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ExamsJEE AdvancedMaths

A list-matching question on differential equations: List-I (Differential Equations): (P) Solution of y - x(dy/dx) = y² + (dy/dx) (Q) Solution of (2x - 10y³)(dy/dx) + y = 0 (R) Solution of (sec² y)(dy/dx) + tan y = 1 (i.e., sec² y dy + tan y dx = dx) (S) Solution of sin y (dy/dx) = cos y (1 - x cos y) Match each differential equation in List-I with its general solution in List-II.

  1. P -> 1; Q -> 2; R -> 3; S -> 4
  2. P -> 3; Q -> 1; R -> 4; S -> 2
  3. P -> 1; Q -> 3; R -> 4; S -> 2
  4. P -> 3; Q -> 4; R -> 1; S -> 2

Correct answer: P -> 3; Q -> 1; R -> 4; S -> 2

Solution

These are standard ODE types: (P) y - x(dy/dx) = y² + dy/dx => (dy/dx)(-x-1) = y² - y => standard separable or Bernoulli. (Q) (2x - 10y³)dy/dx + y = 0 => treat as dx/dy: dx/dy = (10y³ - 2x)/y => dx/dy + 2x/y = 10y², linear in x with IF = y². (R) sec² y (dy/dx) + tan y = 1: let t = tan y, dt/dx + t = 1, linear, solution: t = 1 + Ce^-x i.e. tan y = 1 + Ce^-x. (S) sin y dy/dx = cos y(1 - x cos y): let v = sec y, dv/dx = -csc y * cos y * dy/dx... let u = cos y, du/dx = -sin y dy/dx. Rewrite: -du/dx = u(1-xu) => Bernoulli in u. Solution gives x*sec y form. The standard matching for JEE gives: P->3, Q->1, R->4, S->2.

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