Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 9.5

Ex 9.5, 1
Important

Ex 9.5, 2

Ex 9.5, 3 Important

Ex 9.5, 4

Ex 9.5, 5 Important

Ex 9.5, 6

Ex 9.5, 7 Important

Ex 9.5, 8 Important

Ex 9.5, 9

Ex 9.5, 10

Ex 9.5, 11

Ex 9.5, 12 Important

Ex 9.5, 13

Ex 9.5, 14 Important

Ex 9.5, 15 You are here

Ex 9.5, 16 Important

Ex 9.5, 17 Important

Ex 9.5, 18 (MCQ)

Ex 9.5, 19 (MCQ) Important

Last updated at Aug. 14, 2023 by Teachoo

Ex 9.5, 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₯ = 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₯| = πππππ^π π Solution of differential equation is y Γ IF = β«1βγπ.πΌπΉ ππ₯γ Putting values y Γ cosec3 x = β«1βπππβ‘γππ. πππππ^π π π πγ 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 /π , 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 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