# Ex 9.6, 15 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

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 : 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

Ex 9.6

Ex 9.6, 1
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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

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Ex 9.6, 18 (MCQ)

Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.