Class 12 Differential Equations · Long Answer3 Markslinear_differential_equations
Class 12 Maths · Differential Equations · Long Answer #68
Solve (1 + y²) + (x − e^(arctan y))y′ = 0.
✅ Step-by-Step Solution
Final Answer: x = ½e^(arctan y) + Ce^(−arctan y), with locally invertible branches for y(x).
- Treat x as dependent on y: dx/dy + x/(1 + y²) = e^(arctan y)/(1 + y²).
- Use IF e^(arctan y), and set t = arctan y.
- Then xeᵗ = ∫e^(2t)dt + C = ½e^(2t) + C.
- Thus x = ½e^(arctan y) + Ce^(−arctan y). To obtain y(x), restrict to dx/dy ≠ 0, consistent with the original graph equation.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Treat x as dependent on y: dx/dy + x/(1 + y²) = e^(arctan y)/(1 + y²).
- [1 mark] Use IF e^(arctan y), and set t = arctan y.
- [1 mark] Then xeᵗ = ∫e^(2t)dt + C = ½e^(2t) + C.
- [0.5 marks] Thus x = ½e^(arctan y) + Ce^(−arctan y). To obtain y(x), restrict to dx/dy ≠ 0, consistent with the original graph equation.
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