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Question 21 (OR 2
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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

Paper Summary

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

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 24, 2021 by Teachoo

**
Question 21 (OR 2
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**
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 )