CBSE Class 12 Sample Paper for 2019 Boards

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

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

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Question 21 (OR 1st question) Find the particular solution of the following differential equation. cos y dx + (1 + 2π^(βπ₯)) sin y dy = 0; y(0) = π/4 cos y dx + (1 + 2π^(βπ₯)) sin y dy = 0 cos y dx = β (1 + 2π^(βπ₯)) sin y dy ππ₯/((1 + γ2πγ^(βπ₯))) = β sinβ‘π¦/cosβ‘π¦ dy ππ₯/((1 + γ2πγ^(βπ₯))) = β tan y dy ππ₯/((1 + 2/π^π₯ ) ) = β tan y dy ππ₯/(((π^π₯ + 2)/π^π₯ ) ) = β tan y dy (π^π₯ ππ₯)/((π^π₯ + 2) ) = β tan y dy Integrating both sides β«1β(π^π₯ ππ₯)/((π^π₯ + 2) ) = ββ«1βγtanβ‘π¦ ππ¦γ β«1β(π^π₯ ππ₯)/((π^π₯ + 2) ) = β logβ‘γ |secβ‘π¦ |γ +πΆ Let ex + 2 = t ex dx = dt β«1βππ‘/π‘ = β logβ‘γ |secβ‘π¦ |γ +πΆ logβ‘γ |π‘|γ= β logβ‘γ |secβ‘π¦ |γ +πΆ Putting back value of t logβ‘γ |π^π₯+2|γ = β logβ‘γ |secβ‘π¦ |γ +πΆ logβ‘γ |π^π₯+2|γ+ logβ‘γ |secβ‘π¦ |γ=πΆ As log a + log b = log ab logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=πΆ logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=logβ‘πΎ Cancelling log (π^π₯+2)Γsecβ‘π¦ = K (π^π₯+2)Γ1/cosβ‘π¦ = K (π^π₯+2) = K cos y logβ‘γ |π^π₯+2|γ = β logβ‘γ |secβ‘π¦ |γ +πΆ logβ‘γ |π^π₯+2|γ+ logβ‘γ |secβ‘π¦ |γ=πΆ As log a + log b = log ab logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=πΆ logβ‘γ |(π^π₯+2)Γsecβ‘π¦ |γ=logβ‘πΎ Cancelling log (π^π₯+2)Γsecβ‘π¦ = K (π^π₯+2)Γ1/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/β2 3β2 = K K = 3β2 Putting value of K in (1) (π^π₯+2) = K cos y (π^π₯+2) = 3β2 cos y Hence, the particular solution is π^π+π = 3βπ cos y