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)

Find general solution of  dx/dy = (y tan⁡y  - x tan⁡y  - xy) / y tan y

Question 21 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 21 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 21 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 4

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Transcript

Question 21 (OR 2nd question) Find the general solution of the differential equation: 𝑑𝑥/𝑑𝑦=(𝑦 tan⁡𝑦 − 𝑥 tan⁡𝑦 − 𝑥𝑦)/(𝑦 tan⁡𝑦 ) 𝑑𝑥/𝑑𝑦=(𝑦 tan⁡𝑦 − 𝑥 tan⁡𝑦 − 𝑥𝑦)/(𝑦 tan⁡𝑦 ) 𝑑𝑥/𝑑𝑦=(𝑦 tan⁡𝑦)/(𝑦 tan⁡𝑦 )−(𝑥 tan⁡𝑦 )/(𝑦 tan⁡𝑦 )−𝑥𝑦/(𝑦 tan⁡𝑦 ) 𝑑𝑥/𝑑𝑦=1−𝑥/𝑦−𝑥/tan⁡𝑦 𝑑𝑥/𝑑𝑦+𝑥/𝑦+𝑥/tan⁡𝑦 =1 𝑑𝑥/𝑑𝑦+𝑥(1/𝑦+1/tan⁡𝑦 )=1 Differential equation is of the form 𝑑𝑥/𝑑𝑦 + P1 x = Q1 where P1 = 1/𝑦+1/tan⁡𝑦 & Q1 = 1 Now, IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗 IF = e^∫1▒〖(1/𝑦 + 1/tan⁡𝑦 )𝑑𝑦" " 〗 IF = e^(∫1▒〖1/𝑦 𝑑𝑦〗 +∫1▒〖1/tan⁡𝑦 𝑑𝑦〗) IF = e^(∫1▒〖1/𝑦 𝑑𝑦〗 +∫1▒〖cot⁡𝑦 𝑑𝑦〗) IF = e^(log⁡𝑦 + log⁡sin⁡𝑦 ) IF = e^〖log 〗⁡〖(𝑦 sin⁡𝑦)〗 IF = y sin y Solution is x(IF) = ∫1▒〖(𝑄×𝐼𝐹)𝑑𝑦+𝐶 〗 x (y sin y) =∫1▒〖1×𝑦 sin⁡𝑦 〗 𝑑𝑦+𝐶 xy sin y =∫1▒〖𝑦 sin⁡𝑦 〗 𝑑𝑦+𝐶 (As ∫1▒cot⁡𝑥 𝑑𝑥=log⁡sin⁡𝑥 ) (As log a + log b = log ab) We know that ∫1▒〖𝑓(𝑦) 𝑔⁡(𝑦) 〗 𝑑𝑦=𝑓(𝑦) ∫1▒𝑔(𝑦) 𝑑𝑦−∫1▒(𝑓^′ 𝑦∫1▒𝑔(𝑦) 𝑑𝑦) 𝑑𝑦 Putting f(y) = y and g(y) = sin y xy sin y =𝑦" " ∫1▒sin⁡𝑦 𝑑𝑦−∫1▒(𝑑(𝑦)/𝑑𝑦 ∫1▒〖sin⁡𝑦 𝑑𝑦〗) 𝑑𝑦 xy sin y =−𝑦 cos⁡𝑦 − ∫1▒〖−cos⁡𝑦 𝑑𝑦〗 xy sin y =−𝑦 cos⁡𝑦+ ∫1▒〖cos⁡𝑦 𝑑𝑦〗 xy sin y =−𝑦 cos⁡𝑦+sin⁡𝑦+𝐶 xy sin y =sin⁡𝑦−𝑦 cos⁡𝑦+𝐶 x = (𝒔𝒊𝒏⁡𝒚 − 𝒚 𝒄𝒐𝒔⁡𝒚 + 𝑪" " )/(𝒚 𝐬𝐢𝐧⁡𝒚 )

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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.