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Question 21 (OR 2
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nd
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question)
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Find the general solution of the differential equation:

dx/dy = (y tanβ‘y - x tanβ‘y - xy) / (y tanβ‘y)

Last updated at Nov. 1, 2018 by Teachoo

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Question 21 (OR 2
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**
nd
**
**
question)
**

Find the general solution of the differential equation:

dx/dy = (y tanβ‘y - x tanβ‘y - xy) / (y tanβ‘y)

Transcript

Question 21 (OR 2nd question) Find the general solution of the differential equation: ππ₯/ππ¦=(π¦ tanβ‘π¦ β π₯ tanβ‘π¦ β π₯π¦)/(π¦ tanβ‘π¦ ) ππ₯/ππ¦=(π¦ tanβ‘π¦ β π₯ tanβ‘π¦ β π₯π¦)/(π¦ tanβ‘π¦ ) ππ₯/ππ¦=(π¦ tanβ‘π¦)/(π¦ tanβ‘π¦ )β(π₯ tanβ‘π¦ )/(π¦ tanβ‘π¦ )βπ₯π¦/(π¦ tanβ‘π¦ ) ππ₯/ππ¦=1βπ₯/π¦βπ₯/tanβ‘π¦ ππ₯/ππ¦+π₯/π¦+π₯/tanβ‘π¦ =1 ππ₯/ππ¦+π₯(1/π¦+1/tanβ‘π¦ )=1 Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 where P1 = 1/π¦+1/tanβ‘π¦ & Q1 = 1 Now, IF = π^β«1βγπ_1 ππ¦γ IF = e^β«1βγ(1/π¦ + 1/tanβ‘π¦ )ππ¦" " γ IF = e^(β«1βγ1/π¦ ππ¦γ +β«1βγ1/tanβ‘π¦ ππ¦γ) IF = e^(β«1βγ1/π¦ ππ¦γ +β«1βγcotβ‘π¦ ππ¦γ) IF = e^(logβ‘π¦ + logβ‘sinβ‘π¦ ) IF = e^logβ‘γπ¦ sinβ‘π¦ γ IF = y sin y (As β«1βcotβ‘π₯ ππ₯=logβ‘sinβ‘π₯ ) (As log a + log b = log ab) (As log a + log b = log ab) We know that β«1βγπ(π¦) πβ‘(π¦) γ ππ¦=π(π¦) β«1βπ(π¦) ππ¦ββ«1β(π^β² π¦β«1βπ(π¦) ππ¦) ππ¦ Putting f(y) = y and g(y) = sin y xy sin y =π¦" " β«1βsinβ‘π¦ ππ¦ββ«1β(π(π¦)/ππ¦ β«1βγsinβ‘π¦ ππ¦γ) ππ¦ xy sin y =βπ¦ cosβ‘π¦ β β«1βγβcosβ‘π¦ ππ¦γ xy sin y =βπ¦ cosβ‘π¦+ β«1βγcosβ‘π¦ ππ¦γ xy sin y =βπ¦ cosβ‘π¦+sinβ‘π¦+πΆ xy sin y =sinβ‘π¦βπ¦ cosβ‘π¦+πΆ x = (πππβ‘π β π πππβ‘π + πͺ" " )/(π π¬π’π§β‘π )

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4 (Or 1st)

Question 4 (Or 2nd)

Question 5

Question 6

Question 7

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st)

Question 10 (Or 2nd)

Question 11

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st)

Question 13 (Or 2nd)

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17

Question 18

Question 19

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd) You are here

Question 22

Question 23

Question 24 (Or 1st)

Question 24 (Or 2nd)

Question 25

Question 26 (Or 1st)

Question 26 (Or 2nd)

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28

Question 29

Class 12

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.