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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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