1. Chapter 9 Class 12 Differential Equations
2. Serial order wise

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 : ππ¦/ππ₯β3π¦ cotβ‘γπ₯=sinβ‘γ2π₯;π¦=2γ γ when π₯= π/2 ππ¦/ππ₯β3π¦ cotβ‘γπ₯=sinβ‘2π₯ γ ππ¦/ππ₯ + (β3 cot x) y = (sin 2x) Comparing with ππ¦/ππ₯ + Py = Q P = β3 cot x & Q = sin 2x Finding Integrating factor (IF) IF = e^β«1βπππ₯ = e^β«1βγβ3 cotβ‘γπ₯ ππ₯γ γ = e^(β3β«1βγπππ‘ π₯ ππ₯γ) = e^(3 logβ‘|sinβ‘π₯ | ) = e^logβ‘γ|sinβ‘π₯ |^(β3) γ = e^logβ‘γ1/|sin^3β‘π₯ | γ = e^logβ‘γ|πππ ππ^3 π₯|^3 γ = πππ ππ^3 π₯ β΄ I.F = πππ ππ^3 π₯ Solution of differential equation y Γ IF = β«1βγπ.πΌπΉ ππ₯γ Putting values. y Γ cosec3 x = β«1βsinβ‘γ2π₯. πππ ππ^3 π₯ ππ₯γ y cosec3x = β«1β(2 sinβ‘γπ₯ cosβ‘π₯ γ)/sin^3β‘π₯ dx y cosec3x = β«1β(2 cosβ‘π₯)/sin^2β‘π₯ dx Let I = 2β«1βcosβ‘π₯/sin^2β‘π₯ ππ₯ Put t = sin x Diff w.r.t x ππ‘/ππ₯ = cos x dx = ππ‘/cosβ‘π₯ β΄ I = 2β«1βcosβ‘π₯/(π‘^2 ) ππ‘/(πππ  π₯) I = 2β«1βππ‘/(π‘^2 ) = 2 γπ‘/(β1 )γ^(β1)+π =β2/π‘+π Put value t = sin x = (β2)/sinβ‘π₯ +π =β2 πππ ππ π₯+π Put value of I in (2) β΄ y cosec3 x = β2 cosec x + c y = (β2 πππ ππ π₯)/(πππ ππ^2 π₯) + π/(πππ ππ^3 π₯) y = 2 sin2 x + C sin3 x Putting x = π/2 , y = 2 in (3) 2 = β2 Sin2 π/2 + C Sin3 π/2 2 = β2 (1)2 + C(1)3 2 = β2 + C C = 2 + 2 C = 4 Put value of C in (3) y = β2 sin2 x + C sin3 x y = β2 sin2 x + 4 sin3 x y = 4 sin3 x β 2 sin2 x