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Ex 9.3

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Ex 9.3, 14 You are here

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Ex 9.3, 23 (MCQ)

Last updated at Aug. 11, 2023 by Teachoo

Ex 9.3, 14 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition : ππ¦/ππ₯=π¦ tanβ‘ π₯;π¦=1 When π₯=0 ππ¦/ππ₯=π¦ tanβ‘ x π π/π=γπππ§ π± ππ±γβ‘ Integrating both sides β«1βππ¦/π¦ = β«1βtanβ‘γπ₯ ππ₯γ log |π| = log |π¬ππβ‘π |+πππβ‘π log |π¦| = log (c sec x) y = c sec x (β΅ log π + log b = log πb) Put x = 0 and y = 1 1 = C Sec 0 1 = C Γ 1 C = 1 Put value of C in (1) y = 1 Γ sec x y = sec x