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NDA Mathematics Differential Equations 2025 NDA 2025 (Phase 2)

NDA Mathematics Question (2025) — Solution

Question

If k is an arbitrary constant, then what is the general solution of the equation (x + y)^2 dy dx = k^2?

Options

  1. A. y + x = (x + c) + k
  2. B. x + y = k ( y - c k )
  3. C. x - y = k ( y - c k )
  4. D. y - x = (x + c) + k

Answer

B. x + y = k ( y - c k )

Step-by-step solution

Let x + y = v Differentiating with respect to x, we get 1 + dy dx = dv dx dy dx = dv dx - 1 Substituting in the given differential equation: v^2 ( dv dx - 1 ) = k^2 v^2 dv dx = v^2 + k^2 v^2 v^2 + k^2 dv = dx ( 1 - k^2 v^2 + k^2 ) dv = dx Integrating both sides: ( 1 - k^2 v^2 + k^2 ) dv = dx v - k ^ -1 ( v k ) = x + c Substituting v = x + y: x + y - k ^ -1 ( x + y k ) = x + c y - c = k ^ -1 ( x + y k ) y - c k = ^ -1 ( x + y k ) x + y = k ( y - c k ) Answer: x + y = k ( y - c k )

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