# 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)

Ex 9.5, 19 (MCQ) Important

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+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