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

We have, d y d x = tan - 1 x ⇒ d y = ( tan - 1 x ) d x Integrating on both sides, we get ⇒ ∫ d y = ∫ tan - 1 x d x ⇒ y = ∫ 1 × tan - 1 x d x ⇒ y = tan - 1 x ∫ 1 d x - ∫ d d x tan - 1 x ∫ 1 d x d x ⇒ y = x tan - 1 x - ∫ x 1 + x 2 d x ⇒ y = x tan - 1 x - 1 2 ∫ 2 x 1 + x 2 d x ⇒ y = x tan - 1 x - 1 2 log 1 + x 2 + C Hence, y = x tan - 1 x - 1 2 log 1 + x 2 + C is the solution to the given differential equation.

Solve the following differential equation d y d x - x sin 2 x = 1 x log x

Given d y d x - x sin 2 x = 1 x log x ⇒ d y d x = 1 x log x + x sin 2 x ⇒ d y = 1 x log x + x sin 2 x d x Integrating on both sides, we get ⇒ ∫ d y = ∫ 1 x log x d x + ∫ x sin 2 x d x ⇒ y = I 1 + l 2 . . . . . . . . . . . ( i ) ⇒ I 1 = ∫ 1 x log x d x Let log x = t Differentiate it, ⇒ 1 x d x = d t Now, I 1 = ∫ d t t = log | t | + C 1 ⇒ I 1 = log | log x | + C 1 And, I 2 = ∫ x sin 2 x d x = ∫ x ( 1 - cos 2 x ) 2 d x = 1 2 ∫ ( x - x cos 2 x ) d x = 1 2 ∫ x d x - 1 2 ∫ x cos 2 x d x = 1 2 x 2 2 - 1 2 x ∫ cos 2 x d x - ∫ 1 × ∫ cos 2 x d x d x + C 2 = x 2 4 - 1 2 x sin 2 x 2 - ∫ sin 2 x 2 d x + C 2 = x 2 4 - 1 2 x sin 2 x 2 + cos 2 x 4 + C 2 I 2 = x 2 4 - x sin 2 x 4 - cos 2 x 8 + C 2 Put the value of I 1 and I 2 in equation ( i ) , ⇒ y = I 1 + l 2 ⇒ y = log log x + C 1 + x 2 4 - x sin 2 x 4 - cos 2 x 8 + C 2 ⇒ y = log log x + x 2 4 - x sin 2 x 4 - cos 2 x 8 + C

Solve the following differential equation: 1 - x 4 d y = x d x

Given 1 - x 4 d y = x d x ⇒ d y = x d x 1 - x 4 Integrating on both sides, we get ⇒ ∫ d y = ∫ x d x 1 - x 4 Let x 2 = t Hence 2 x d x = d t ⇒ x d x = d t 2 ⇒ ∫ d y = ∫ d t 2 1 - t 2 ⇒ y = 1 2 sin - 1 ( t ) + C ⇒ y = 1 2 sin - 1 ( x 2 ) + C

Solve the following initial value problem: sin d y d x = k ; y ( 0 ) = 1

Given: sin d y d x = k ; y ( 0 ) = 1 ⇒ d y d x = sin - 1 k ⇒ d y = sin - 1 k d x ⇒ ∫ d y = ∫ sin - 1 k d x ⇒ y = x sin - 1 k + c ......(i) Given that y 0 = 1 , it means at x = 0 , y = 1 Put x = 0 , y = 1 in equation (i), 1 = 0 + c ⇒ c = 1 Put c = 1 in equation (i), y = x sin - 1 k + 1 ⇒ y - 1 = x sin - 1 k

Solve the following initial value problem: e d y d x = x + 1 ; y 0 = 3 .

We have, e d y d x = x + 1 Taking log e on both sides, we get d y d x log e e = log e x + 1 ⇒ d y d x = log e x + 1 ⇒ d y = log e x + 1 d x Integrating both sides, we get ∫ d y = ∫ log e x + 1 d x ⇒ y = ∫ 1 × log e ( x + 1 ) d x ⇒ y = log e x + 1 ∫ 1 d x - ∫ d d x log e x + 1 ∫ 1 d x d x ⇒ y = x log e x + 1 - ∫ x x + 1 d x ⇒ y = x log e x + 1 - ∫ 1 - 1 x + 1 d x ⇒ y = x log e x + 1 - x + log e x + 1 + C ....(1) It is given that y 0 = 3 , ∴ 3 = 0 × log e 0 + 1 - 0 + log e 0 + 1 + C ⇒ C = 3 Substituting the value of C in equation (1) we get y = x log e x + 1 - x + log e x + 1 + 3 Hence y = x + 1 log e x + 1 - x + 3 is the solution to the given differential equation.

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

We have, x d y d x + 1 = 0 ⇒ d y d x = - 1 x ⇒ d y = - 1 x d x Integrating both sides, we get ⇒ ∫ d y = ∫ - 1 x d x ⇒ y = - log e x + C .......(1) It is given that y - 1 = 0 . ∴ 0 = - log e - 1 + C ⇒ C = 0 Substituting the value of C in (1), we get y = - log e x Hence, y = - log e x is the solution to the given differential equation.

Solve the following initial value problem: x x 2 - 1 d y d x = 1 ; y 2 = 0 .

Given: x x 2 - 1 d y d x = 1 ⇒ d y = d x x x 2 - 1 ⇒ d y = d x x x - 1 x + 1 Integrate both sides, ⇒ ∫ d y = ∫ d x x x - 1 x + 1 Let 1 x x - 1 x + 1 = A x + B x - 1 + C x + 1 1 = A x - 1 x + 1 + B x x + 1 + C x x - 1 . . . . . . . i Put x = 0 in equation i , 1 = A - 1 1 ⇒ A = - 1 Put x = 1 in equation i , 1 = 2 B ⇒ B = 1 2 Put x = - 1 in equation i , 1 = 2 C ⇒ C = 1 2 Now, ∫ d y = ∫ d x x x - 1 x + 1 ⇒ ∫ d y = ∫ - 1 x + 1 2 x - 1 + 1 2 x + 1 d x ⇒ y = - ln x + 1 2 ln x - 1 + 1 2 ln x + 1 + c ⇒ y = - 1 2 ln x 2 + 1 2 ln x 2 - 1 + c ⇒ y = 1 2 ln x 2 - 1 x 2 + c . . . . . . . . . ii Given that: y 2 = 0 Applying boundary condition in equation ii , 0 = 1 2 ln 2 2 - 1 2 2 + c ⇒ 0 = 1 2 ln 3 4 + c ⇒ c = - 1 2 ln 3 4 Now, y = 1 2 ln x 2 - 1 x 2 - 1 2 ln 3 4 is the required solution of given differential equation.

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 5: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Differential Equations, Exercise 5: EXERCISE

Attempt the free practice questions on Chapter 3: Differential Equations, Exercise 5: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-2 solutions are prepared by Experienced Embibe Experts.