Solve the differential equation x d y d x + y = x e x

We have, x d y d x + y = x e x The given equation can be written as d y d x + 1 x y = e x This is a linear differential equation and comparing it with d y d x + P y = Q we get, P = 1 x , Q = e x The integrating factor is I F = e ∫ P d x = e ∫ 1 x d x = e log x = x The solution is given by, y × I F = ∫ Q × I F d x + C where C is an arbitrary constant. Hence we get, y x = ∫ x e x d x + C ⇒ y x = x ∫ e x d x - ∫ d ( x ) d x ∫ e x d x d x + C ⇒ y x = x e x - ∫ e x d x + C ⇒ y x = x e x - e x + C ⇒ y = ( x - 1 ) e x + C x is the required solution.

Solve the differential equation x d y d x + y = x log x

The given equation can be written as d y d x + y x = log x This is a linear differential equation and comparing it with d y d x + P y = Q we get, P = 1 x , Q = log x The integrating factor is I F = e ∫ P d x = e ∫ 1 x d x = e log x = x The solution is given by, y × I F = ∫ Q × I F d x + C where C is an arbitrary constant. Hence we get, x y = ∫ x log x + C ⇒ x y = log x ∫ x d x - ∫ d log x d x ∫ x d x d x + C ⇒ x y = log x × x 2 2 - ∫ 1 x × x 2 2 d x + C ⇒ x y = log x × x 2 2 - x 2 4 + C ⇒ 4 x y = 2 x 2 log x - x 2 + C 1 where C 1 = 4 C is the required solution.

Solve the differential equation d y d x + y tan x = x 2 cos 2 x

The given equation is d y d x + y tan x = x 2 cos 2 x This is a linear differential equation and comparing it with d y d x + P y = Q we get, P = tan x , Q = x 2 cos 2 x The integrating factor is I F = e ∫ P d x = e ∫ tan x d x = e log ( sec x ) = sec x The solution is given by, y × I F = ∫ Q × I F d x + C where C is an arbitrary constant. Hence we get, y sec x = ∫ x 2 cos 2 x sec x d x + C ⇒ y sec x = ∫ x 2 cos x d x + C ⇒ y sec x = x 2 ∫ cos x d x - ∫ d d x x 2 ∫ cos x d x d x + C ⇒ y sec x = x 2 sin x - ∫ 2 x sin x d x + C ⇒ y sec x = x 2 sin x - 2 x ∫ sin x d x - ∫ d x d x ∫ sin x d x d x + C ⇒ y sec x = x 2 sin x - 2 - x cos x + ∫ cos x d x + C ⇒ y sec x = x 2 sin x - 2 - x cos x + sin x + C ⇒ y = x 2 sin x cos x + 2 x cos 2 x - 2 sin x cos x + C cos x is the required solution.

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

Given that x d y d x + 2 y = x 2 log x ⇒ d y d x + 2 y x = x log x This is a linear differential equation in the form of d y d x + P y = Q , where P = 2 x , Q = x log x I . F . = e ∫ P d x = e ∫ 2 x d x = e 2 log x = x 2 So, the solution is given by, y × I . F . = ∫ Q × I . F . d x + C ⇒ x 2 y = ∫ x 3 log x d x + C ⇒ x 2 y = log x ∫ x 3 d x - ∫ d d x log x ∫ x 3 d x d x + C ⇒ x 2 y = x 4 log x 4 - ∫ x 3 4 d x + C ⇒ x 2 y = x 4 log x 4 - x 4 16 + C ⇒ y = x 2 log x 4 - x 2 16 + C x 2 ⇒ y = x 2 16 4 log x - 1 + C x 2 is the general solution of the given differential equation.

Solve the following initial value problem:- x d y d x - y = x + 1 e - x , y 1 = 0

Given that x d y d x - y = x + 1 e - x ⇒ d y d x - 1 x y = x + 1 x e - x This is a linear differential equation in the form of d y d x + P y = Q , where P = - 1 x and Q = x + 1 x e - x ∴ I . F . = e ∫ P d x = e - ∫ 1 x d x = e - log x = 1 x Solution of the linear differential equation is given by y × ( I . F ) = ∫ Q × ( I . F ) d x + C ⇒ 1 x × y = ∫ 1 x + 1 x 2 e - x d x + C Let 1 x e - x = t ⇒ - 1 x e - x - 1 x 2 e - x d x = d t ⇒ 1 x + 1 x 2 e - x d x = - d t ∴ 1 x y = ∫ - d t + C ⇒ y x = - t + C ⇒ y x = - e - x x + C ⇒ y = - e - x + C x Given that y 1 = 0 ∴ 0 = - e - 1 + C ⇒ C = e - 1 ∴ y = - e - x + x e - 1 ⇒ y = x e - 1 - e - x Hence, y = x e - 1 - e - x is the required solution.

Solve the following initial value problem:- d y d x + 2 y tan x = sin x ; y = 0 when x = π 3

Given that d y d x + 2 y tan x = sin x This is a linear differential equation in the form of d y d x + P y = Q , where P = 2 tan x and Q = sin x ∴ I . F . = e ∫ P d x = e 2 ∫ tan x d x = e 2 log sec x = sec 2 x Solution of the linear differential equation is given by y × ( I . F ) = ∫ Q × ( I . F ) d x + C ⇒ y sec 2 x = ∫ tan x sec x d x + C ⇒ y sec 2 x = sec x + C Given that y π 3 = 0 ∴ 0 sec π 3 2 = sec π 3 + C ⇒ C = - 2 ∴ y sec 2 x = sec x - 2 ⇒ y = cos x - 2 cos 2 x is the required solution.

E M B I B E

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE

R D Sharma Solutions for Chapter: Differential Equations, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Differential Equations, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 3: Differential Equations, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-2 solutions are prepared by Experienced Embibe Experts.