# Misc 7 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

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 Important

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important You are here

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13 (MCQ)

Misc 14 (MCQ) Important

Misc 15 (MCQ)

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 7 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₯) π π/(π + π)^π = (βπ^π π π)/(π + πππ) Integrating both sides β«1βππ¦/γ(1 + π¦)γ^2 = β«1β(ππ₯ ππ₯)/γ1 + πγ^2π₯ 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«1βπ π/γπ + πγ^π = ββ«1βγ(π π )/(π + π^π ) 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 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 /π. 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.