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

Example 22 - Chapter 9 Class 12 Differential Equations - Part 2
Example 22 - Chapter 9 Class 12 Differential Equations - Part 3 Example 22 - Chapter 9 Class 12 Differential Equations - Part 4


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.