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Ex 9.5
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 You are here
Ex 9.5, 15
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, 14 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : (1+𝑥^2 ) 𝑑𝑦/𝑑𝑥+2𝑥𝑦=1/(1+𝑥^2 ) ;𝑦=0 when 𝑥=1 (1 + x2) 𝑑𝑦/𝑑𝑥 + 2xy = 1/(1 + 𝑥2) Divide both sides by (1+𝑥2) 𝑑𝑦/𝑑𝑥 + 2𝑥𝑦/(1 + 𝑥^2 ) = 1/((1 + 𝑥2).(1 + 𝑥2)) 𝒅𝒚/𝒅𝒙 + (𝟐𝒙𝒚/(𝟏 + 𝒙^𝟐 ))y = 𝟏/((𝟏 + 𝒙𝟐) ) Comparing with 𝑑𝑦/𝑑𝑥 + Py = Q P = 𝟐𝒙/(𝟏 + 𝒙^𝟐 ) & Q = 𝟏/(𝟏 + 𝒙𝟐)𝟐 Find Integrating factor IF = 𝑒^∫1▒〖𝑝 𝑑𝑥〗 IF = 𝒆^∫1▒〖𝟐𝒙/(𝟏 + 𝒙^𝟐 ) 𝒅𝒙〗 Let 𝟏+𝒙^𝟐 = t Diff . w.r.t. x 2x = 𝑑/𝑑𝑥 t dx = 𝑑𝑡/2𝑥 IF = e^(∫1▒2𝑥/𝑡 " " 𝑑𝑡/2𝑥) IF = e^∫1▒〖 𝑑𝑡/𝑡〗 IF = e^𝑙𝑜𝑔|𝑡| IF = t IF = 1 + x2 Solution of the differential equation is y × I.F = ∫1▒〖𝑄×𝐼.𝐹 𝑑𝑥〗 Putting values y × (1 + x2) = ∫1▒𝟏/(𝟏 + 𝒙^𝟐 )^𝟐 "(1 + x2)".dx y × (1 + x2) = ∫1▒1/((1 + 𝑥^2 ) )dx y (1 + x2) = tan^(−1)〖 𝑥+𝑐〗 Given that y = 0 when x = 1 Putting y = 0 and x = 1 in (1) y (1 + x2) = tan−1 x + c 0(1 + 12) = tan−1 (1)+ c 0 = 𝜋/4 + C C = − 𝝅/𝟒 Putting value of C in (2) y (1 + x2) = tan-1 x + c y (1 + x2) = tan-1 x − 𝝅/𝟒