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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

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 (π‘₯^3+π‘₯^2+π‘₯+1) 𝑑𝑦/𝑑π‘₯=2π‘₯^2+π‘₯ 𝑑𝑦 = (2π‘₯^2 + π‘₯)/(π‘₯^3 + π‘₯^2 + π‘₯ + 1) 𝑑π‘₯ Integrating both sides ∫1▒〖𝑑𝑦=∫1β–’(2π‘₯^2 + π‘₯)/(π‘₯3 + π‘₯2 + π‘₯ + 1)γ€— dx y = ∫1β–’(2π‘₯^2 + π‘₯)/( (π‘₯ + 1)(π‘₯^2 + 1)) dx Rough x = βˆ’1 is a solution of x3 + x2 + x + 1 as (-1)2 + (-1)2 + (βˆ’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 (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 Putting x = 0 0 = A (0 + 1) + (B(0) + C)(0 + 1) 0 = A + C(1) A = βˆ’C Since A = 1/2 ∴ C = (βˆ’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 = (βˆ’1)/2 3 = 2 Γ— 1/2 + 2B + 2 (βˆ’1/2 ) 3 = 2B B = 3/2 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 dt = 2x dx ∴ dx = 𝑑𝑑/2π‘₯ So, ∫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 back 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 Putting 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 (∡ log 1 = 0 & tan = 0) (∡ alog x = log xπ‘Ž) y = 1/4 [log⁑〖 (π‘₯+1)^2 γ€—+log⁑〖(π‘₯^2+1)^3 γ€— ] "βˆ’ " 1/2 " tanβˆ’1 x + 1 " y = 𝟏/πŸ’ π’π’π’ˆβ‘γ€–γ€– [(𝒙+𝟏)γ€—^𝟐 (𝒙^𝟐+𝟏)^πŸ‘] γ€— "βˆ’ " 𝟏/𝟐 " tanβˆ’1 x + 1 " ( As log π‘Ž + log b = log π‘Žb )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.