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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 March 23, 2023 by Teachoo
Ex 9.4, 11 Find a particular solution satisfying the given condition : (𝑥^3+𝑥^2+𝑥+1) 𝑑𝑦/𝑑𝑥=2𝑥^2+𝑥; 𝑦=1 when 𝑥=0 (𝑥^3+𝑥^2+𝑥+1) 𝑑𝑦/𝑑𝑥=2𝑥^2+𝑥 𝑑𝑦 = (2𝑥^2 + 𝑥)/(𝑥^3 + 𝑥^2 + 𝑥 + 1) 𝑑𝑥 Integrating both sides ∫1▒〖𝑑𝑦=∫1▒(2𝑥^2 + 𝑥)/(𝑥3 + 𝑥2 + 𝑥 + 1)〗 dx y = ∫1▒(2𝑥^2 + 𝑥)/( (𝑥 + 1)(𝑥^2 + 1)) dx Rough x = −1 is a solution of x3 + x2 + x + 1 as (-1)2 + (-1)2 + (−1) + 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 (2𝑥^2 +𝑥)/((𝑥 + 1)(𝑥^2+1)) = 𝐴/(𝑥 + 1)+(𝐵𝑥 + 𝐶)/(𝑥^2 + 1) (2𝑥^2 +𝑥)/((𝑥 + 1)(𝑥^2+1)) = (𝐴(𝑥^2+1) + (𝐵𝑥 + 𝑐)(𝑥 + 1))/((𝑥 + 1)(𝑥^2 + 1)) 2𝑥^2 + x = A (𝑥^2+ 1) + (Bx + C) (x + 1) Putting x = −1 2(−1)2 − 1 = A ((−1)2 + 1) + (B(−1) + C)(−1 + 1) 2 − 1 = A(2) + (–B + C)(0) 1 = 2A A = 𝟏/𝟐 Putting x = 0 0 = A (0 + 1) + (B(0) + C)(0 + 1) 0 = A + C(1) A = −C Since A = 1/2 ∴ C = (−1)/2 Putting x = 1 2(1) + 1 = A (12 + 1) + (B(1) + C) (1 + 1) 3 = 2A + 2B + 2C Putting A = 1/2, C = (−1)/2 3 = 2 × 1/2 + 2B + 2 (−1/2 ) 3 = 2B B = 3/2 Hence, (2𝑥^2 + 𝑥)/((𝑥 + 1)(𝑥^2+1)) = 1/(2(𝑥 + 1)) + (3/2 𝑥 − 1/2)/(𝑥^2 + 1) = 1/(2(𝑥 + 1)) + (3𝑥 −1)/(〖2(𝑥〗^(2 )+ 1)) Now, our equation becomes y = ∫1▒〖(2𝑥^2 + 𝑥)/((𝑥 + 1)(𝑥^2 + 1)) 𝑑𝑥〗 y = ∫1▒〖1/(2(𝑥 + 1))+3𝑥/(2(𝑥^(2 )+1)) − 1/2(𝑥^2 + 1) 𝑑𝑥〗 y = ∫1▒〖1/(2(𝑥 + 1)) 𝑑𝑥〗+∫1▒〖3𝑥/(2(𝑥^(2 )+1)) 𝑑𝑥〗−∫1▒〖 1/2(𝑥^2 + 1) 𝑑𝑥〗 y = 1/2 log (x + 1) +∫1▒〖3𝑥/(2(𝑥^(2 )+1)) 𝑑𝑥〗− 1/2 tan−1 x Integrating ∫1▒〖𝟑𝒙/(𝟐(𝒙^𝟐 + 𝟏)) 𝒅𝒙〗 Put t = x2 + 1 dt = 2x dx ∴ dx = 𝑑𝑡/2𝑥 So, ∫1▒〖3𝑥/(2(𝑥^2 + 1)) 𝑑𝑥〗 = 3/2 ∫1▒𝑥/𝑡×𝑑𝑡/2𝑥 = 3/4 ∫1▒𝑑𝑡/𝑡 = 3/4 log |𝑡|+𝑐 Putting back value of t ∫1▒〖3𝑥/(2(𝑥^2+1)) 𝑑𝑥〗 = 3/4 log (x2 + 1) + C Now, From (1) y = 1/2 log (x + 1) +∫1▒〖3𝑥/(2(𝑥^(2 )+1)) 𝑑𝑥〗− 1/2 tan−1 x y = 1/2 log (x + 1) +3/4 " log (x2 + 1)"− 1/2 tan−1 x + C Putting x = 0 and y = 1 1 = 1/2 log (0 + 1) + 3/4 log (0 + 1) − 1/2 tan−1 0 + C 1 = 1/2 log (1) + 3/4 log (1) − 1/2 tan−1 0 + C 1 = 0 + 0 − 0 + C 1 = C Putting value of C in (1) y = 1/2 log (x + 1) + 3/4 log (x2 + 1) − 1/2 tan−1 x + 1 y = 2/4 log (x + 1) + 3/4 log (x2 + 1) − 1/2 tan−1 x + 1 y = 1/4 log (x + 1)2 + 1/4 log (x2 + 1)3 − 1/2 tan−1 x + 1 (∵ log 1 = 0 & tan = 0) (∵ alog x = log x𝑎) y = 1/4 [log〖 (𝑥+1)^2 〗+log〖(𝑥^2+1)^3 〗 ] "− " 1/2 " tan−1 x + 1 " y = 𝟏/𝟒 𝒍𝒐𝒈〖〖 [(𝒙+𝟏)〗^𝟐 (𝒙^𝟐+𝟏)^𝟑] 〗 "− " 𝟏/𝟐 " tan−1 x + 1 " ( As log 𝑎 + log b = log 𝑎b )