Show that y = sin x is the solution of the initial value problem d 2 y d x 2 + y = 0 , y ( 0 ) = 0 , y ' ( 0 ) = 1 .

The given function is y = sin x . . . . . . i Differentiating it with respect to x , ⇒ d y d x = cos x . . . . . . . . . . ( ii ) Again, Differentiating it with respect to x , ⇒ d 2 y d x 2 = - sin x ⇒ d 2 y d x 2 = - y ⇒ d 2 y d x 2 + y = 0 Thus, y = sin x satisfies the differential equation d 2 y d x 2 + y = 0 Put x = 0 in the equation ( i ) , ⇒ y 0 = sin 0 = 0 Put x = 0 in equation ( ii ) , y ' 0 = cos 0 = 1 ∴ y ' 0 = 1 Hence, y = sin x is the solution of the given initial value problem.

Show that y = e x + e - x is the solution of the initial value problem d 2 y d x 2 - y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 0 .

The given function is y = e x + e - x . . . . . . . . ( i ) Differentiating it with respect to x , d y d x = e x - e - x . . . . . . . . . . ( ii ) Again, differentiating it with respect to x , d 2 y d x 2 = e x + e - x d 2 y d x 2 = y [Using equation ( i ) ] d 2 y d x 2 - y = 0 Thus, y = e x + e - x satisfies the differential equation d 2 y d x 2 - y = 0 . Put x = 0 , in the equation ( i ) , y 0 = e 0 + e 0 ⇒ y 0 = 2 Put x = 0 , in the equation ( ii ) y ' 0 = e 0 - e 0 ⇒ y ' 0 = 0 Hence, y = e x + e - x is the solution of the given initial value problem.

Show that y = e x + e 2 x is the solution of the initial value problem d 2 y d x 2 - 3 d y d x + 2 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 3 .

The given function is y = e x + e 2 x . . . . . . . . . . ( i ) Differentiating it with respect to x , d y d x = e x + 2 e 2 x . . . . . . . . . . . ( ii ) Again, differentiating it with respect to x , ⇒ d 2 y d x 2 = e x + 4 e 2 x We have to prove that d 2 y d x 2 - 3 d y d x + 2 y = 0 Consider L . H . S . = d 2 y d x 2 - 3 d y d x + 2 y = e x + 4 e 2 x - 3 e x + 2 e 2 x + 2 e x + e 2 x = e x + 4 e 2 x - 3 e x - 6 e 2 x + 2 e x + 2 e 2 x = 0 = R . H . S . Hence, d 2 y d x 2 - 3 d y d x + 2 y = 0 Thus, y = e x + 2 e 2 x satisfies the differential equation d 2 y d x 2 - 3 d y d x + 2 y = 0 . Put x = 0 in the equation ( i ) , y 0 = e 0 + e 0 ⇒ y 0 = 1 + 1 = 2 Put x = 0 in the equation ( ii ) , y ' 0 = e 0 + 2 e 0 ⇒ y ' 0 = 1 + 2 = 3 Hence, y = e x + 2 e 2 x is the solution of the given initial value problem.

Show that y = x e x + e x is the solution of the initial value problem d 2 y d x 2 - 2 d y d x + y = 0 , y ( 0 ) = 1 , y ' ( 0 ) = 2 .

The given function is y = x e x + e x . . . . . . ( i ) Differentiating it with respect to x , d y d x = x d d x e x + e x d d x ( x ) + d e x d x ⇒ d y d x = x e x + e x ( 1 ) + e x ⇒ d y d x = x e x + 2 e x . . . . . . . . . . . ( ii ) Again, differentiating it with respect to x , d 2 y d x 2 = x d d x e x + e x d d x ( x ) + d d x 2 e x ⇒ d 2 y d x 2 = x e x + e x + 2 e x ⇒ d 2 y d x 2 = x e x + 3 e x . . . . . . . . . . . iii We have to prove that: d 2 y d x 2 - 2 d y d x + y = 0 Consider, L . H . S . = d 2 y d x 2 - 2 d y d x + y = x e x + 3 e x - 2 x e x + 2 e x + x e x + e x = 2 x e x + 4 e x - 2 x e x - 4 e x = 0 = R . H . S . Thus, y = x e x + e x satisfies the differential equation d 2 y d x 2 - 2 d y d x + y = 0 . Put x = 0 in the equation ( i ) , y 0 = 0 + e 0 = 1 So, y ( 0 ) = 1 Put x = 0 in the equation ( ii ) , y ' 0 = 0 + 2 e 0 = 2 So, y ' 0 = 2 Hence, y = x e x + e x is the solution of the given initial value problem.

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 4: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Differential Equations, Exercise 4: EXERCISE

Attempt the free practice questions on Chapter 3: Differential Equations, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-2 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Differential Equations, Exercise 4: EXERCISE with Hints & Solutions

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 1

Show that $y = \log x$ is the solution of the initial value problem $x \frac{d y}{d x} = 1, y (1) = 0$ .

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 2

Show that $y = e^{x}$ is the solution of the initial value problem $\frac{d y}{d x} = y, y (0) = 1$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 3

Show that $y = \sin x$ is the solution of the initial value problem $\frac{d^{2} y}{d x^{2}} + y = 0, y (0) = 0, y^{'} (0) = 1$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 5

Show that $y = e^{- x} + 2$ is the solution of the initial value problem $\frac{d y}{d x} + y = 2, y (0) = 3$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 6

Show that $y = \sin x + \cos x$ is the solution of the initial value problem $\frac{d^{2} y}{d x^{2}} + y = 0, y (0) = 1, y^{'} (0) = 1$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 7

Show that $y = e^{x} + e^{- x}$ is the solution of the initial value problem $\frac{d^{2} y}{d x^{2}} - y = 0, y (0) = 2, y^{'} (0) = 0$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 8

Show that $y = e^{x} + e^{2 x}$ is the solution of the initial value problem $\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + 2 y = 0, y (0) = 2, y^{'} (0) = 3$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 3 - Differential Equations>EXERCISE>Q 9

Show that $y = x e^{x} + e^{x}$ is the solution of the initial value problem $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 0, y (0) = 1, y^{'} (0) = 2$ .

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