Solve the differential equation 1 + x 2 d y d x + y = e tan - 1 x , given that y = 1 when x = 0 .

We have, 1 + x 2 d y d x + y = e tan - 1 x ⇒ d y d x + y 1 + x 2 = e tan - 1 x 1 + x 2 Comparing it with standard equation, d y d x + P y = Q , where P = 1 1 + x 2 and Q = e tan - 1 x 1 + x 2 Now, integrating factor I.F = e ∫ P d x = e ∫ 1 1 + x 2 d x = e tan - 1 x So, the solution is given by: y I . F = ∫ Q x I . F d x + c ⇒ y e tan - 1 x = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x + c . . . 1 Let I = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x Putting e tan - 1 x = t ⇒ 1 1 + x 2 · e tan - 1 x d x = d t Thus, I = ∫ t d t = t 2 2 + c ⇒ I = e tan - 1 x 2 2 + c Considering 1 , we get: y e tan - 1 x = e tan - 1 x 2 2 + c It is given that y = 1 , x = 0 ⇒ 1 · e tan - 1 0 = e tan - 1 0 2 2 + c ⇒ 1 · e tan - 1 tan 0 = e tan - 1 tan 0 2 2 + c ⇒ 1 · e 0 = e 0 2 2 + c ∵ tan - 1 tan A = A ⇒ 1 = 1 4 + c ⇒ c = 3 4 Thus, the solution is y e tan - 1 x = e tan - 1 x 2 2 + 3 4 .

Find the integrating factor of the linear equation d y d x + P y = Q and hence obtain the general solution of the equation.

A differential equation is linear if the dependent variable y and its derivative appear only in first degree. The general form of a linear differential equation of first order is d y d x + P y = Q . . . 1 where P and Q are functions of x (or constants) This type of differential equations are solved when they are multiplied by a factor, which is called the integrating factor, because by multiplication of this factor the LHS of differential equation 1 becomes exact differential of some function. Multiplying both sides of 1 by e ∫ P d x , we get e ∫ P d x d y d x + P y = Q e ∫ P d x ⇒ d d x y e ∫ P d x = Q e ∫ P d x On integrating both sides w.r.t x , we get y e ∫ P d x = ∫ Q e ∫ P d x + C . . . 2 This is the required solution, where C is the constant of integration. Here, e ∫ P d x is called the integrating factor. The solution 2 in short may also be written as y · I . F . = ∫ Q · I . F . d x + C

Show that the general solution of the differential equation d y d x + y 2 + y + 1 x 2 + x + 1 = 0 is ( x + y + 1 ) = A ( 1 − x − y − 2 x y ) , where A is an arbitrary constant.

Given differential equation is d y d x + y 2 + y + 1 x 2 + x + 1 = 0 ⇒ d y d x = - y 2 + y + 1 x 2 + x + 1 ⇒ d y y 2 + y + 1 = - d x x 2 + x + 1 ⇒ d y y 2 + y + 1 + d x x 2 + x + 1 = 0 Integrating both sides, we get: ∫ d y y 2 + y + 1 + ∫ d x x 2 + x + 1 = C ⇒ ∫ d y y + 1 2 2 + 3 2 2 + ∫ d x x + 1 2 2 + 3 2 2 = C ⇒ 2 3 t a n - 1 y + 1 2 3 2 + 2 3 t a n - 1 x + 1 2 3 2 = C ⇒ t a n - 1 2 y + 1 3 + t a n - 1 2 x + 1 3 = 3 C 2 ⇒ t a n - 1 2 y + 1 3 + 2 x + 1 3 1 - 2 y + 1 3 · 2 x + 1 3 = 3 C 2 ⇒ t a n - 1 2 x + 2 y + 2 3 1 - 4 x y + 2 x + 2 y + 1 3 = 3 C 2 ⇒ t a n - 1 2 3 x + y + 1 3 - 4 x y - 2 x - 2 y - 1 = 3 C 2 ⇒ t a n - 1 3 x + y + 1 2 1 - x - y - 2 x y = 3 C 2 ⇒ 3 x + y + 1 2 1 - x - y - 2 x y = tan ⁡ 3 C 2 = B , where B = tan ⁡ 3 C 2 ⇒ x + y + 1 = 2 B 3 1 - x - y - 2 x y ⇒ x + y + 1 = A 1 - x - y - 2 x y , where A = 2 B 3 Hence proved

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2019

Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: Manipur Board-2018

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: Manipur Board-2018

Attempt the free practice questions on Chapter 9: Differential Equations, Exercise 1: Manipur 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.

Questions from Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: Manipur Board-2018 with Hints & Solutions

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2018>Q 2

Write the steps for solving a homogeneous differential equation.

HARD

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2018>Q 3

Solve the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$ , given that $y = 1$ when $x = 0$ .

EASY

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2019>Q 4

What is meant by the general solution of a differential equation?

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2019>Q 5

Find the differential equation of the family of curves $y = e^{x} (A \cos x + B \sin x)$ , where $A$ and $B$ are arbitrary constants.

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2019>Q 6

Find the integrating factor of the linear equation $\frac{d y}{d x} + P y = Q$ and hence obtain the general solution of the equation.

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2020>Q 8

What is meant by a linear differential equation of the first order?

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2020>Q 9

What is meant by a homogeneous differential equation? Describe how it can be reduced to a form in which the variables are separable.

HARD

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Differential Equations>Manipur Board-2020>Q 10

Show that the general solution of the differential equation $\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$ is $(x + y + 1) = A (1 - x - y - 2 x y)$ , where $A$ is an arbitrary constant.

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