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

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

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

Given differential equation is x log x d y d x + y = 2 log x . So, d y d x + y x log x = 2 x Which is a linear differential equation in y . On Comparing it with d y d x + P y = Q , We get P = 1 x log x , Q = 2 x IF = e ∫ P d x = e ∫ 1 x log x d x = e log log x ∵ ∫ f ' x f x d x = log f x + c = log x So, the integrating factor, IF = log x The complete solution is y · IF = ∫ Q · IF d x ⇒ y · log x = ∫ 2 x log x d x ⇒ y log x = 2 ∫ log x · 1 x d x ⇒ y log x = 2 log x 2 2 + c ∵ ∫ f x n f ' x d x = f x n + 1 n + 1 + c ⇒ y log x = log x 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 = cos x

Given differential equation is 1 + x 2 d y d x + 2 x y = cos x . So, d y d x + 2 x 1 + x 2 y = cos 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 = cos x 1 + x 2 IF = e ∫ P 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 = 1 + x 2 So, the integrating factor, IF = 1 + x 2 The complete solution is y · IF = ∫ Q · IF d x ⇒ y · 1 + x 2 = ∫ cos x 1 + x 2 1 + x 2 d x ⇒ y 1 + x 2 = ∫ cos x d x ⇒ y 1 + x 2 = sin x + 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 = 4 x 2

Given differential equation is 1 + x 2 d y d x + 2 x y = 4 x 2 . So, d y d x + 2 x 1 + x 2 y = 4 x 2 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 = 4 x 2 1 + x 2 IF = e ∫ P 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 = 1 + x 2 So, the integrating factor, IF = 1 + x 2 The complete solution is y · IF = ∫ Q · IF d x ⇒ y · 1 + x 2 = ∫ 4 x 2 1 + x 2 1 + x 2 d x ⇒ y 1 + x 2 = ∫ 4 x 2 d x ⇒ y 1 + x 2 = 4 x 3 3 + c Which is the required solution of given differential equation.

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

Given differential equation is x 2 - 1 d y d x + 2 x y = 1 x 2 - 1 . So, d y d x + 2 x x 2 - 1 y = 1 x 2 - 1 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 x 2 - 1 , Q = 1 x 2 - 1 2 IF = e ∫ P d x = e ∫ 2 x x 2 - 1 d x = e log x 2 - 1 ∵ ∫ f ' x f x d x = log f x + c = x 2 - 1 The complete solution is y · IF = ∫ Q · IF d x ⇒ y · x 2 - 1 = ∫ 1 x 2 - 1 2 x 2 - 1 d x ⇒ y x 2 - 1 = ∫ 1 x 2 - 1 d x ⇒ y x 2 - 1 = 1 2 log x - 1 x + 1 + c ∵ ∫ 1 x 2 - a 2 d x = 1 2 a log x - a x + a + c Which is the required solution of given differential equation.

Solve the following differential equation : d y d x - y tan x = e x sec x

Given differential equation is d y d x - y tan x = e x sec 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 = e x sec x IF = e ∫ P d x = e - ∫ tan x d x = e - ∫ sin x cos x d x = e log cos x = cos x So, the integrating factor, IF = cos x . The complete solution is y · IF = ∫ Q · IF d x ⇒ y · cos x = ∫ e x sec x · cos x d x ⇒ y cos x = ∫ e x d x ⇒ y cos x = e x + k Which is the required solution of given differential equation.

Solve the differential equation : d y d x + y cot x = x 2 cot x + 2 x

The given equation is d y d x + y cot x = x 2 cot x + 2 x This is a linear differential equation and comparing it with d y d x + P y = Q , we get P = cot x , Q = x 2 cot x + 2 x Integrating factor, IF = e ∫ P d x = e ∫ cot x d x = e log sin x ⇒ IF = sin x The solution is given by y · IF = ∫ Q · IF d x ⇒ y sin x = ∫ x 2 cot x + 2 x sin x d x ⇒ y sin x = ∫ x 2 cos x d x + 2 ∫ x sin x d x [ ∵ cot x = cos x sin x ] Let, I = ∫ x 2 cos x d x & J = ∫ x sin x d x ⇒ y sin x = I + 2 J ....(1) Using integration by parts for I , we get ⇒ y sin x = x 2 ∫ cos x d x + ∫ d d x x 2 ∫ cos x d x d x ⇒ I = x 2 sin x - ∫ 2 x sin x d x + c ⇒ I = x 2 sin x - 2 J + c ⇒ I + 2 J = x 2 sin x + c Eq(1) becomes, y sin x = x 2 sin x + c Which is the required solution.

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

Given differential equation is 1 + x 2 d y d x + y = e tan - 1 x . So, d y d x + 1 1 + x 2 y = e tan - 1 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 = 1 1 + x 2 , Q = e tan - 1 x 1 + x 2 IF = e ∫ P d x = e ∫ 1 1 + x 2 d x = e tan - 1 x So, the integrating factor, IF = e tan - 1 x The complete solution is y · IF = ∫ Q · IF d x ⇒ y · e tan - 1 x = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x ⇒ y e tan - 1 x = ∫ e tan - 1 x 2 1 + x 2 d x Put tan - 1 x = t ⇒ d x 1 + x 2 = d t ⇒ y e tan - 1 x = ∫ e 2 t d t ⇒ y e tan - 1 x = e 2 t 2 + k ⇒ y e tan - 1 x = e 2 tan - 1 x 2 + k Which is the required solution of given differential equation.

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

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