# Misc 11 - 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

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11 You are here

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 11 Find a particular solution of the differential equation ππ¦/ππ₯+π¦ cotβ‘γπ₯=4π₯ πππ ππ π₯ (π₯β 0) ,γ given that π¦=0 when π₯=π/2Given ππ¦/ππ₯+π¦ cotβ‘γπ₯=4π₯ πππ ππ π₯ γ This of the form ππ¦/ππ₯+ππ¦=π where P = cot x & Q = 4x cosec x IF = π^β«1βπππ₯ IF = π^β«1βππ¨πβ‘γπ π πγ IF = π^(logβ‘(sinβ‘γπ₯)γ ) IF = sin x Solution is y (IF) = β«1βγ(πΓπΌ.πΉ)ππ₯+π γ y sin x = β«1βγππ πππππ π πππβ‘π π π+π γ y sin x = β«1βγ4π₯ 1/sinβ‘π₯ sinβ‘π₯ ππ₯+π γ y sin x = β«1βγ4π₯ ππ₯+π γ y sin x = (4π₯^2)/2+π y sin x = 2x2 + C Given that π=π when π=π /π Put x = π/2 & y = 0 in (1) 0 Γ sin π/2 = 2 (π/2)^2+πΆ 0 = 2 (γπ/4γ^2 ) + C 0 = γπ/2γ^2 + C C = γβπ γ^π/π Putting value of C in (1) y sin x = 2x2 + c y sin x = 2π^π β γπ /πγ^π