Example 22 - Particular solution dy/dx + y cot x = 2x + x2 cot x - Examples

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Example 22 Find the particular solution of the differential equation 𝑑𝑦/𝑑π‘₯+𝑦 cot⁑〖π‘₯=2π‘₯+π‘₯^2 cot⁑π‘₯(π‘₯β‰ 0) γ€— given that 𝑦=0 π‘€β„Žπ‘’π‘› π‘₯=πœ‹/2 𝑑𝑦/𝑑π‘₯+𝑦 cot⁑〖π‘₯=2π‘₯+π‘₯^2 cot⁑π‘₯ γ€— Differential equation is of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = cot x & Q = 2x + x2 cot x IF = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— IF = 𝑒^∫1β–’γ€–cot⁑π‘₯ 𝑑π‘₯γ€— IF = 𝑒^"log sin x " IF = sin x Solution is y (IF) =∫1β–’γ€–(𝑄×𝐼𝐹) 𝑑π‘₯+𝑐〗 y sin x ∫1β–’γ€–sin⁑π‘₯=(2π‘₯+π‘₯^(2 ) cot⁑π‘₯ ) 𝑑π‘₯+𝑐〗 y sin x = ∫1β–’γ€–(2π‘₯ sin⁑π‘₯+π‘₯^(2 ) sin⁑〖π‘₯ cot⁑π‘₯ γ€— ) 𝑑π‘₯+𝑐〗 y sin⁑π‘₯ = ∫1β–’γ€–2π‘₯ sin⁑π‘₯ 𝑑π‘₯+γ€— ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ cot⁑π‘₯ 𝑑π‘₯+γ€— 𝐢 y sin⁑π‘₯ = 2∫1β–’γ€–sin⁑π‘₯ (π‘₯)𝑑π‘₯+𝑓π‘₯^2 sin⁑π‘₯ cot⁑〖π‘₯ 𝑑π‘₯+𝑐〗 γ€— y sin⁑π‘₯ = 2 [sin⁑〖π‘₯ ∫1β–’γ€–π‘₯ 𝑑π‘₯βˆ’γ€— ∫1β–’γ€–[cos⁑〖π‘₯ ∫1β–’γ€–π‘₯ 𝑑π‘₯γ€— γ€— ] 𝑑π‘₯γ€— γ€— ] + ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— cos⁑π‘₯/sin⁑π‘₯ dx + C y sin x = 2 sin x ["sin x " [π‘₯^2/2]" βˆ’2" ∫1β–’γ€–γ€–cos x〗⁑[π‘₯^2/2] 𝑑π‘₯+∫1β–’π‘₯^2 γ€— cos⁑π‘₯ " dx + c" ] y sin x = x2sin x βˆ’ ∫1β–’π‘₯^2 cos x dx + ∫1β–’π‘₯^2 cos x dx + c y sin x = x2 sin x + c Given that y = 0 when x = πœ‹/2 Putting π‘₯=πœ‹/2 and y = 0 in (1) (0) sin πœ‹/2=(πœ‹/2)^2 sin⁑〖(πœ‹/2)+Cγ€— 0 =πœ‹^2/4 (1)+C γ€–βˆ’πœ‹γ€—^2/4=C Putting value in C in (1) y sin x = π‘₯^2 sin⁑〖π‘₯ βˆ’γ€— πœ‹^2/4 Dividing whole by sin x (𝑦 sin⁑π‘₯)/sin⁑π‘₯ =(π‘₯^2 sin⁑π‘₯)/sin⁑π‘₯ βˆ’πœ‹^2/(4 sin⁑π‘₯ ) π’š=𝒙^πŸβˆ’π…^𝟐/γ€–πŸ’ 𝐬𝐒𝐧〗⁑𝒙

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