# Ex 9.4, 11

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.4, 11find a particular solution satisfying the given condition :(π₯^3+π₯^2+π₯+1) ππ¦/ππ₯=2π₯^2+π₯;π¦=1 When π₯=0 Integrating by partial fractions, using formula (2π₯^2 +π₯)/((π₯ + 1)(π₯^2+1)) = π΄/(π₯ + 1)+(π΅π₯ + πΆ)/(π₯^2 + 1) (2π₯^2 +π₯)/((π₯ + 1)(π₯^2+1)) = (π΄(π₯^2+1) + (π΅π₯ + π)(π₯ + 1))/((π₯ + 1)(π₯^2 + 1)) 2π₯^2 + x = A (π₯^2+ 1) + (Bx + C) (x + 1) Putting x = β1 2(β1)2 β 1 = A ((β1)2 + 1) + (B(β1) + C)(β1 + 1) 2 β 1 = A(2) + (βB + C)(0) 1 = 2A A = 1/2 Hence, (2π₯^2 + π₯)/((π₯ + 1)(π₯^2+1)) = 1/(2(π₯ + 1)) + (3/2 π₯ β 1/2)/(π₯^2 + 1) = 1/(2(π₯ + 1)) + (3π₯ β1)/(γ2(π₯γ^(2 )+ 1)) Now, our equation becomes y = β«1βγ(2π₯^2 + π₯)/((π₯ + 1)(π₯^2 + 1)) ππ₯γ y = β«1βγ1/(2(π₯ + 1))+3π₯/(2(π₯^(2 )+1)) β 1/2(π₯^2 + 1) ππ₯γ y = β«1βγ1/(2(π₯ + 1)) ππ₯γ+β«1βγ3π₯/(2(π₯^(2 )+1)) ππ₯γββ«1βγ 1/2(π₯^2 + 1) ππ₯γ y = 1/2 log (x + 1) +β«1βγ3π₯/(2(π₯^(2 )+1)) ππ₯γβ 1/2 tanβ1 x Integrating β«1βγππ/(π(π^π + π)) π πγ Put t = x2 + 1 ππ‘/2 = 2x dx = ππ‘/2π₯ β«1βγ3π₯/(2(π₯^2 + 1)) ππ₯γ = 3/2 β«1βπ₯/π‘Γππ‘/2π₯ = 3/4 β«1βππ‘/π‘ = 3/4 log |π‘|+π Putting value of t β«1βγ3π₯/(2(π₯^2+1)) ππ₯γ = 3/4 log (x2 + 1) + C Now, From (1) y = 1/2 log (x + 1) +β«1βγ3π₯/(2(π₯^(2 )+1)) ππ₯γβ 1/2 tanβ1 x y = 1/2 log (x + 1) +3/4 " log (x2 + 1)"β 1/2 tanβ1 x + C Put x = 0 and y = 1 1 = 1/2 log (0 + 1) + 3/4 log (0 + 1) β 1/2 tanβ1 0 + C 1 = 1/2 log (1) + 3/4 log (1) β 1/2 tanβ1 0 + C 1 = 0 + 0 β 0 + C 1 = C Putting value of C in (1) y = 1/2 log (x + 1) + 3/4 log (x2 + 1) β 1/2 tanβ1 x + 1 y = 2/4 log (x + 1) + 3/4 log (x2 + 1) β 1/2 tanβ1 x + 1 y = 1/4 log (x + 1)2 + 1/4 log (x2 + 1)3 β 1/2 tanβ1 x + 1 y = 1/4 ["log (x + 1)2 + log (x2 + 1)3 " ]"β " 1/2 " tanβ1 x + 1 " y = π/π "log" [" (x + 1)2" "(x2 + 1)3 " ]" β" π/π " tanβ1 x + 1 "

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.