Class 12 Differential Equations · Very Long Answer5 Marksseparable_differential_equations

Class 12 Maths · Differential Equations · Very Long Answer #70

Solve 3eˣtan y dx + (2 − eˣ)sec²y dy = 0, y(0) = π/4.

✅ Step-by-Step Solution

Final Answer: y = arctan((2 − eˣ)³), x ∈ ℝ.

Put u = tan y, so u′ = sec²y y′. The equation becomes (2 − eˣ)u′ + 3eˣu = 0.
For x ≠ ln 2 and u ≠ 0, separate du/u = −3eˣdx/(2 − eˣ).
Integrate ln|u| = 3ln|2 − eˣ| + C, so u = A(2 − eˣ)³.
At x = 0, u = 1 gives A = 1. Choose the arctangent branch through π/4.
Thus y = arctan((2 − eˣ)³). Direct substitution verifies its smooth extension through x = ln 2. Because that point is singular, other global continuations can use a different constant on x > ln 2; the displayed branch is one valid continuation.

[1 mark] Put u = tan y, so u′ = sec²y y′. The equation becomes (2 − eˣ)u′ + 3eˣu = 0.
[1 mark] For x ≠ ln 2 and u ≠ 0, separate du/u = −3eˣdx/(2 − eˣ).
[1 mark] Integrate ln|u| = 3ln|2 − eˣ| + C, so u = A(2 − eˣ)³.
[1 mark] At x = 0, u = 1 gives A = 1. Choose the arctangent branch through π/4.
[1 mark] Thus y = arctan((2 − eˣ)³). Direct substitution verifies its smooth extension through x = ln 2. Because that point is singular, other global continuations can use a different constant on x > ln 2; the displayed branch is one valid continuation.

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