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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 ) 𝑑𝑦/𝑑π‘₯+2π‘₯𝑦=1/(1+π‘₯^2 ) ;𝑦=0 when π‘₯=1 (1 + x2) 𝑑𝑦/𝑑π‘₯ + 2xy = 1/(1 + π‘₯2) Divide both sides by (1+π‘₯2) 𝑑𝑦/𝑑π‘₯ + 2π‘₯𝑦/(1 + π‘₯^2 ) = 1/((1 + π‘₯2).(1 + π‘₯2)) π’…π’š/𝒅𝒙 + (πŸπ’™π’š/(𝟏 + 𝒙^𝟐 ))y = 𝟏/((𝟏 + π’™πŸ) ) Comparing with 𝑑𝑦/𝑑π‘₯ + Py = Q P = πŸπ’™/(𝟏 + 𝒙^𝟐 ) & Q = 𝟏/(𝟏 + π’™πŸ)𝟐 Find Integrating factor IF = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— IF = 𝒆^∫1β–’γ€–πŸπ’™/(𝟏 + 𝒙^𝟐 ) 𝒅𝒙〗 Let 𝟏+𝒙^𝟐 = t Diff . w.r.t. x 2x = 𝑑/𝑑π‘₯ t dx = 𝑑𝑑/2π‘₯ IF = e^(∫1β–’2π‘₯/𝑑 " " 𝑑𝑑/2π‘₯) IF = e^∫1β–’γ€– 𝑑𝑑/𝑑〗 IF = e^π‘™π‘œπ‘”|𝑑| IF = t IF = 1 + x2 Solution of the differential equation is y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹 𝑑π‘₯γ€— Putting values y Γ— (1 + x2) = ∫1β–’πŸ/(𝟏 + 𝒙^𝟐 )^𝟐 "(1 + x2)".dx y Γ— (1 + x2) = ∫1β–’1/((1 + π‘₯^2 ) )dx y (1 + x2) = tan^(βˆ’1)⁑〖 π‘₯+𝑐〗 Given that y = 0 when x = 1 Putting y = 0 and x = 1 in (1) y (1 + x2) = tanβˆ’1 x + c 0(1 + 12) = tanβˆ’1 (1)+ c 0 = πœ‹/4 + C C = βˆ’ 𝝅/πŸ’ Putting value of C in (2) y (1 + x2) = tan-1 x + c y (1 + x2) = tan-1 x βˆ’ 𝝅/πŸ’

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo