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

Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations - Part 4 Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations - Part 5 Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations - Part 6 Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations - Part 7

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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 (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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.