**
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) = Ο/4

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) You are here

Question 21 (Or 2nd) Important

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 Oct. 1, 2019 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) = Ο/4

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