If p and q are respectively degree and order of the differential equation y = e d y d x , then write the relation between p and q .

Given differential equation is y = e d y d x Take log on both sides, then log y = log e d y d x ⇒ log y = d y d x log e ⇒ log y = d y d x ⇒ d y d x - log y = 0 The order of a differential equation is the order of the highest derivative involved in the differential equation. So, the order of the given differential equation is 1 , as we have d y d x is in the equation. The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions. The power of d y d x is 1 . So, the degree of the differential equation is 1 . Hence, p = 1 and q = 1 Thus, the relation between p and q is p = q .

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

Given differential equation is y 2 + x 2 d y d x = x y d y d x ⇒ y 2 = x y d y d x - x 2 d y d x ⇒ y 2 = d y d x x y - x 2 ⇒ y 2 x y - x 2 = d y d x → 1 This is a homogenous differential equation, because the numerator and denominator degree are having same degree. So it can be solved by using y = v x → 2 Now, differentiate y with respect to x , then d y d x = v d d x x + x d d x v = v + x d v d x → 3 Substitute 3 and 2 in 1 , then we get: v + x d v d x = v x 2 x v x - x 2 ⇒ v + x d v d x = v 2 v - 1 ⇒ x d v d x = v 2 v - 1 - v ⇒ x d v d x = v 2 - v 2 + v v - 1 ⇒ x d v d x = v v - 1 ⇒ v - 1 v d v = d x x ⇒ 1 - 1 v d v = d x x Applying integration on both sides, then ∫ 1 - 1 v d v = ∫ d x x ⇒ v - log v = log x + log c ⇒ y x = log y x + log c x ⇒ y x = log y x · c x ⇒ y x = log y c ⇒ y c = e y x ⇒ y = K e y x , which is the required general solution, where K = 1 c

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Differential Equations>Odisha Board-2018

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

Author:Embibe Experts

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

Attempt the free practice questions on Chapter 11: Differential Equations, Exercise 1: Odisha 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: Odisha Board-2018 with Hints & Solutions

MEDIUM

12th Odisha Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Differential Equations>Odisha Board-2018>Q 1

If $p and q$ are respectively degree and order of the differential equation $y = e^{\frac{d y}{d x}}$ , then write the relation between $p and q$ .

MEDIUM

12th Odisha Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Differential Equations>Odisha Board-2018>Q 2

Form the differential equation whose general solution is $y = a \sin t + b e^{t}$ .

HARD

12th Odisha Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Differential Equations>Odisha Board-2018>Q 3

Solve the following differential equation:

$(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d y = 0$ .

HARD

12th Odisha Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Differential Equations>Odisha Board-2018>Q 4

Solve: $(4 x + 6 y + 5) d x - (2 x + 3 y + 4) d y = 0$ .

EASY

12th Odisha Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Differential Equations>Odisha Board-2019>Q 5

Write the order of the differential equation whose solution is given by $y = (c_{1} + c_{2}) \cos (x + c_{3}) + c_{4} e^{x + c_{5}}$ , where $c_{1}, c_{2}, c_{3}, c_{4} and c_{5}$ are arbitrary constants.