RD Sharma Class 12 Ex 22.4 Solutions Chapter 22 Differential Equations

Here we provide RD Sharma Class 12 Ex 22.4 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.4 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter22
Exercise22.4
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 22.4 Solutions Chapter 22 Differential Equations

Question 1. For each of the following initial value problems verify that the accompanying function is a solution: x(dy/dx) = 1, y(1) = 0

Function: y = logx

Solution:

We have,

y = logx          -(1)

On differentiating eq(1) w.r.t x,

dy/dx = (1/x)

x(dy/dx) = 1

Thus, y = logx satisfy the given differential equation.

If x = 1, y = log(1) = 0 

So, y(1) = 0

Question 2. For each of the following initial value problems verify that the accompanying function is a solution: (dy/dx) = y, y(0) = 0

Function: y = e

Solution:

We have,

 y = ex          -(1)

On differentiating eq(1) w.r.t x

dy/dx = ex 

(dy/dx) = y

Thus, y = ex satisfy the given differential equation.

If x = 0, y = e= 1 

So, y(0) = 1

Question 3. For each of the following initial value problems verify that the accompanying function is a solution: (d2y/dx2) + y = 0, y(0) = 0, y'(0) = 1

Function: y = sinx 

Solution:

We have,

y = sinx           -(1)

On differentiating eq(1) w.r.t x,

(dy/dx) = cosx          -(2)

Again, differentiating eq(2) w.r.t x,

d2y/dx= -sinx

d2y/dx+ sinx = 0

Thus, y = sinx satisfy the given differential equation.

If x = 0, y(0) = sin(0) = 0 

y'(0) = cos(0) = 1

Question 4. For each of the following initial value problems verify that the accompanying function is a solution: d2y/dx– (dy/dx) = 0, y(0) = 2, y'(0) = 1

Function: y = e+ 1

Solution:

We have,

y = e+ 1           -(1)

On differentiating eq(1) w.r.t x,

(dy/dx) = ex           -(2)

Again differentiating eq(2) w.r.t x,

d2y/dx= ex         

d2y/dx– e= 0        

d2y/dx– (dy/dx) = 0

Thus, y = e+ 1 satisfy the given differential equation.

If x = 0, y(0) = e+ 1, y(0) = 1 + 1 = 2

y'(0) = e= 1

Question 5. For each of the following initial value problems verify that the accompanying function is a solution: (dy/dx) + y = 2

Function: y = e-x + 2

Solution:

We have,

 y = e-x + 2           -(1)

On differentiating eq(i) w.r.t x,

(dy/dx) = -e-x

(dy/dx) + e-x = 0

(dy/dx) + (y – 2) = 0

(dy/dx) + y = 2

Thus, y = e-x + 2 satisfy the given differential equation.

If x = 0, y(0) = e-0 + 2 = 1 + 2 = 3

Question 6. For each of the following initial value problems verify that the accompanying function is a solution: (d2y/dx2) + y = 0, y(0) = 1, y'(0) = 1

Function: y = sinx + cosx

Solution:

We have,

 y = sinx + cosx           -(1)

On differentiating eq(i) w.r.t x,

dy/dx = cosx – sinx           -(2)

Again differentiating eq(ii) w.r.t x,

d2y/dx= -sinx – cosx

d2y/dx= -(sinx + cosx)

(d2y/dx2) + y = 0 

Thus, y = sinx + cosx satisfy the given differential equation.

If x = 0, y(0) = sin0 + cos0 = 1

y'(0) = cos0 – sin0 = 1

Question 7. For each of the following initial value problems verify that the accompanying function is a solution: (d2y/dx2) – y = 0, y(0) = 2, y'(0) = 0

Function: y = e+ e-x 

Solution:

We have,

 y = e+ e-x           -(1)

On differentiating eq(i) w.r.t x,

dy/dx = e– e-x           -(2)

Again differentiating eq(2) w.r.t x,

d2y/dx= e+ e-x

d2y/dx= y

d2y/dx– y = 0

Thus, y = e+ e-x satisfy the given differential equation.

If x = 0, y(0) = e+ e-0 = 1 + 1 = 2

y'(0) = e– e-0 = 0

Question 8. For each of the following initial value problems verify that the accompanying function is a solution: (d2y/dx2) – 3(dy/dx) + 2y = 0, y(0) = 2, y'(0) = 3

Function: y = e+ e2x

Solution:

We have,

y = e+ e2x           -(1)

On differentiating eq(1) w.r.t x,

dy/dx = e+ 2e2x           -(2)

Again differentiating equation(2) w.r.t x,

d2y/dx= e+ 4e2x

d2y/dx= 3(e+ 2e2x) – 2(ex + e2x)

(d2y/dx2) = 3(dy/dx) – 2y

(d2y/dx2) – 3(dy/dx) + 2y = 0

Thus, y = e+ e2x satisfy the given differential equation.

If x = 0, y(0) = e+ e= 1 + 1 = 2

y'(0) = e+ 2e= 1 + 2 = 3

Question 9. For each of the following initial value problems verify that the accompanying function is a solution: (d2y/dx2) – 2(dy/dx) + y = 0, y(0) = 1, y'(0) = 2

Function: y = xe+ e

Solution:

We have,

y = xe+ ex            -(1)

On differentiating eq(1) w.r.t x,

dy/dx = xe+ e+ e 

dy/dx = xe+ 2ex           -(2)

Again differentiating eq(2) w.r.t x,

d2y/dx= xe+ e+ 2e            

d2y/dx= xe+ e+ 2e+ xe+ e– xe– ex    

d2y/dx= 2(xe+ ex) – (xe+ ex)

(d2y/dx2) = 2(dy/dx) – y

(d2y/dx2) – 2(dy/dx) + y = 0

Thus, y = xe+ ex satisfy the given differential equation.

If x = 0, y(0) = 0e+ e= 1

y'(0) = 0e+ 2e= 2

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