





Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Ex 9.4
Ex 9.4, 2
Ex 9.4, 3
Ex 9.4, 4 Important
Ex 9.4, 5
Ex 9.4, 6
Ex 9.4, 7 Important
Ex 9.4, 8
Ex 9.4, 9 Important
Ex 9.4, 10 Important
Ex 9.4, 11 Important You are here
Ex 9.4, 12
Ex 9.4, 13
Ex 9.4, 14
Ex 9.4, 15 Important
Ex 9.4, 16
Ex 9.4, 17 Important
Ex 9.4, 18
Ex 9.4, 19 Important
Ex 9.4, 20 Important
Ex 9.4, 21
Ex 9.4, 22 Important
Ex 9.4, 23 (MCQ)
Last updated at Aug. 20, 2021 by Teachoo
Ex 9.4, 11 Find a particular solution satisfying the given condition : (π₯^3+π₯^2+π₯+1) ππ¦/ππ₯=2π₯^2+π₯; π¦=1 when π₯=0 (π₯^3+π₯^2+π₯+1) ππ¦/ππ₯=2π₯^2+π₯ ππ¦ = (2π₯^2 + π₯)/(π₯^3 + π₯^2 + π₯ + 1) ππ₯ Integrating both sides β«1βγππ¦=β«1β(2π₯^2 + π₯)/(π₯3 + π₯2 + π₯ + 1)γ dx y = β«1β(2π₯^2 + π₯)/( (π₯ + 1)(π₯^2 + 1)) dx Rough x = β1 is a solution of x3 + x2 + x + 1 as (-1)2 + (-1)2 + (β1) + 1 = 0 Hence (x + 1) is one of its factors. So, we can write x3 + x2 + x + 1 = (x + 1) (x2 + 1) 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 = π/π Putting x = 0 0 = A (0 + 1) + (B(0) + C)(0 + 1) 0 = A + C(1) A = βC Since A = 1/2 β΄ C = (β1)/2 Putting x = 1 2(1) + 1 = A (12 + 1) + (B(1) + C) (1 + 1) 3 = 2A + 2B + 2C Putting A = 1/2, C = (β1)/2 3 = 2 Γ 1/2 + 2B + 2 (β1/2 ) 3 = 2B B = 3/2 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 dt = 2x dx β΄ dx = ππ‘/2π₯ So, 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 back 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 Putting 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 (β΅ log 1 = 0 & tan = 0) (β΅ alog x = log xπ) y = 1/4 [logβ‘γ (π₯+1)^2 γ+logβ‘γ(π₯^2+1)^3 γ ] "β " 1/2 " tanβ1 x + 1 " y = π/π πππβ‘γγ [(π+π)γ^π (π^π+π)^π] γ "β " π/π " tanβ1 x + 1 " ( As log π + log b = log πb )