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

Transcript

Misc 9 Find the particular solution of the differential equation (1 + 𝑒^2π‘₯) dy + (1 + 𝑦^2) ex dx = 0, given that y = 1 when x = 0. Given (1 + e2x) dy + (1 + y2)𝑒^π‘₯ dx = 0 (1 + e2x) dy = βˆ’(1 + y2)𝑒^π‘₯ dx 𝑑𝑦/𝑑π‘₯ = (βˆ’(1 + 𝑦^2 ).𝑒^π‘₯)/(1 + 𝑒2π‘₯) 𝑑𝑦/(1 + 𝑦)^2 = (βˆ’π‘’^π‘₯ 𝑑π‘₯)/(1 + 𝑒2π‘₯) Integrating both sides ∫1▒𝑑𝑦/γ€–(1 + 𝑦)γ€—^2 = ∫1β–’(𝑒π‘₯ 𝑑π‘₯)/γ€–1 + 𝑒〗^2π‘₯ …(1) Let t = ex Diff w.r.t.x 𝑑𝑑/𝑑π‘₯=𝑒^π‘₯ 𝑑𝑑/𝑒π‘₯= 𝑑π‘₯ ∴ Our equation becomes ∫1▒𝑑𝑦/γ€–1 + 𝑦〗^2 = βˆ’βˆ«1β–’γ€–(𝑒π‘₯ )/(1 + 𝑑^2 ) (𝑑𝑑 )/(𝑒π‘₯ )γ€— ∫1▒𝑑𝑦/γ€–1 + 𝑦〗^2 = βˆ’βˆ«1β–’γ€–(𝑑𝑑 )/(1 + 𝑑^2 ) γ€— tan^(βˆ’1)⁑𝑦=βˆ’tan^(βˆ’1)⁑𝑑+𝐢 Putting back value of t = ex 〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘π’š=βˆ’γ€–π’•π’‚π’γ€—^(βˆ’πŸ)⁑(𝒆^𝒙 )+π‘ͺ (As ∫1▒𝑑π‘₯/γ€–1 + π‘₯γ€—^2 =tan^(βˆ’1)⁑π‘₯) …(2) Given that y = 1 when x = 0 Put y = 1 and x = 0 in equation (2) tan^(βˆ’1)⁑〖(1)γ€—=βˆ’tan^(βˆ’1)⁑(𝒆^𝟎 )+𝐢 tan^(βˆ’1)⁑1=βˆ’tan^(βˆ’1)⁑𝟏+𝐢 tan^(βˆ’1)⁑1+tan^(βˆ’1)⁑1=𝐢 2 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑𝟏=𝐢 2 Γ— 𝝅/πŸ’=𝐢 2 Γ— πœ‹/2=𝐢 C = πœ‹/2. Putting value of C in (2) tan^(βˆ’1)⁑𝑦=βˆ’tan^(βˆ’1)⁑(𝑒^π‘₯ )+𝐢 tan^(βˆ’1)⁑𝑦=βˆ’tan^(βˆ’1)⁑(𝑒^π‘₯ )+" " πœ‹/2 〖𝒕𝒂𝒏〗^(βˆ’πŸ)β‘π’š+〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑(𝒆^𝒙 )=" " 𝝅/𝟐 is the required particular solution.

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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.