides with respect to x: d^2y dx^2 = d dx (e^ -2y ) = -2e^ -2y dy dx Substituting dy dx = e^ -2y : d^2y dx^2 = -2e^ -2y (e^ -2y ) = -2e^ -4y From dy dx = e^ -2y , we get: dx dy = 1 e^ -2y = e^ 2y Squaring both sides: ( dx dy )^2 = (e^ 2y )^2 = e^ 4y Now, multiplying the two results: d^2y dx^2 ( dx dy )^2 = (-2e^ -4y )(e^ 4y ) = -2e^0 = -2 Answer: -2