Class 12 Differential Equations · Short Answer2 Marksfirst_order_differential_equations
Class 12 Maths · Differential Equations · Short Answer #39
Solve cos(y′) = a, −1 ≤ a ≤ 1, with y(0) = 1.
✅ Step-by-Step Solution
Final Answer: y = 1 + mx, where m = 2kπ ± arccos a, k ∈ ℤ.
- For a differentiable solution, y′ takes values in {2kπ ± arccos a : k ∈ ℤ}. By the intermediate-value property of derivatives, it is constant on a connected interval.
- Write y′ = m for one such value. Integration gives y = mx + C, and y(0) = 1 gives C = 1.
- Therefore all branches are y = 1 + (2kπ ± arccos a)x. The principal branch alone is y = 1 + x arccos a.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] For a differentiable solution, y′ takes values in {2kπ ± arccos a : k ∈ ℤ}. By the intermediate-value property of derivatives, it is constant on a connected interval.
- [1 mark] Write y′ = m for one such value. Integration gives y = mx + C, and y(0) = 1 gives C = 1.
- [0.5 marks] Therefore all branches are y = 1 + (2kπ ± arccos a)x. The principal branch alone is y = 1 + x arccos a.
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