Class 12 Differential Equations · Long Answer3 Marksseparable_differential_equations
Class 12 Maths · Differential Equations · Long Answer #66
Solve y′ = cos(x + y) + sin(x + y).
✅ Step-by-Step Solution
Final Answer: ln|1 + tan((x + y)/2)| = x + C on applicable charts; also y = −x + π + 2kπ and y = −x − π/2 + 2kπ, k ∈ ℤ.
- Put z = x + y, so z′ = 1 + cos z + sin z. First retain equilibria with 1 + cos z + sin z = 0: z = π + 2kπ or −π/2 + 2kπ.
- For non-equilibrium branches, dz/(1 + cos z + sin z) = dx. Put t = tan(z/2), with dz = 2dt/(1 + t²).
- The integral reduces to ∫dt/(1 + t) = x + C, hence ln|1 + t| = x + C.
- Thus ln|1 + tan((x + y)/2)| = x + C locally; include both equilibrium families y = −x + π + 2kπ and y = −x − π/2 + 2kπ. Change tangent charts when necessary.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Put z = x + y, so z′ = 1 + cos z + sin z. First retain equilibria with 1 + cos z + sin z = 0: z = π + 2kπ or −π/2 + 2kπ.
- [1 mark] For non-equilibrium branches, dz/(1 + cos z + sin z) = dx. Put t = tan(z/2), with dz = 2dt/(1 + t²).
- [1 mark] The integral reduces to ∫dt/(1 + t) = x + C, hence ln|1 + t| = x + C.
- [0.5 marks] Thus ln|1 + tan((x + y)/2)| = x + C locally; include both equilibrium families y = −x + π + 2kπ and y = −x − π/2 + 2kπ. Change tangent charts when necessary.
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