**
Question 21 (OR 1
**
**
st
**
**
question)
**

Find the particular solution of the following differential equation.

cos y dx + (1 + 2e
^{
-x
}
) sin y dy = 0; y(0) = π/4

Last updated at Nov. 1, 2018 by Teachoo

**
Question 21 (OR 1
**
**
st
**
**
question)
**

Find the particular solution of the following differential equation.

cos y dx + (1 + 2e
^{
-x
}
) sin y dy = 0; y(0) = π/4

Transcript

Question 21 (OR 1st question) Find the particular solution of the following differential equation. cos y dx + (1 + 2π^(βπ₯)) sin y dy = 0; y(0) = π/4 cos y dx + (1 + 2π^(βπ₯)) sin y dy = 0 cos y dx = β (1 + 2π^(βπ₯)) sin y dy ππ₯/((1 + γ2πγ^(βπ₯))) = β sinβ‘π¦/cosβ‘π¦ dy ππ₯/((1 + γ2πγ^(βπ₯))) = β tan y dy ππ₯/((1 + 2/π^π₯ ) ) = β tan y dy ππ₯/(((π^π₯ + 2)/π^π₯ ) ) = β tan y dy (π^π₯ ππ₯)/((π^π₯ + 2) ) = β tan y dy Integrating both sides β«1β(π^π₯ ππ₯)/((π^π₯ + 2) ) = ββ«1βγtanβ‘π¦ ππ¦γ β«1β(π^π₯ ππ₯)/((π^π₯ + 2) ) = β logβ‘γ |secβ‘π¦ |γ +πΆ Let ex + 2 = t ex dx = dt β«1βππ‘/π‘ = β logβ‘γ |secβ‘π¦ |γ +πΆ logβ‘γ |π‘|γ= β logβ‘γ |secβ‘π¦ |γ +πΆ Putting back value of t logβ‘γ |π^π₯+2|γ = β logβ‘γ |secβ‘π¦ |γ +πΆ logβ‘γ |π^π₯+2|γ+ logβ‘γ |secβ‘π¦ |γ=πΆ As log a + log b = log ab logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=πΆ logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=logβ‘πΎ Cancelling log (π^π₯+2)Γsecβ‘π¦ = K (π^π₯+2)Γ1/cosβ‘π¦ = K (π^π₯+2) = K cos y logβ‘γ |π^π₯+2|γ = β logβ‘γ |secβ‘π¦ |γ +πΆ logβ‘γ |π^π₯+2|γ+ logβ‘γ |secβ‘π¦ |γ=πΆ As log a + log b = log ab logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=πΆ logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=logβ‘πΎ Cancelling log (π^π₯+2)Γsecβ‘π¦ = K (π^π₯+2)Γ1/cosβ‘π¦ = K (π^π₯+2) = K cos y We need to find the particular solution for y(0) = π/4 Putting x = 0, y = π/4 in (1) (π^π₯+2) = K cos y (π^0+2) = K cos π/4 (1+2) = K cos 45Β° 3 = K Γ 1/β2 3β2 = K K = 3β2 Putting value of K in (1) (π^π₯+2) = K cos y (π^π₯+2) = 3β2 cos y Hence, the particular solution is π^π+π = 3βπ cos y

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) You are here

Question 21 (Or 2nd)

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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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.