Question 21 (OR 2 nd question)

Find the general solution of the differential equation:

  dx/dy = (y tan⁑y  - x tan⁑y  - xy) / (y tan⁑y)

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  1. Class 12
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Transcript

Question 21 (OR 2nd question) Find the general solution of the differential equation: 𝑑π‘₯/𝑑𝑦=(𝑦 tan⁑𝑦 βˆ’ π‘₯ tan⁑𝑦 βˆ’ π‘₯𝑦)/(𝑦 tan⁑𝑦 ) 𝑑π‘₯/𝑑𝑦=(𝑦 tan⁑𝑦 βˆ’ π‘₯ tan⁑𝑦 βˆ’ π‘₯𝑦)/(𝑦 tan⁑𝑦 ) 𝑑π‘₯/𝑑𝑦=(𝑦 tan⁑𝑦)/(𝑦 tan⁑𝑦 )βˆ’(π‘₯ tan⁑𝑦 )/(𝑦 tan⁑𝑦 )βˆ’π‘₯𝑦/(𝑦 tan⁑𝑦 ) 𝑑π‘₯/𝑑𝑦=1βˆ’π‘₯/π‘¦βˆ’π‘₯/tan⁑𝑦 𝑑π‘₯/𝑑𝑦+π‘₯/𝑦+π‘₯/tan⁑𝑦 =1 𝑑π‘₯/𝑑𝑦+π‘₯(1/𝑦+1/tan⁑𝑦 )=1 Differential equation is of the form 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 where P1 = 1/𝑦+1/tan⁑𝑦 & Q1 = 1 Now, IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗 IF = e^∫1β–’γ€–(1/𝑦 + 1/tan⁑𝑦 )𝑑𝑦" " γ€— IF = e^(∫1β–’γ€–1/𝑦 𝑑𝑦〗 +∫1β–’γ€–1/tan⁑𝑦 𝑑𝑦〗) IF = e^(∫1β–’γ€–1/𝑦 𝑑𝑦〗 +∫1β–’γ€–cot⁑𝑦 𝑑𝑦〗) IF = e^(log⁑𝑦 + log⁑sin⁑𝑦 ) IF = e^γ€–log 〗⁑〖(𝑦 sin⁑𝑦)γ€— IF = y sin y Solution is x(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑𝑦+𝐢 γ€— x (y sin y) =∫1β–’γ€–1×𝑦 sin⁑𝑦 γ€— 𝑑𝑦+𝐢 xy sin y =∫1▒〖𝑦 sin⁑𝑦 γ€— 𝑑𝑦+𝐢 (As ∫1β–’cot⁑π‘₯ 𝑑π‘₯=log⁑sin⁑π‘₯ ) (As log a + log b = log ab) We know that ∫1▒〖𝑓(𝑦) 𝑔⁑(𝑦) γ€— 𝑑𝑦=𝑓(𝑦) ∫1▒𝑔(𝑦) π‘‘π‘¦βˆ’βˆ«1β–’(𝑓^β€² π‘¦βˆ«1▒𝑔(𝑦) 𝑑𝑦) 𝑑𝑦 Putting f(y) = y and g(y) = sin y xy sin y =𝑦" " ∫1β–’sin⁑𝑦 π‘‘π‘¦βˆ’βˆ«1β–’(𝑑(𝑦)/𝑑𝑦 ∫1β–’γ€–sin⁑𝑦 𝑑𝑦〗) 𝑑𝑦 xy sin y =βˆ’π‘¦ cos⁑𝑦 βˆ’ ∫1β–’γ€–βˆ’cos⁑𝑦 𝑑𝑦〗 xy sin y =βˆ’π‘¦ cos⁑𝑦+ ∫1β–’γ€–cos⁑𝑦 𝑑𝑦〗 xy sin y =βˆ’π‘¦ cos⁑𝑦+sin⁑𝑦+𝐢 xy sin y =sinβ‘π‘¦βˆ’π‘¦ cos⁑𝑦+𝐢 x = (π’”π’Šπ’β‘π’š βˆ’ π’š π’„π’π’”β‘π’š + π‘ͺ" " )/(π’š π¬π’π§β‘π’š )

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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.