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Misc 9 - Find particular solution: (1 + e2x)dy + (1 + y2) ex

Misc 9 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 9 - Chapter 9 Class 12 Differential Equations - Part 3 Misc 9 - Chapter 9 Class 12 Differential Equations - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.