Solve: d y d x + 1 = e x + y .

The given equation is d y d x + 1 = e x + y . . . . . 1 which is not variable separable Let x + y = v ⇒ 1 + d y d x = d v d x Then, equation 1 becomes, d v d x = e v ⇒ e - v d v = d x Integrating both sides, we get ⇒ ∫ e - v d v = ∫ d x ⇒ - e - v = x + C ⇒ - 1 = e v x + C ⇒ x + C e x + y + 1 = 0 , which is the required solution and C is a arbitrary constant.

Find the general solution of d y d x = e x + y .

Given that, the differential equation is d y d x = e x + y ⇒ d y d x = e x · e y ⇒ d y e y = e x d x [Separating variables] ⇒ e - y d y = e x d x Apply integration on both sides, then ∫ e - y d y = ∫ e x d x ⇒ - e - y = e x + c ⇒ e - y = - e x - c , which is the required general solution.

Solve the differential equation: cos ⁡ x · d y d x + y sin ⁡ x = sec 2 ⁡ x

Given differential equation is cos x · d y d x + y sin x = sec 2 x . So, d y d x + y tan x = sec 3 x which is a linear differential equation in y . On Comparing it with d y d x + P y = Q , we get P = tan x , Q = sec 3 x I . F . = e ∫ P d x = e ∫ tan x d x = e log sec x = sec x So, the integrating factor, I . F . = sec x The complete solution is y · I . F = ∫ Q · I . F d x ⇒ y · sec x = ∫ sec 3 x · sec x d x ⇒ y sec x = ∫ sec 2 x · sec 2 x d x ⇒ y sec x = ∫ 1 + tan 2 x sec 2 x d x ⇒ y sec x = ∫ sec 2 x + tan 2 x sec 2 x d x ⇒ y sec x = tan x + tan 3 x 3 + k ∵ ∫ f x n f ' x d x = f x n + 1 n + 1 + c which is the required solution of given differential equation.

Solve the differential equation d y d x = 2 y + x + 1 2 x + 4 y + 3

Given, d y d x = 2 y + x + 1 2 x + 4 y + 3 = x + 2 y + 1 2 x + 2 y + 3 Let v = x + 2 y , then d v d x = 1 + 2 d y d x Substituting x + 2 y = v in d y d x , we get d v d x = 1 + 2 ( v + 1 ) 2 v + 3 = 2 v + 3 + 2 v + 2 2 v + 3 ⇒ d v d x = 4 v + 5 2 v + 3 ⇒ 2 v + 3 4 v + 5 d v = d x Integrating on both sides, we get ∫ 2 v + 3 4 v + 5 d v = ∫ d x ⇒ 1 2 ∫ 4 v + 6 4 v + 5 d v = ∫ d x ⇒ 1 2 ∫ 4 v + 5 + 1 4 v + 5 d v = ∫ d x ⇒ 1 2 ∫ 1 + 1 4 v + 5 d v = ∫ d x ⇒ 1 2 v + log 4 v + 5 4 = x + c Substituting v = x + 2 y . ⇒ 1 2 x + 2 y + log 4 x + 2 y + 5 8 = x + c ⇒ 4 x + 8 y - 8 x + log ( 4 x + 8 y + 5 ) = 8 c = c ' Let 8 c = c ' Therefore, the required solution is 8 y - 4 x + log ( 4 x + 8 y + 5 ) = c ' .

Solve the differential equation sin − 1 ⁡ d y d x = x + y

Given that sin - 1 d y d x = x + y ⇒ d y d x = sin x + y ........(i) Let x + y = t ⇒ d t d x = 1 + d y d x ........(ii) From (i) and (ii) we get d t d x - 1 = sin t ⇒ d t 1 + sin t = d x Integrating both sides, we get ∫ d t 1 + sin t = ∫ d x ⇒ ∫ 1 1 + sin ⁡ t × 1 - sin t 1 - sin t d t = ∫ d x ⇒ ∫ 1 - sin t 1 - sin 2 t d t = ∫ d x ⇒ ∫ ( 1 - sin t ) d t cos 2 t = ∫ d x ⇒ ∫ sec 2 t d t - ∫ tan t sec t d t = ∫ d x ⇒ tan t - sec t = x + c ⇒ tan x + y - sec x + y = x + c Hence, the required solution is tan ⁡ ( x + y ) − sec ⁡ ( x + y ) = x + c .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Calculus>Chapter 8 - Differential Equations>Telangana Board-2018

Embibe Experts Solutions for Exercise 1: Telangana Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 1: Telangana Board-2018

Attempt the free practice questions from Exercise 1: Telangana Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.