Find the particular solutions of the differential equation d y d x = 2 x y , given that y = 1 when x = 0 .

Given that, d y d x = 2 x y ⇒ 1 y d y = 2 x d x On integrating, we have, log y = x 2 + c . . . 1 Given y = 1 when x = 0 ⇒ log 1 = 0 + c [using 1 ] ⇒ c = 0 ∴ from 1 gives , log y = x 2 y = e x 2 ∵ y > 0 Which gives the required particular solution.

Find the particular solution of the differential equation 2 x 2 d y d x - 2 x y + y 2 = 0 , given that y = e when x = e .

Given that, 2 x 2 d y d x - 2 x y + y 2 = 0 d y d x = 2 x y - y 2 2 x 2 . . . ( 1 ) Also , dy dx = y x - 1 2 y x 2 = ϕ y x ∴ ϕ x , y is homogeneous function of degree 0 Put, y = v x ⇒ d y d x = v + x d v d x From equation ( 1 ) we have, v + x d v d x = 2 v x 2 - v 2 x 2 2 x 2 = 2 v - v 2 2 x d v d x = 2 v - v 2 2 - v = 2 v - v 2 - 2 v 2 = - v 2 2 2 v 2 d v = - d x x Integrating on both sides, we get, ∫ 2 v 2 d v = - ∫ d x x + log c - 2 v + log x = C - 2 x y + log x = C . . . 2 When x = e , y = e ∴ from 2 , - 2 e e + log e = C - 2 + 1 = C ⇒ C = - 1 Thus, equation 2 becomes, - 2 x y + log x = - 1 ⇒ 1 + log x = 2 x y y = 2 x 1 + log x Which is the required solution.

Find the particular solutions of the following differential equations ( x 2 + 1 ) d y d x - 2 x y = ( x 4 + 2 x 2 + 1 ) cos x , given that y = 0 when x = 0 .

Given differential equation is, ( x 2 + 1 ) d y d x - 2 x y = ( x 4 + 2 x 2 + 1 ) cos x d y d x - 2 x 1 + x 2 y = x 4 + 2 x 2 + 1 x 2 + 1 cos x Which is linear differential equation of first order in y and is of the form, d y d x + P y = Q Where P = - 2 x 1 + x 2 and Q = x 4 + 2 x 2 + 1 x 2 + 1 cos x ∴ I . F = e ∫ P d y = e ∫ - 2 x d x 1 + x 2 = e - log 1 + x 2 = 1 1 + x 2 And solution is given by, y × I . F = ∫ Q × I . F d x + c y 1 1 + x 2 = ∫ x 2 + 1 x 2 + 1 cos x = ∫ cos x d x + c y 1 + x 2 = sin x + c . . . 1 Given y = 0 when x = 0 ∴ from 1 , we get , 0 = 0 + c ⇒ c = 0 ∴ from 1 , we have , y 1 + x 2 = sinx y = 1 + x 2 sin x Which gives the required solution.

Find the particular solution of the differential equation 1 - y 2 d x = ( sin - 1 y - x ) d y , given that y = 0 when x = 0 .

Given differential equation is, 1 - y 2 d x = ( sin - 1 y - x ) d y d x d y + x 1 - y 2 = sin - 1 y 1 - y 2 Which is linear differential equation of first order in x and is of the form, d x d y + P x = Q Where P = 1 1 - y 2 and Q = sin - 1 y 1 - y 2 ∴ I . F = e ∫ P d y = e ∫ 1 1 - y 2 d y = e sin - 1 y And solution is given by, x e ∫ P d y = ∫ Q e ∫ P d y d y + c x e sin - 1 y = ∫ sin - 1 y 1 - y 2 e sin - 1 y d y + c Put, sin - 1 y = t ⇒ 1 1 - y 2 d y = d t x e sin - 1 y = ∫ t e t d t + c = t - 1 e t + c x e sin - 1 y = sin - 1 y - 1 e sin - 1 y + c . . . 1 ∴ from 1 , 0 = 0 - 1 + c ⇒ c = 1 ∴ from 1 , we get , x e sin - 1 y = sin - 1 y - 1 e sin - 1 y + 1 x = sin - 1 y - 1 + e - sin - 1 y Which gives the required solution.

Find the particular solution of the differential equation x e y x - y sin y x + x sin y x d y d x = 0 , given that y = 0 when x = 1 .

Given that, d y d x = y sin y x - x e y x x sin y x d y d x = y x sin y x - e y x sin y x = ϕ y x . . . 1 Thus equation 1 be a homogeneous differential equation Put y = v x ⇒ d y d x = v + x d v d x ∴ from 1 , v + x dv dx = vsinv - e v sinv x d v d x = v sin v - e v sin v - v x d v d x = v sin v - e v - v sin v sin v x d v d x = - e v sin v ∫ sin v e - v d v = - ∫ d x x + c . . . 2 Let I = ∫ sin v e - v d v = - sin v e - v + ∫ cos v e - v d v ∴ I = - sin v e - v + - cos v e - v - ∫ sin v e - v d v 2 I = - sin v + cos v e - v ∴ from 2 , we get , - sin v + cos v e - v 2 = - log x + c e - y x sin y x + cos y x = 2 log x + A . . . 3 Where, A = - 2 c When x = 1 , y = 0 ∴ from 3 , we have , 1 0 + 1 = 2 × 0 + A ⇒ A = 1 Thus equation 3 becomes, e - y x sin y x + cos y x = 2 log x + 1 Which is the required solution.

E M B I B E

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST

M L Aggarwal Solutions for Chapter: Differential Equations, Exercise 9: CHAPTER TEST

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Differential Equations, Exercise 9: CHAPTER TEST

Attempt the free practice questions on Chapter 9: Differential Equations, Exercise 9: CHAPTER TEST with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Differential Equations, Exercise 9: CHAPTER TEST with Hints & Solutions

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 14

In a bank principal increases at the rate of r% per year. Find the value of r if Rs. $100$ double itself in $10$ years.
(given $\log_{e} 2 = 0.6931$ )

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 17

Solve the differential equation $(x + \tan y) d y = \sin 2 y d x .$

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 18

Find the particular solutions of the differential equation
$\frac{d y}{d x} = 2 x y$ , given that $y = 1$ when $x = 0$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 18

Find the particular solution of the differential equation
$2 x^{2} \frac{d y}{d x} - 2 x y + y^{2} = 0$ , given that $y = e$ when $x = e$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 18

Find the particular solution of the differential equation
$(x y - x^{2}) d x - x^{2} d y = 0$ , given that $y = 1$ when $x = 1$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 18

Find the particular solutions of the following differential equations
$(x^{2} + 1) \frac{d y}{d x} - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$ , given that $y = 0$ when $x = 0$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 18

Find the particular solution of the differential equation
$\sqrt{1 - y^{2}} d x = (\sin^{- 1} y - x) d y$ , given that $y = 0$ when $x = 0$ .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 9 - Differential Equations>CHAPTER TEST>Q 18

Find the particular solution of the differential equation
$x e^{\frac{y}{x}} - y \sin (\frac{y}{x}) + x \sin (\frac{y}{x}) \frac{d y}{d x} = 0$ , given that $y = 0$ when $x = 1$ .

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