Examples

Chapter 9 Class 12 Differential Equations
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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 where P = cot x & Q = 2x + x2 cot x IF = π^β«1βγπ ππ₯γ IF = π^β«1βγcotβ‘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sinβ‘π₯Γ(2π₯+π₯^(2 ) cotβ‘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 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 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₯^2 cos x dx + β«1βγπ₯^2 sinβ‘π₯ γ cotβ‘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 β¦(1) 0 =π^2/4 (1)+C γβπγ^2/4=C Putting value in C in (1) y sin x = x2 sin x + C y sin x = π₯^2 sinβ‘γπ₯ βγ π^2/4 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