2 marks, −0.667MathematicsDifference and differential equationsDifference equationsModerateabout 2 min

IIT JAM 2022 Question 16: Difference and differential equations

Let a second order difference equation be $y_{n+2} + 4y_n = 4y_{n+1}$ , $n = 2, 3, 4, \ldots$ , with $y_0 = 1$ , $y_1 = 4$ . Then the general solution is

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Answer and worked solution

Answer: (B)

Write the equation as $y_{n+2} - 4y_{n+1} + 4y_n = 0$ . The characteristic equation $r^2 - 4r + 4 = 0$ gives $(r-2)^2 = 0$ , a repeated root $r = 2$ , so $y_n = (c_1 + c_2 n)\,2^n$ .

From $y_0 = 1$ : $c_1 = 1$ . From $y_1 = 4$ : $(1 + c_2)\cdot 2 = 4$ , so $c_2 = 1$ . Hence $y_n = (1 + n)\,2^n$ , option (B).

The trap: (A)

(1 + n^2)\,2^n

A repeated root contributes a factor linear in $n$ , not quadratic; option (A) over-corrects with $n^2$ .

Plain-text version

Let a second order difference equation be yₙ₊₂ + 4yₙ = 4yₙ₊₁, n = 2, 3, 4, …, with y₀ = 1, y₁ = 4. Then the general solution is (A) (1 + n²) 2ⁿ (B) (1 + n) 2ⁿ (C) (1 + 1/n)2ⁿ (D) (n² + n + 1) 2ⁿ

Also written as: Let a second order difference equation be y_(n+2) + 4y_n = 4y_(n+1), n = 2, 3, 4,..., with y_0 = 1, y_1 = 4. Then the general solution is

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What this question tests

This question is on Difference equations, part of Difference and differential equations in Mathematics.

Solving equations that link a variable to its past values.

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IIT JAM 2021 · Q45Consider xₙ = ((n+1)/n)xₙ₋₁ for n = 1, 2, 3, … with x₀ = 2. Then x₁₀₀ - x₅₀ equals _ (in integer).

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