Evaluate 1 + x x 2 + y 2 d x + x 2 + y 2 - 1 y d y = 0 .

Given, 1 + x x 2 + y 2 d x + x 2 + y 2 - 1 y d y = 0 d x + x x 2 + y 2 d x + y x 2 + y 2 d y - y d y = 0 By rearranging the equation we get, d x - y d y + x x 2 + y 2 d x + y x 2 + y 2 d y = 0 d x - y d y + x 2 + y 2 x d x + y d y = 0 d x - y d y + 1 2 x 2 + y 2 d ( x 2 + y 2 ) = 0 By integrating both the sides of the equation we get, ∫ d x - ∫ y d y + 1 2 ∫ x 2 + y 2 d ( x 2 + y 2 ) = ∫ 0 x - y 2 2 + 1 2 · 2 3 ( x 2 + y 2 ) 3 2 = C The required solution is x - y 2 2 + 1 3 ( x 2 + y 2 ) 3 2 + C .

Solve d 2 y d x 2 = x .

Given, d 2 y d x 2 = x Integrating the equation with respect to x we get, ∫ d 2 y d x 2 = ∫ x d x d y d x = x 2 2 + C 1 Again integrating the equation with respect to x we get, ∫ d y d x = ∫ x 2 2 + ∫ C 1 y = x 3 6 + C 1 x + C 2

Evaluate d 2 y d x 2 = sin 2 x + x e x .

Given, d 2 y d x 2 = sin 2 x + x e x d 2 y d x 2 = 1 - cos 2 x 2 + x e x ( ∵ 1 - cos 2 x = 2 sin 2 x ) Integrating both the sides of the equation with respect to x we get, ∫ d 2 y d x 2 = ∫ 1 - cos 2 x 2 d x + ∫ x e x d x d y d x = 1 2 x - sin 2 x 2 + ∫ x e x d x d y d x = 1 2 x - 1 4 sin 2 x + x e x - e x + C 1 Integrating both the sides of the equation again with respect to x we get, y = 1 2 · x 2 2 + 1 4 · cos 2 x 2 + ∫ x e x d x - e x + C 1 x + C 2 y = x 2 4 + cos 2 x 8 + x e x - e x - e x + C 1 x + C 2 y = 1 8 2 x 2 + cos 2 x + x e x - 2 e x + C 1 x + C 2 Hence, this is the required equation.

Solve the differential equation d 2 y d x 2 = 6 x + sec 2 x .

Given, that d 2 y d x 2 = 6 x + sec 2 x Integrating both the sides of the equation with respect to x , we get ∫ d 2 y d x 2 = 6 ∫ x d x + ∫ sec 2 x d x d y d x = 6 · x 2 2 + tan x + C 1 Integrating both the sides of the equation again with respect to x , we get ∫ d y d x = 6 ∫ x 2 2 d x + ∫ tan x d x + ∫ C 1 d x y = 3 · x 3 3 + log sec x + C 1 x + C 2 y = x 3 + log sec x + C 1 x + C 2 Hence, this is the required solution.

Solve the differential equation d 2 y d x 2 = cos x - sin x .

Given, d 2 y d x 2 = cos x - sin x Integrating both the sides of the equation with respect to x we get, ∫ d 2 y d x 2 = ∫ cos x - ∫ sin x d y d x = sin x + cos x + C 1 Integrating both the sides of the equation again with respect to x we get, ∫ d y d x = ∫ sin x + ∫ cos x + ∫ C 1 y = - cos x + sin x + C 1 x + C 2 Hence, this is the required solution.

Solve the differential equation d 2 y d x 2 = log x . Given, y = 1 and d y d x = - 1 , when x = 1 .

Given, d 2 y d x 2 = log x Integrating both the sides of the equation with respect to x we get, ∫ d 2 y d x 2 = ∫ log x d y d x = ∫ log x · 1 d x d y d x = x log x - ∫ 1 x · x d x + C 1 d y d x = x log x - x + C 1 ... ( i ) Given, When x = 1 , d y d x = - 1 . Therefore, from ( i ) we get, - 1 = 1 × log 1 - 1 + C 1 C 1 = 0 Substituting the value of C 1 in equation ( i ) we get, d y d x = x log x - x ... ( i i ) Integrating both the sides of the equation again with respect to x , we get, ∫ d y d x d x = ∫ ( x log x - x ) d x y = ∫ x log x d x - x 2 2 + C 2 y = log x · ∫ x d x - ∫ d d x log x · ∫ x d x d x - x 2 2 + C 2 y = x 2 2 log x - ∫ 1 x · x 2 2 d x - x 2 2 + C 2 y = x 2 2 log x - x 2 4 - x 2 2 + C 2 When x = 1 , y = 1 . ∴ 1 = - 1 4 - 1 2 + C 2 C 2 = 7 4 Substituting the value of C 2 in equation ( i i ) we get, y = x 2 log x 2 - 3 4 x 2 + 7 4 Hence, this is the final equation.

Solve the differential equation d 2 y d x 2 = e x sin x + cos x for y . Given, y = 1 and d y d x = 0 , when x = 0 .

Given, that d 2 y d x 2 = e x sin x + cos x Integrating both the sides of the equation with respect to x we get, ∫ d 2 y d x 2 = ∫ e x sin x + cos x d y d x = e x . sin x + C 1 ... ( i ) (since we know that ∫ e x ( f ( x ) + f ' ( x ) ) d x = e x f ( x ) + C ) Integrating both the sides of the equation again with respect to x we get, ∫ d y d x = ∫ e x . sin x + C 1 y = ∫ e x sin x d x + C 1 x + C 2 ... ( i i ) Let I = ∫ sin x e x d x . I = sin x . e x - ∫ cos x · e x d x I = e x sin x - cos x . e x - ∫ - sin x e x d x I = e x sin x - e x cos x - I 2 I = e x ( sin x - cos x ) I = e x 2 ( sin x - cos x ) Substituting the value of I in equation ( i i ) , we get y = e x 2 ( sin x - cos x ) + C 1 x + C 2 ... ( i i i ) When x = 0 , d y d x = 0 and y = 1 . Therefore, from (i) we get, 0 = e 0 sin 0 + C 1 ⇒ C 1 = 0 Substituting the value of C 1 in equation ( i i i ) we get, 1 = e 0 2 ( sin 0 - cos 0 ) + 0 + C 2 ⇒ C 2 = 3 2 Substituting the values of C 1 and C 2 in equation ( i i i ) we get, y = e x 2 ( sin x - cos x ) + 3 2

Solve the differential equation d 2 y d x 2 = x + sin x . Given, y = 0 and d y d x = - 1 , when x = 0 .

Given, d 2 y d x 2 = x + sin x Integrating both the sides of the equation with respect to x we get, d y d x = x 2 2 - cos x + C 1 ... ( i ) Integrating both the sides of the equation again with respect to x we get, y = x 3 6 - sin x + C 1 x + C 2 ... ( i i ) When x = 0 , d y d x = - 1 and y = 0 . Therefore from ( i ) we get, - 1 = 0 - cos 0 + C 1 C 1 = 0 Substituting the value of C 1 in equation ( i i ) we get, 0 = 0 - sin 0 + 0 + C 2 0 = C 2 C 2 = 0 Substituting values of C 1 and C 2 in equation ( i i ) we get, y = x 3 6 - sin x

E M B I B E

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE

K. C. Sinha Solutions for Chapter: Applications of Differential Equations, Exercise 2: EXERCISE

Author: K. C. Sinha

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Applications of Differential Equations, Exercise 2: EXERCISE

Attempt the free practice questions on Chapter 6: Applications of Differential Equations, Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from K. C. Sinha Solutions for Chapter: Applications of Differential Equations, Exercise 2: EXERCISE with Hints & Solutions

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 6

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

EASY

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 7

Solve $\frac{d^{2} y}{d x^{2}} = x$ .

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 8

Evaluate $\frac{d^{2} y}{d x^{2}} = \sin^{2} x + x e^{x}$ .

EASY

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 9

Solve the differential equation $\frac{d^{2} y}{d x^{2}} = 6 x + \sec^{2} x$ .

EASY

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 10

Solve the differential equation $\frac{d^{2} y}{d x^{2}} = \cos x - \sin x$ .

MEDIUM

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 11

Solve the differential equation $\frac{d^{2} y}{d x^{2}} = \log x$ . Given, $y = 1$ and $\frac{d y}{d x} = - 1$ , when $x = 1$ .

MEDIUM

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 6 - Applications of Differential Equations>EXERCISE>Q 12

Solve the differential equation $\frac{d^{2} y}{d x^{2}} = e^{x} (\sin x + \cos x)$ for $y$ . Given, $y = 1$ and $\frac{d y}{d x} = 0$ , when $x = 0$ .