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Ex 9.6
Ex 9.6, 2
Ex 9.6, 3 Important
Ex 9.6, 4
Ex 9.6, 5 Important
Ex 9.6, 6
Ex 9.6, 7 Important
Ex 9.6, 8 Important
Ex 9.6, 9
Ex 9.6, 10 Deleted for CBSE Board 2022 Exams
Ex 9.6, 11 Deleted for CBSE Board 2022 Exams
Ex 9.6, 12 Important Deleted for CBSE Board 2022 Exams
Ex 9.6, 13
Ex 9.6, 14 Important
Ex 9.6, 15 You are here
Ex 9.6, 16 Important
Ex 9.6, 17 Important
Ex 9.6, 18 (MCQ)
Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams
Last updated at Dec. 11, 2019 by Teachoo
Ex 9.6, 15 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : ππ¦/ππ₯β3π¦ cotβ‘γπ₯=sinβ‘γ2π₯;π¦=2γ γ when π₯= π/2 ππ¦/ππ₯β3π¦ cotβ‘γπ₯=sinβ‘2π₯ γ ππ¦/ππ₯ + (β3 cot x) y = (sin 2x) Comparing with ππ¦/ππ₯ + Py = Q P = β3 cot x & Q = sin 2x Finding Integrating factor (IF) IF = e^β«1βπππ₯ β¦(1) = e^β«1βγβ3 cotβ‘γπ₯ ππ₯γ γ = e^(β3β«1βγπππ‘ π₯ ππ₯γ) = e^(β3 logβ‘|sinβ‘π₯ | ) = e^logβ‘γ|sinβ‘π₯ |^(β3) γ = e^logβ‘γ1/|sin^3β‘π₯ | γ = e^logβ‘|πππ ππ^3 π₯| = πππ ππ^3 π₯ β΄ I.F = πππ ππ^3 π₯ Solution of differential equation is y Γ IF = β«1βγπ.πΌπΉ ππ₯γ Putting values y Γ cosec3 x = β«1βsinβ‘γ2π₯. πππ ππ^3 π₯ ππ₯γ y cosec3x = β«1β(2 sinβ‘γπ₯ cosβ‘π₯ γ)/sin^3β‘π₯ dx y cosec3x = β«1β(2 cosβ‘π₯)/sin^2β‘π₯ dx y cosec3x = 2β«1βγcosβ‘π₯/π ππβ‘π₯ Γ1/sinβ‘π₯ γ dx y cosec3x = 2β«1βγcotβ‘π₯ πππ ππ π₯γ dx y cosec3x = 2 (βcosec x) + C y = (β2 πππ ππ π₯)/(πππ ππ^3 π₯) + πΆ/(πππ ππ^3 π₯) y = (β2)/(πππ ππ^2 π₯) + πΆ/(πππ ππ^3 π₯) y = β2 sin2 x + C sin3 x Putting x = π/2 , y = 2 in (2) 2 = β2 sin2 π/2 + C sin3 π/2 2 = β2 (1)2 + C(1)3 2 = β2 + C C = 2 + 2 C = 4 β¦(2) Put value of C in (3) y = β2 sin2 x + C sin3 x y = β2 sin2 x + 4 sin3 x y = 4 sin3 x β 2 sin2 x