Ex 9.6, 15 - Find particular solution: dy/dx - 3y cot x = sin 2x - Ex 9.6

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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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‘₯ = e^∫1β–’γ€–βˆ’3 cot⁑〖π‘₯ 𝑑π‘₯γ€— γ€— = e^(βˆ’3∫1β–’γ€–π‘π‘œπ‘‘ π‘₯ 𝑑π‘₯γ€—) = e^(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‘₯|^3 γ€— = π‘π‘œπ‘ π‘’π‘^3 π‘₯ ∴ I.F = π‘π‘œπ‘ π‘’π‘^3 π‘₯ Solution of differential equation 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 Let I = 2∫1β–’cos⁑π‘₯/sin^2⁑π‘₯ 𝑑π‘₯ Put t = sin x Diff w.r.t x 𝑑𝑑/𝑑π‘₯ = cos x dx = 𝑑𝑑/cos⁑π‘₯ ∴ I = 2∫1β–’cos⁑π‘₯/(𝑑^2 ) 𝑑𝑑/(π‘π‘œπ‘  π‘₯) I = 2∫1▒𝑑𝑑/(𝑑^2 ) = 2 〖𝑑/(βˆ’1 )γ€—^(βˆ’1)+𝑐 =βˆ’2/𝑑+𝑐 Put value t = sin x = (βˆ’2)/sin⁑π‘₯ +𝑐 =βˆ’2 π‘π‘œπ‘ π‘’π‘ π‘₯+𝑐 Put value of I in (2) ∴ y cosec3 x = βˆ’2 cosec x + c y = (βˆ’2 π‘π‘œπ‘ π‘’π‘ π‘₯)/(π‘π‘œπ‘ π‘’π‘^2 π‘₯) + 𝑐/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) y = 2 sin2 x + C sin3 x Putting x = πœ‹/2 , y = 2 in (3) 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 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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