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Ex 9.3, 14 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition : 𝑑𝑦/𝑑π‘₯=𝑦 tan⁑ π‘₯;𝑦=1 When π‘₯=0 𝑑𝑦/𝑑π‘₯=𝑦 tan⁑ x π’…π’š/π’š=γ€–π­πšπ§ 𝐱 𝐝𝐱〗⁑ Integrating both sides ∫1▒𝑑𝑦/𝑦 = ∫1β–’tan⁑〖π‘₯ 𝑑π‘₯γ€— log |π’š| = log |π¬πžπœβ‘π’™ |+π’π’π’ˆβ‘π’„ log |𝑦| = log (c sec x) y = c sec x (∡ log π‘Ž + log b = log π‘Žb) 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.