Examples

Chapter 9 Class 12 Differential Equations
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### Transcript

Example 17 Find the particular solution of the differential equation ππ¦/ππ₯+π¦ cotβ‘γπ₯=2π₯+π₯^2 cotβ‘π₯(π₯β 0) γ given that π¦=0 π€βππ π₯=π/2 ππ¦/ππ₯+π¦ cotβ‘γπ₯=2π₯+π₯^2 cotβ‘π₯ γ Differential equation is of the form ππ/ππ+π·π=πΈ where P = cot x & Q = 2x + x2 cot x IF = π^β«1βγπ ππ₯γ IF = π^β«1βγππ¨π­β‘π ππγ IF = γπ^logβ‘sinβ‘π₯ γ^" " IF = sin x Solution is y (IF) =β«1βγ(πΓπΌπΉ) ππ₯+πγ y sin x = β«1βγπ¬π’π§β‘πΓ(ππ+π^(π ) ππ¨π­β‘π ) ππγ + C y sin x = β«1βγ(2π₯ sinβ‘π₯+π₯^(2 ) sinβ‘γπ₯ cotβ‘π₯ γ ) ππ₯γ + C y sinβ‘π₯ = β«1βγ2π₯ sinβ‘π₯ ππ₯+γ β«1βγπ₯^2 sinβ‘π₯ cotβ‘π₯ ππ₯+γ πΆ y sinβ‘π₯ = 2β«1βγπ¬π’π§β‘π (π) ππγ+β«1βγπ₯^2 sinβ‘π₯ cotβ‘π₯ ππ₯+γ πΆ Integrating by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = sin x & g (x) = π₯ y sinβ‘π₯ = 2 [π¬π’π§β‘π β«1βγπ ππβγ β«1βγ[πππβ‘γπ β«1βγπ ππγ γ ] ππγ] + β«1βγπ₯^2 sinβ‘π₯ γ cotβ‘π₯ dx + C y sinβ‘π₯ = 2 [sinβ‘π₯ [π₯^2/2]ββ«1βγπππβ‘γπ γ [π₯^2/2]ππγ] + β«1βγπ₯^2 sinβ‘π₯ γ cotβ‘π₯ dx y sin x = x2sin x β β«1βπ^π cos x dx + β«1βγπ^π πππβ‘π γ πππβ‘π dx + C y sin x = x2sin x β β«1βπ₯^2 cos x dx + β«1βγπ₯^2 sinβ‘π₯ γΓcosβ‘π₯/sinβ‘π₯ dx + C y sin x = x2sin x β β«1βπ₯^2 cos x dx + β«1βγπ₯^2 cosβ‘π₯ γ dx + C y sin x = x2 sin x + C Given that y = 0 when x = π/2 Putting π=π/π and y = 0 in (1) (0) sin π/2=(π/2)^2 sinβ‘γ(π/2)+Cγ 0 =π^2/4 (1)+C γβπγ^π/π=π Putting value in C in (1) y sin x = x2 sin x + C y sin x = π^π πππβ‘γπ βγ π^π/π Dividing both sides by sin x (π¦ sinβ‘π₯)/sinβ‘π₯ =(π₯^2 sinβ‘π₯)/sinβ‘π₯ βπ^2/(4 sinβ‘π₯ ) π=π^πβπ^π/γπ π¬π’π§γβ‘π where sinβ‘γπ₯β 0γ y sinβ‘π₯ = 2 [π¬π’π§β‘π β«1βγπ ππβγ β«1βγ[πππβ‘γπ β«1βγπ ππγ γ ] ππγ] + β«1βγπ₯^2 sinβ‘π₯ γ cotβ‘π₯ dx + C

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.