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Ex 9.4, 11 - Find particular solution: (x3 + x2 + x + 1) dy/dx - Variable separation - Equation given

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.4, 11 find a particular solution satisfying the given condition : (π‘₯^3+π‘₯^2+π‘₯+1) 𝑑𝑦/𝑑π‘₯=2π‘₯^2+π‘₯;𝑦=1 When π‘₯=0 Integrating by partial fractions, using formula (2π‘₯^2 +π‘₯)/((π‘₯ + 1)(π‘₯^2+1)) = 𝐴/(π‘₯ + 1)+(𝐡π‘₯ + 𝐢)/(π‘₯^2 + 1) (2π‘₯^2 +π‘₯)/((π‘₯ + 1)(π‘₯^2+1)) = (𝐴(π‘₯^2+1) + (𝐡π‘₯ + 𝑐)(π‘₯ + 1))/((π‘₯ + 1)(π‘₯^2 + 1)) 2π‘₯^2 + x = A (π‘₯^2+ 1) + (Bx + C) (x + 1) Putting x = βˆ’1 2(βˆ’1)2 βˆ’ 1 = A ((βˆ’1)2 + 1) + (B(βˆ’1) + C)(βˆ’1 + 1) 2 βˆ’ 1 = A(2) + (–B + C)(0) 1 = 2A A = 1/2 Hence, (2π‘₯^2 + π‘₯)/((π‘₯ + 1)(π‘₯^2+1)) = 1/(2(π‘₯ + 1)) + (3/2 π‘₯ βˆ’ 1/2)/(π‘₯^2 + 1) = 1/(2(π‘₯ + 1)) + (3π‘₯ βˆ’1)/(γ€–2(π‘₯γ€—^(2 )+ 1)) Now, our equation becomes y = ∫1β–’γ€–(2π‘₯^2 + π‘₯)/((π‘₯ + 1)(π‘₯^2 + 1)) 𝑑π‘₯γ€— y = ∫1β–’γ€–1/(2(π‘₯ + 1))+3π‘₯/(2(π‘₯^(2 )+1)) βˆ’ 1/2(π‘₯^2 + 1) 𝑑π‘₯γ€— y = ∫1β–’γ€–1/(2(π‘₯ + 1)) 𝑑π‘₯γ€—+∫1β–’γ€–3π‘₯/(2(π‘₯^(2 )+1)) 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€– 1/2(π‘₯^2 + 1) 𝑑π‘₯γ€— y = 1/2 log (x + 1) +∫1β–’γ€–3π‘₯/(2(π‘₯^(2 )+1)) 𝑑π‘₯γ€—βˆ’ 1/2 tanβˆ’1 x Integrating ∫1β–’γ€–πŸ‘π’™/(𝟐(𝒙^𝟐 + 𝟏)) 𝒅𝒙〗 Put t = x2 + 1 𝑑𝑑/2 = 2x dx = 𝑑𝑑/2π‘₯ ∫1β–’γ€–3π‘₯/(2(π‘₯^2 + 1)) 𝑑π‘₯γ€— = 3/2 ∫1β–’π‘₯/𝑑×𝑑𝑑/2π‘₯ = 3/4 ∫1▒𝑑𝑑/𝑑 = 3/4 log |𝑑|+𝑐 Putting value of t ∫1β–’γ€–3π‘₯/(2(π‘₯^2+1)) 𝑑π‘₯γ€— = 3/4 log (x2 + 1) + C Now, From (1) y = 1/2 log (x + 1) +∫1β–’γ€–3π‘₯/(2(π‘₯^(2 )+1)) 𝑑π‘₯γ€—βˆ’ 1/2 tanβˆ’1 x y = 1/2 log (x + 1) +3/4 " log (x2 + 1)"βˆ’ 1/2 tanβˆ’1 x + C Put x = 0 and y = 1 1 = 1/2 log (0 + 1) + 3/4 log (0 + 1) βˆ’ 1/2 tanβˆ’1 0 + C 1 = 1/2 log (1) + 3/4 log (1) βˆ’ 1/2 tanβˆ’1 0 + C 1 = 0 + 0 βˆ’ 0 + C 1 = C Putting value of C in (1) y = 1/2 log (x + 1) + 3/4 log (x2 + 1) βˆ’ 1/2 tanβˆ’1 x + 1 y = 2/4 log (x + 1) + 3/4 log (x2 + 1) βˆ’ 1/2 tanβˆ’1 x + 1 y = 1/4 log (x + 1)2 + 1/4 log (x2 + 1)3 βˆ’ 1/2 tanβˆ’1 x + 1 y = 1/4 ["log (x + 1)2 + log (x2 + 1)3 " ]"βˆ’ " 1/2 " tanβˆ’1 x + 1 " y = 𝟏/πŸ’ "log" [" (x + 1)2" "(x2 + 1)3 " ]" βˆ’" 𝟏/𝟐 " tanβˆ’1 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.
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