Solve the following differential equation : ( x + y + 1 ) d y d x = 1

Given differential equation is ( x + y + 1 ) d y d x = 1 . So, ( x + y + 1 ) = d x d y d x d y - x = y + 1 Which is a linear differential equation in x . On Comparing it with d x d y + P x = Q , We get P = - 1 , Q = y + 1 IF = e ∫ P d y = e ∫ - d y = e - y So, the integrating factor, IF = e - y The complete solution is x · IF = ∫ Q · IF d y ⇒ x · e - y = ∫ y + 1 · e - y d y ⇒ x e - y = - y + 1 e - y + ∫ e - y d y [Integration by parts] ⇒ x e - y = - y + 1 e - y - e - y ⇒ x e - y = - e - y y + 2 + c Which is the required solution of given differential equation.

Solve the following differential equation : x ( x - 1 ) d y d x - y = x 3 ( x - 1 ) 3

Given differential equation is x ( x - 1 ) d y d x - y = x 3 ( x - 1 ) 3 . So, d y d x - y x ( x - 1 ) = x 2 ( x - 1 ) 2 Which is a linear differential equation in y . On Comparing it with d y d x + P y = Q , We get P = - 1 x ( x - 1 ) , Q = x 2 ( x - 1 ) 2 IF = e ∫ P d x = e ∫ - 1 x ( x - 1 ) d x e ∫ 1 x - 1 x - 1 d x = e log x - log ( x - 1 ) = e log x x - 1 = x x - 1 So, the integrating factor, IF = x x - 1 The complete solution is y · IF = ∫ Q · IF d x ⇒ y · x x - 1 = ∫ x 2 x - 1 2 · x x - 1 d x ⇒ y x x - 1 = ∫ x 3 x - 1 d x ⇒ y x x - 1 = ∫ x 4 - x 3 d x ⇒ x y x - 1 = x 5 5 - x 4 4 + c Which is the required solution of given differential equation.

Solve the following differential equation : x + 2 y 3 d y d x = y

Given differential equation is x + 2 y 3 d y d x = y . So, d x d y = x + 2 y 3 y = x y + 2 y 2 d x d y - x y = 2 y 2 Which is a linear differential equation in x . On Comparing it with d x d y + P x = Q , We get P = - 1 y , Q = 2 y 2 IF = e ∫ P d y = e ∫ - 1 y d y = e - log y = e log 1 y So, the integrating factor, IF = 1 y The complete solution is x · IF = ∫ Q · IF d y ⇒ x · 1 y = ∫ 2 y 2 · 1 y d y ⇒ x y = ∫ 2 y d y ⇒ x y = y 2 + c Which is the required solution of given differential equation.

Solve the following differential equation : 1 - x 2 d y d x + 2 x y = x 1 - x 2

Given differential equation is 1 - x 2 d y d x + 2 x y = x 1 - x 2 . So, d y d x + 2 x 1 - x 2 y = x 1 - x 2 Which is a linear differential equation in y . On Comparing it with d y d x + P y = Q , We get P = 2 x 1 - x 2 , Q = x 1 - x 2 IF = e ∫ P d x = e ∫ 2 x 1 - x 2 d x = e - ∫ - 2 x 1 - x 2 d x = e - log 1 - x 2 ∵ ∫ f ' x f x d x = log f x + c = e log 1 1 - x 2 So, the integrating factor, IF = 1 1 - x 2 The complete solution is y · IF = ∫ Q · IF d x ⇒ y · 1 1 - x 2 = ∫ x 1 - x 2 1 1 - x 2 d x ⇒ y 1 1 - x 2 = ∫ x 1 - x 2 3 2 d x Put 1 - x 2 = t ⇒ - 2 x d x = d t So, y 1 1 - x 2 = ∫ - 1 2 t - 3 2 d t ⇒ y 1 1 - x 2 = - 1 2 · t - 1 2 - 1 2 + c = 1 t + c ⇒ y 1 1 - x 2 = 1 1 - x 2 + c Which is the required solution of given differential equation.

Solve the following differential equation : x ( x - 1 ) d y d x - ( x - 2 ) y = x 3 ( 2 x - 1 )

Given differential equation is x ( x - 1 ) d y d x - ( x - 2 ) y = x 3 ( 2 x - 1 ) . So, d y d x - x - 2 x ( x - 1 ) y = x 2 ( 2 x - 1 ) ( x - 1 ) Which is a linear differential equation in y . On Comparing it with d y d x + P y = Q , We get P = - x - 2 x ( x - 1 ) , Q = x 2 ( 2 x - 1 ) ( x - 1 ) IF = e ∫ P d x = e ∫ 2 - x x ( x - 1 ) d x Consider, 2 - x x ( x - 1 ) = A x + B x - 1 ⇒ 2 - x = A ( x - 1 ) + B x Now, x = 0 ⇒ 2 = - A ⇒ A = - 2 and x = 1 ⇒ B = 1 So, 2 - x x ( x - 1 ) = - 2 x + 1 x - 1 Integrating on both sides, we get ∫ 2 - x x ( x - 1 ) d x = - 2 ∫ d x x + ∫ d x x - 1 = - 2 log x + log ( x - 1 ) = log x - 1 x 2 So, IF = e ∫ 2 - x x ( x - 1 ) d x = e log x - 1 x 2 = x - 1 x 2 The integrating factor, IF = x - 1 x 2 . The complete solution is y · IF = ∫ Q · IF d x ⇒ y · x - 1 x 2 = ∫ x 2 ( 2 x - 1 ) ( x - 1 ) x - 1 x 2 d x ⇒ y x - 1 x 2 = ∫ ( 2 x - 1 ) d x ⇒ y x - 1 x 2 = x 2 - x + c Which is the required solution of given differential equation.

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

Given, d y d x x 2 y 3 + x y = 1 ⇒ d x d y = x y + x 2 y 3 ⇒ d x d y - x y = x 2 y 3 This equation is a bernoulli's equation. Dividing by x 2 , 1 x 2 d x d y - y x = y 3 Put z = - 1 x ⇒ d z d y = 1 x 2 d x d y ⇒ d z d y + z y = y 3 This is a linear equation in z . On Comparing it with d z d y + P z = Q , We get P = y , Q = y 3 IF = e ∫ P d y = e ∫ y d y = e y 2 2 The integrating factor, IF = e y 2 2 . The complete solution is z · IF = ∫ Q · IF d y ⇒ z · e y 2 2 = ∫ y 3 · e y 2 2 d y Consider ∫ y 3 · e y 2 2 d y Put y 2 2 = t ⇒ y d y = d t ∫ y 3 · e y 2 2 d y = ∫ 2 t e t d t = 2 t e t - ∫ e t d t [Integration by parts method] = 2 t e t - e t + c = 2 e t t - 1 + c = 2 e y 2 2 y 2 2 - 1 + c So, z · e y 2 2 = 2 e y 2 2 y 2 2 - 1 + c Replacing z = - 1 x ⇒ - 1 x e y 2 2 = 2 e y 2 2 y 2 2 - 1 + c ⇒ e y 2 2 y 2 + 1 x - 2 + c = 0 Which is the required solution of given differential equation.

Solve the following differential equation : d y d x + x sin 2 y = x 3 cos 2 y

Given, d y d x + x sin 2 y = x 3 cos 2 y ⇒ 1 cos 2 y d y d x + 2 sin y cos y cos 2 y x = x 3 ⇒ sec 2 y d y d x + x 2 tan y = x 3 This equation can be reducible to linear equation. Put z = tan y ⇒ d z d x = sec 2 y d y d x d z d x + 2 z x = x 3 This is a linear equation in z . On Comparing it with d z d x + P z = Q , We get P = 2 x , Q = x 3 IF = e ∫ P d x = e ∫ 2 x d x = e x 2 The integrating factor, IF = e x 2 . The complete solution is z · IF = ∫ Q · IF d x ⇒ z · e x 2 = ∫ x 3 · e x 2 d x Consider ∫ x 3 · e x 2 d x Put x 2 = t ⇒ 2 x d x = d t ∫ x 3 · e x 2 d x = 1 2 ∫ t e t d t = 1 2 t e t - ∫ e t d t [Integration by parts method] = 1 2 t e t - e t + c = 1 2 e t t - 1 + c = 1 2 e x 2 x 2 - 1 + c So, z e x 2 = 1 2 e x 2 x 2 - 1 + c Replacing z = tan y ⇒ tan y e x 2 = 1 2 e x 2 x 2 - 1 + c Which is the required solution of given differential equation.

Solve the following differential equation : y 2 + x - 1 y d y d x = 0

Given differential equation is y 2 + x - 1 y d y d x = 0 . So, x - 1 y d y d x = - y 2 ⇒ d x d y = x - 1 y - y 2 = - x y 2 + 1 y 3 ⇒ d x d y + 1 y 2 x = 1 y 3 Which is a linear differential equation in x . On Comparing it with d x d y + P x = Q , We get P = 1 y 2 , Q = 1 y 3 IF = e ∫ P d y = e ∫ 1 y 2 d y = e - 1 y So, the integrating factor, IF = e - 1 y The complete solution is x · IF = ∫ Q · IF d y ⇒ x · e - 1 y = ∫ 1 y 3 · e - 1 y d y Consider, ∫ 1 y 3 · e - 1 y d y Put - 1 y = z ⇒ 1 y 2 d y = d z ⇒ ∫ 1 y 3 · e - 1 y d y = - ∫ z e z d z [Integration by parts] = - z e z - ∫ e z d z = - z e z - e z + c = - e z z - 1 + c Replacing z = - 1 y = - e - 1 y - 1 y - 1 + c So, x e - 1 y = ∫ 1 y 3 e - 1 y d y = e - 1 y 1 + y y + c ⇒ x y = y + 1 + c y e 1 y Which is the required solution of given differential equation.

E M B I B E

Intermediate Second Year Mathematics Paper 2B>Calculus>Chapter 8 - Differential Equations>Exercise 8(e)

Telangana Board Solutions for Exercise 5: Exercise 8(e)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 5: Exercise 8(e)

Attempt the practice questions from Exercise 5: Exercise 8(e) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2B solutions are prepared by Experienced Embibe Experts.