Question 21 (OR 2 nd question)
Find the general solution of the differential equation:
dx/dy = (y tany - x tany - xy) / (y tany)
CBSE Class 12 Sample Paper for 2019 Boards
Question 1 Important
Question 2
Question 3
Question 4 (Or 1st) Important
Question 4 (Or 2nd)
Question 5
Question 6
Question 7 Important
Question 8 (Or 1st) Important
Question 8 (Or 2nd)
Question 9
Question 10 (Or 1st) Important
Question 10 (Or 2nd)
Question 11 Important
Question 12 (Or 1st)
Question 12 (Or 2nd)
Question 13 (Or 1st) Important
Question 13 (Or 2nd)
Question 14 Important
Question 15
Question 16 (Or 1st)
Question 16 (Or 2nd) Important
Question 17
Question 18
Question 19 Important
Question 20 Important
Question 21 (Or 1st)
Question 21 (Or 2nd) Important You are here
Question 22
Question 23 Important
Question 24 (Or 1st)
Question 24 (Or 2nd) Important
Question 25
Question 26 (Or 1st) Important
Question 26 (Or 2nd)
Question 27 (Or 1st) Important
Question 27 (Or 2nd) Important
Question 28
Question 29 Important
CBSE Class 12 Sample Paper for 2019 Boards
Last updated at April 16, 2024 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)
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 )