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Misc 11 Find a particular solution of the differential equation 𝑑𝑦/𝑑π‘₯+𝑦 cot⁑〖π‘₯=4π‘₯ π‘π‘œπ‘ π‘’π‘ π‘₯ (π‘₯β‰ 0) ,γ€— given that 𝑦=0 when π‘₯=πœ‹/2Given 𝑑𝑦/𝑑π‘₯+𝑦 cot⁑〖π‘₯=4π‘₯ π‘π‘œπ‘ π‘’π‘ π‘₯ γ€— This of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = cot x & Q = 4x cosec x IF = 𝑒^∫1▒𝑃𝑑π‘₯ IF = 𝒆^∫1β–’πœπ¨π­β‘γ€–π’™ 𝒅𝒙〗 IF = 𝑒^(log⁑(sin⁑〖π‘₯)γ€— ) IF = sin x Solution is y (IF) = ∫1β–’γ€–(𝑄×𝐼.𝐹)𝑑π‘₯+𝑐 γ€— y sin x = ∫1β–’γ€–πŸ’π’™ 𝒄𝒐𝒔𝒆𝒄 𝒙 π’”π’Šπ’β‘π’™ 𝒅𝒙+𝒄 γ€— y sin x = ∫1β–’γ€–4π‘₯ 1/sin⁑π‘₯ sin⁑π‘₯ 𝑑π‘₯+𝑐 γ€— y sin x = ∫1β–’γ€–4π‘₯ 𝑑π‘₯+𝑐 γ€— y sin x = (4π‘₯^2)/2+𝑐 y sin x = 2x2 + C Given that π’š=𝟎 when 𝒙=𝝅/𝟐 Put x = πœ‹/2 & y = 0 in (1) 0 Γ— sin πœ‹/2 = 2 (πœ‹/2)^2+𝐢 0 = 2 (γ€–πœ‹/4γ€—^2 ) + C 0 = γ€–πœ‹/2γ€—^2 + C C = γ€–βˆ’π…γ€—^𝟐/𝟐 Putting value of C in (1) y sin x = 2x2 + c y sin x = 2𝒙^𝟐 βˆ’ 〖𝝅/πŸγ€—^𝟐

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.