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Last updated at Dec. 11, 2019 by Teachoo

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Misc 9 Find the particular solution of the differential equation (1 + π^2π₯) dy + (1 + π¦^2) ex dx = 0, given that y = 1 when x = 0. Given (1 + e2x) dy + (1 + y2)π^π₯ dx = 0 (1 + e2x) dy = β(1 + y2)π^π₯ dx ππ¦/ππ₯ = (β(1 + π¦^2 ).π^π₯)/(1 + π2π₯) ππ¦/(1 + π¦)^2 = (βπ^π₯ ππ₯)/(1 + π2π₯) Integrating both sides β«1βππ¦/γ(1 + π¦)γ^2 = β«1β(ππ₯ ππ₯)/γ1 + πγ^2π₯ β¦(1) Let t = ex Diff w.r.t.x ππ‘/ππ₯=π^π₯ ππ‘/ππ₯= ππ₯ β΄ Our equation becomes β«1βππ¦/γ1 + π¦γ^2 = ββ«1βγ(ππ₯ )/(1 + π‘^2 ) (ππ‘ )/(ππ₯ )γ β«1βππ¦/γ1 + π¦γ^2 = ββ«1βγ(ππ‘ )/(1 + π‘^2 ) γ tan^(β1)β‘π¦=βtan^(β1)β‘π‘+πΆ Putting back value of t = ex γπππγ^(βπ)β‘π=βγπππγ^(βπ)β‘(π^π )+πͺ (As β«1βππ₯/γ1 + π₯γ^2 =tan^(β1)β‘π₯) β¦(2) Given that y = 1 when x = 0 Put y = 1 and x = 0 in equation (2) tan^(β1)β‘γ(1)γ=βtan^(β1)β‘(π^π )+πΆ tan^(β1)β‘1=βtan^(β1)β‘π+πΆ tan^(β1)β‘1+tan^(β1)β‘1=πΆ 2 γπππγ^(βπ)β‘π=πΆ 2 Γ π /π=πΆ 2 Γ π/2=πΆ C = π/2. Putting value of C in (2) tan^(β1)β‘π¦=βtan^(β1)β‘(π^π₯ )+πΆ tan^(β1)β‘π¦=βtan^(β1)β‘(π^π₯ )+" " π/2 γπππγ^(βπ)β‘π+γπππγ^(βπ)β‘(π^π )=" " π /π is the required particular solution.

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Chapter 9 Class 12 Differential Equations

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About the Author

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.