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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.6, 15 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : 𝑑𝑦/𝑑π‘₯βˆ’3𝑦 cot⁑〖π‘₯=sin⁑〖2π‘₯;𝑦=2γ€— γ€— when π‘₯= πœ‹/2 𝑑𝑦/𝑑π‘₯βˆ’3𝑦 cot⁑〖π‘₯=sin⁑2π‘₯ γ€— 𝑑𝑦/𝑑π‘₯ + (βˆ’3 cot x) y = (sin 2x) Comparing with 𝑑𝑦/𝑑π‘₯ + Py = Q P = βˆ’3 cot x & Q = sin 2x Finding Integrating factor (IF) IF = e^∫1▒𝑝𝑑π‘₯ …(1) = e^∫1β–’γ€–βˆ’3 cot⁑〖π‘₯ 𝑑π‘₯γ€— γ€— = e^(βˆ’3∫1β–’γ€–π‘π‘œπ‘‘ π‘₯ 𝑑π‘₯γ€—) = e^(βˆ’3 log⁑|sin⁑π‘₯ | ) = e^log⁑〖|sin⁑π‘₯ |^(βˆ’3) γ€— = e^log⁑〖1/|sin^3⁑π‘₯ | γ€— = e^log⁑|π‘π‘œπ‘ π‘’π‘^3 π‘₯| = π‘π‘œπ‘ π‘’π‘^3 π‘₯ ∴ I.F = π‘π‘œπ‘ π‘’π‘^3 π‘₯ Solution of differential equation is y Γ— IF = ∫1▒〖𝑄.𝐼𝐹 𝑑π‘₯γ€— Putting values y Γ— cosec3 x = ∫1β–’sin⁑〖2π‘₯. π‘π‘œπ‘ π‘’π‘^3 π‘₯ 𝑑π‘₯γ€— y cosec3x = ∫1β–’(2 sin⁑〖π‘₯ cos⁑π‘₯ γ€—)/sin^3⁑π‘₯ dx y cosec3x = ∫1β–’(2 cos⁑π‘₯)/sin^2⁑π‘₯ dx y cosec3x = 2∫1β–’γ€–cos⁑π‘₯/𝑠𝑖𝑛⁑π‘₯ Γ—1/sin⁑π‘₯ γ€— dx y cosec3x = 2∫1β–’γ€–cot⁑π‘₯ π‘π‘œπ‘ π‘’π‘ π‘₯γ€— dx y cosec3x = 2 (βˆ’cosec x) + C y = (βˆ’2 π‘π‘œπ‘ π‘’π‘ π‘₯)/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) + 𝐢/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) y = (βˆ’2)/(π‘π‘œπ‘ π‘’π‘^2 π‘₯) + 𝐢/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) y = βˆ’2 sin2 x + C sin3 x Putting x = πœ‹/2 , y = 2 in (2) 2 = βˆ’2 sin2 πœ‹/2 + C sin3 πœ‹/2 2 = βˆ’2 (1)2 + C(1)3 2 = βˆ’2 + C C = 2 + 2 C = 4 …(2) Put value of C in (3) y = βˆ’2 sin2 x + C sin3 x y = βˆ’2 sin2 x + 4 sin3 x y = 4 sin3 x βˆ’ 2 sin2 x

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.