# Misc 9 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by

Last updated at Dec. 11, 2019 by

Transcript

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.

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Deleted for CBSE Board 2022 Exams

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6

Misc 7 Important

Misc 8

Misc 9 Important You are here

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16 (MCQ)

Misc 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Misc 18 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

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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 10 years. He provides courses for Maths and Science at Teachoo.