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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 9.3, 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‘₯ π’…π’š = (πŸπ’™^𝟐 + 𝒙)/(𝒙^πŸ‘ + 𝒙^𝟐 + 𝒙 + 𝟏) 𝒅𝒙 Integrating both sides ∫1▒〖𝑑𝑦=∫1β–’(2π‘₯^2 + π‘₯)/(π‘₯3 + π‘₯2 + π‘₯ + 1)γ€— dx y = ∫1β–’(πŸπ’™^𝟐 + 𝒙)/( (𝒙 + 𝟏)(𝒙^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 (πŸπ’™^𝟐 +𝒙)/((𝒙 + 𝟏)(𝒙^𝟐+𝟏)) = 𝑨/(𝒙 + 𝟏)+(𝑩𝒙 + π‘ͺ)/(𝒙^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’πŸ)/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’πŸ)/𝟐 3 = 2 Γ— 1/2 + 2B + 2 (βˆ’1/2 ) 3 = 2B B = πŸ‘/𝟐 Hence, (πŸπ’™^𝟐 + 𝒙)/((𝒙 + 𝟏)(𝒙^𝟐+𝟏)) = 1/(2(π‘₯ + 1)) + (3/2 π‘₯ βˆ’ 1/2)/(π‘₯^2 + 1) = 𝟏/(𝟐(𝒙 + 𝟏)) + (πŸ‘π’™ βˆ’πŸ)/(γ€–πŸ(𝒙〗^(𝟐 )+ 𝟏)) Now, our equation becomes y = ∫1β–’γ€–(2π‘₯^2 + π‘₯)/((π‘₯ + 1)(π‘₯^2 + 1)) 𝑑π‘₯γ€— y = ∫1β–’γ€–πŸ/(𝟐(𝒙 + 𝟏))+πŸ‘π’™/(𝟐(𝒙^(𝟐 )+𝟏)) βˆ’ 𝟏/𝟐(𝒙^𝟐 + 𝟏) 𝒅𝒙〗 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€–πŸ‘π’™/(𝟐(𝒙^(𝟐 )+𝟏)) π’…π’™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€–πŸ‘π’™/(𝟐(𝒙^𝟐 + 𝟏)) 𝒅𝒙〗 = 3/2 ∫1β–’π‘₯/𝑑×𝑑𝑑/2π‘₯ = 3/4 ∫1▒𝑑𝑑/𝑑 = πŸ‘/πŸ’ log |𝒕|+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) + πŸ‘/πŸ’ 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⁑〖 (π‘₯+1)^2 γ€—+log⁑〖(π‘₯^2+1)^3 γ€— ] "βˆ’ " 1/2 " tanβˆ’1 x + 1 " As log π‘Ž + log b = log π‘Žb 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 "

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Davneet Singh

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.