# Ex 9.5, 14 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.5

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Ex 9.5, 14 Important You are here

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Ex 9.5, 18 (MCQ)

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Last updated at April 16, 2024 by Teachoo

Ex 9.5, 14 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : (1+π₯^2 ) ππ¦/ππ₯+2π₯π¦=1/(1+π₯^2 ) ;π¦=0 when π₯=1 (1 + x2) ππ¦/ππ₯ + 2xy = 1/(1 + π₯2) Divide both sides by (1+π₯2) ππ¦/ππ₯ + 2π₯π¦/(1 + π₯^2 ) = 1/((1 + π₯2).(1 + π₯2)) π π/π π + (πππ/(π + π^π ))y = π/((π + ππ) ) Comparing with ππ¦/ππ₯ + Py = Q P = ππ/(π + π^π ) & Q = π/(π + ππ)π Find Integrating factor IF = π^β«1βγπ ππ₯γ IF = π^β«1βγππ/(π + π^π ) π πγ Let π+π^π = t Diff . w.r.t. x 2x = π/ππ₯ t dx = ππ‘/2π₯ IF = e^(β«1β2π₯/π‘ " " ππ‘/2π₯) IF = e^β«1βγ ππ‘/π‘γ IF = e^πππ|π‘| IF = t IF = 1 + x2 Solution of the differential equation is y Γ I.F = β«1βγπΓπΌ.πΉ ππ₯γ Putting values y Γ (1 + x2) = β«1βπ/(π + π^π )^π "(1 + x2)".dx y Γ (1 + x2) = β«1β1/((1 + π₯^2 ) )dx y (1 + x2) = tan^(β1)β‘γ π₯+πγ Given that y = 0 when x = 1 Putting y = 0 and x = 1 in (1) y (1 + x2) = tanβ1 x + c 0(1 + 12) = tanβ1 (1)+ c 0 = π/4 + C C = β π /π Putting value of C in (2) y (1 + x2) = tan-1 x + c y (1 + x2) = tan-1 x β π /π