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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

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⁑π‘₯ γ€—^" " IF = sin x Solution is y (IF) =∫1β–’γ€–(𝑄×𝐼𝐹) 𝑑π‘₯+𝑐〗 y sin x = ∫1β–’γ€–sin⁑π‘₯Γ—(2π‘₯+π‘₯^(2 ) cot⁑π‘₯ ) 𝑑π‘₯γ€— + C y sin x = ∫1β–’γ€–(2π‘₯ sin⁑π‘₯+π‘₯^(2 ) sin⁑〖π‘₯ cot⁑π‘₯ γ€— ) 𝑑π‘₯γ€— + 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∫1▒〖𝐬𝐒𝐧⁑𝒙 (𝒙) 𝒅𝒙〗+∫1β–’γ€–π‘₯^2 sin⁑π‘₯ cot⁑π‘₯ 𝑑π‘₯+γ€— 𝐢 y sin⁑π‘₯ = 2 [𝐬𝐒𝐧⁑𝒙 ∫1▒〖𝒙 π’…π’™βˆ’γ€— ∫1β–’γ€–[𝒄𝒐𝒔⁑〖𝒙 ∫1▒〖𝒙 𝒅𝒙〗 γ€— ] 𝒅𝒙〗] + ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— cot⁑π‘₯ dx + C Integrating by parts with ∫1▒〖𝑓(π‘₯) 𝑔(π‘₯) 𝑑π‘₯=𝑓(π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯ βˆ’βˆ«1β–’γ€–[𝑓^β€² (π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯] 𝑑π‘₯γ€—γ€—γ€—γ€— Take f (x) = sin x & g (x) = π‘₯ y sin⁑π‘₯ = 2 [𝐬𝐒𝐧⁑𝒙 ∫1▒〖𝒙 π’…π’™βˆ’γ€— ∫1β–’γ€–[𝒄𝒐𝒔⁑〖𝒙 ∫1▒〖𝒙 𝒅𝒙〗 γ€— ] 𝒅𝒙〗] + ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— cot⁑π‘₯ dx + C y sin⁑π‘₯ = 2 [sin⁑π‘₯ [π‘₯^2/2]βˆ’βˆ«1▒〖𝒄𝒐𝒔⁑〖𝒙 γ€— [π‘₯^2/2]𝒅𝒙〗] + ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— cot⁑π‘₯ dx y sin x = x2sin x βˆ’ ∫1β–’π‘₯^2 cos x dx + ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— cot⁑π‘₯ dx + C y sin x = x2sin x βˆ’ ∫1β–’π‘₯^2 cos x dx + ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€—Γ—cos⁑π‘₯/sin⁑π‘₯ dx + C y sin x = x2sin x βˆ’ ∫1β–’π‘₯^2 cos x dx + ∫1β–’γ€–π‘₯^2 cos⁑π‘₯ γ€— dx + C y sin x = x2 sin x + C Given that y = 0 when x = πœ‹/2 Putting 𝒙=𝝅/𝟐 and y = 0 in (1) (0) sin πœ‹/2=(πœ‹/2)^2 sin⁑〖(πœ‹/2)+Cγ€— …(1) 0 =πœ‹^2/4 (1)+C γ€–βˆ’πœ‹γ€—^2/4=C Putting value in C in (1) y sin x = x2 sin x + C y sin x = π‘₯^2 sin⁑〖π‘₯ βˆ’γ€— πœ‹^2/4 Dividing both sides 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 where sin⁑〖π‘₯β‰ 0γ€—

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.