Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations

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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 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 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/2 = 2x dx = 𝑑𝑡/2𝑥 ∫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 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 Put 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 y = 1/4 ["log (x + 1)2 + log (x2 + 1)3 " ]"− " 1/2 " tan−1 x + 1 " y = 𝟏/𝟒 "log" [" (x + 1)2" "(x2 + 1)3 " ]" −" 𝟏/𝟐 " tan−1 x + 1 "