Miscellaneous

Chapter 9 Class 12 Differential Equations
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### Transcript

Misc 11 Find a particular solution of the differential equation ππ¦/ππ₯+π¦ cotβ‘γπ₯=4π₯ πππ ππ π₯ (π₯β 0) ,γ given that π¦=0 when π₯=π/2Given ππ¦/ππ₯+π¦ cotβ‘γπ₯=4π₯ πππ ππ π₯ γ This of the form ππ¦/ππ₯+ππ¦=π where P = cot x & Q = 4x cosec x IF = π^β«1βπππ₯ IF = π^β«1βππ¨π­β‘γπ ππγ IF = π^(logβ‘(sinβ‘γπ₯)γ ) IF = sin x Solution is y (IF) = β«1βγ(πΓπΌ.πΉ)ππ₯+π γ y sin x = β«1βγππ πππππ π πππβ‘π ππ+π γ y sin x = β«1βγ4π₯ 1/sinβ‘π₯ sinβ‘π₯ ππ₯+π γ y sin x = β«1βγ4π₯ ππ₯+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+πΆ 0 = 2 (γπ/4γ^2 ) + C 0 = γπ/2γ^2 + C C = γβπγ^π/π Putting value of C in (1) y sin x = 2x2 + c y sin x = 2π^π β γπ/πγ^π