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Ex 9.6, 13 - Find particular solution: dy/dx + 2y tan x = sin x - Ex 9.6

Ex 9.6, 13 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.6, 13 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.6, 13 - Chapter 9 Class 12 Differential Equations - Part 4


Transcript

Ex 9.6, 13 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : 𝑑𝑦/𝑑π‘₯+2𝑦 tan⁑〖π‘₯=sin⁑〖π‘₯;𝑦=0γ€— γ€— when π‘₯= πœ‹/3 𝑑𝑦/𝑑π‘₯+2𝑦 tan⁑〖π‘₯=sin⁑π‘₯ γ€— Differential equation is of the form 𝑑𝑦/𝑑π‘₯ + Py = Q 𝑑𝑦/𝑑π‘₯ + 2y tan x = sin x Where P = 2 tan x & Q = sin x Finding Integrating factor IF = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— IF = 𝑒^∫1β–’γ€–2 tan⁑π‘₯ 𝑑π‘₯γ€— IF = e2 log sec x IF = 𝑒^log⁑sec^2⁑π‘₯ IF = sec2 x Solution is y (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 y (sec2 x) = ∫1β–’γ€–sin⁑π‘₯ sec^2⁑π‘₯ 𝑑π‘₯+𝑐〗 y sec2 x = ∫1β–’γ€–sin⁑π‘₯ 1/cos^2⁑π‘₯ γ€— dx + C y sec2 x = ∫1β–’γ€–sin⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Γ—1/π‘π‘œπ‘ β‘π‘₯ γ€— dx + C y sec2 x = ∫1β–’tan⁑〖π‘₯ sec⁑〖π‘₯ γ€— γ€— dx + C y sec2 x = sec⁑"x + C " y = sec⁑〖π‘₯ γ€—/sec^2⁑π‘₯ + 𝑐/sec^2⁑π‘₯ y = cos x + C cos2 x Putting x = πœ‹/3 & y = 0 0 = cos πœ‹/3 + C cos2 πœ‹/3 0 = 1/2 + C (1/2)^2 (βˆ’1)/2 = C (1/4) (βˆ’4)/2 = C C = βˆ’2 Putting value of C in (1) y = cos x + C cos2 x y = cos x βˆ’ 2 cos2 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.