Question 21 (OR 2 nd question)
Find the general solution of the differential equation:
dx/dy = (y tany - x tany - xy) / (y tany)
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
CBSE Class 12 Sample Paper for 2019 Boards
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CBSE Class 12 Sample Paper for 2019 Boards
Last updated at May 29, 2023 by Teachoo
Question 21 (OR 2 nd question)
Find the general solution of the differential equation:
dx/dy = (y tany - x tany - xy) / (y tany)
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Question 21 (OR 2nd question) Find the general solution of the differential equation: 𝑑𝑥/𝑑𝑦=(𝑦 tan𝑦 − 𝑥 tan𝑦 − 𝑥𝑦)/(𝑦 tan𝑦 ) 𝑑𝑥/𝑑𝑦=(𝑦 tan𝑦 − 𝑥 tan𝑦 − 𝑥𝑦)/(𝑦 tan𝑦 ) 𝑑𝑥/𝑑𝑦=(𝑦 tan𝑦)/(𝑦 tan𝑦 )−(𝑥 tan𝑦 )/(𝑦 tan𝑦 )−𝑥𝑦/(𝑦 tan𝑦 ) 𝑑𝑥/𝑑𝑦=1−𝑥/𝑦−𝑥/tan𝑦 𝑑𝑥/𝑑𝑦+𝑥/𝑦+𝑥/tan𝑦 =1 𝑑𝑥/𝑑𝑦+𝑥(1/𝑦+1/tan𝑦 )=1 Differential equation is of the form 𝑑𝑥/𝑑𝑦 + P1 x = Q1 where P1 = 1/𝑦+1/tan𝑦 & Q1 = 1 Now, IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗 IF = e^∫1▒〖(1/𝑦 + 1/tan𝑦 )𝑑𝑦" " 〗 IF = e^(∫1▒〖1/𝑦 𝑑𝑦〗 +∫1▒〖1/tan𝑦 𝑑𝑦〗) IF = e^(∫1▒〖1/𝑦 𝑑𝑦〗 +∫1▒〖cot𝑦 𝑑𝑦〗) IF = e^(log𝑦 + logsin𝑦 ) IF = e^〖log 〗〖(𝑦 sin𝑦)〗 IF = y sin y Solution is x(IF) = ∫1▒〖(𝑄×𝐼𝐹)𝑑𝑦+𝐶 〗 x (y sin y) =∫1▒〖1×𝑦 sin𝑦 〗 𝑑𝑦+𝐶 xy sin y =∫1▒〖𝑦 sin𝑦 〗 𝑑𝑦+𝐶 (As ∫1▒cot𝑥 𝑑𝑥=logsin𝑥 ) (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▒sin𝑦 𝑑𝑦−∫1▒(𝑑(𝑦)/𝑑𝑦 ∫1▒〖sin𝑦 𝑑𝑦〗) 𝑑𝑦 xy sin y =−𝑦 cos𝑦 − ∫1▒〖−cos𝑦 𝑑𝑦〗 xy sin y =−𝑦 cos𝑦+ ∫1▒〖cos𝑦 𝑑𝑦〗 xy sin y =−𝑦 cos𝑦+sin𝑦+𝐶 xy sin y =sin𝑦−𝑦 cos𝑦+𝐶 x = (𝒔𝒊𝒏𝒚 − 𝒚 𝒄𝒐𝒔𝒚 + 𝑪" " )/(𝒚 𝐬𝐢𝐧𝒚 )