NCERT Class 12 · Ch 9 Exercise 9.3 · Q10
NCERT Class 12 Maths · Chapter 9 Differential Equations · Exercise 9.3 Question 10
Solve eˣtan y dx+(1−eˣ)sec²y dy=0.
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: tan y=C(1−eˣ).
- Put z=tan y, so dz=sec²y dy. The equation is (1−eˣ)z′+eˣz=0.
- Then z′/z=−eˣ/(1−eˣ), so z=C(1−eˣ). Include C=0 directly.
- Hence tan y=C(1−eˣ), on intervals with cos y≠0. The explicit tangent relation can extend through x=0.
- Differentiate the resulting relation to recover the equation on intervals where the original expressions and the required slope are defined.
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