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Ex 9.5

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Ex 9.5, 15 You are here

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Ex 9.5, 18 (MCQ)

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Last updated at May 29, 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₯ 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