If y = e^(2 tan⁻¹x), prove that (1 + x²)y₂ + 2(x − 1)y₁ = 0.

1. 1 · Write what is given and what is to be found. Given: y = e^(2 tan⁻¹x). To prove: (1 + x²)y₂ + 2(x − 1)y₁ = 0.
2. 2 · Write the rules. d/dx e^(u) = e^(u) × u′, and d/dx tan⁻¹x = 1/(1 + x²).
3. 3 · Find y₁. y₁ = e^(2 tan⁻¹x) × 2/(1 + x²) = 2y/(1 + x²).
4. 4 · Clear the fraction. (1 + x²)y₁ = 2y.
5. 5 · Differentiate again. Product rule: 2xy₁ + (1 + x²)y₂ = 2y₁.
6. 6 · Rearrange. (1 + x²)y₂ + 2xy₁ − 2y₁ = 0, that is (1 + x²)y₂ + 2(x − 1)y₁ = 0.
7. 7 · Check and box the answer. The result matches the required form. Hence proved.

Scoring tip: Clear the fraction before differentiating the second time. Using the quotient rule here wastes time and invites slips.