# NCERT Solutions class 12 Maths Exercise 9.6 (Ex 9.6) Chapter 9 Differential Equations

## NCERT Solutions for Class 12 Maths Exercise 9.6 Chapter 9 Differential Equations – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6 (Ex 9.6) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 9 Differential Equations Exercise 9.6 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

# NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.6) Exercise 9.6

In each of the following differential equations given in each Questions 1 to 4, find the general solution:

1.

Ans. Given: Differential equation

Comparing with , we have P = 2 and Q = sin x.

=>  I.F. =

Solution is (I.F.) =

……….(i)

Applying product rule,

=>   I =

Again applying product rule,

=> I =

I =

I =

I =

Putting the value of I in eq. (i),

=>

### 2.

Ans. Given: Differential equation

Comparing with ,

we have P = 2 and Q = .

=> I.F. =

Solution is

=>  (I.F.) =

### 3.

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

=>   I.F. =  = x

Solution is

=> (I.F.) =

### 4.

Ans. Given: Differential equation

Comparing with , we have P =  and Q = .

I.F. =

Solution is

=> (I.F.) =

=>y(secx+tanx)=(secx.tanx+sec2x1)dx+c=>y(secx+tanx)=∫(secx.tanx+sec2x−1)dx+c

### For each of the following differential equations given in Question 5 to 8, find the general solution:

5.

Ans. Given: Differential equation

=>dydx+ycos2x=tanxcos2x=>dydx+ycos2x=tanxcos2x

Comparing with ,

we have P =  and Q = .

=> I.F. =

Solution is

=> (I.F.) =

…….(i)

Putting  and differentiating

Applying product rule,

Putting this value in eq. (i),

=>

divide by etanx, we get

y = (tanx -1) + ce-tanx

### 6.

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

=>  I.F. =

Solution is

=> (I.F.) =

### 7.

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

=> I.F. =

Solution is

=> (I.F.) =

Applying Product rule of Integration,

=>ylogx=2[(logx)x111x.x11dx]+c=>ylogx=2[(logx)x−1−1−∫1x.x−1−1dx]+c

### 8.

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

=> I.F. =

Solution is

=>  (I.F.) =

### For each of the following differential equations given in Question 9 to 12, find the general solution:

9.

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

=> I.F. =

Solution is

=>  (I.F.) =

Applying product rule of Integration,

=

### 10.

Ans. Given: Differential equation

dxdy=x+ydxdy=x+y

dxdyx=ydxdy−x=y

Comparing with , we have P =  and Q = .

I.F. =

Solution is

=> x(I.F.) = Q(I.F.)dy+c∫Q(I.F.)dy+c

Applying product rule of Integration,

=> xe-y

=> xe-y =

x+y+1 = cey

### 11.

Ans. Given: Differential equation

dxdy+1yx=ydxdy+1yx=y

Comparing with , w

e have P =  and Q = .

I.F. =

Solution is x(I.F.) = Q(I.F.)dy+c∫Q(I.F.)dy+c

### 12.

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

=> I.F. =

Solution is x(I.F.) = Q(I.F.)dy+c∫Q(I.F.)dy+c

### For each of the differential equations given in Questions 13 to 15, find a particular solution satisfying the given condition:

13.  when  x=π3x=π3

Ans. Given: Differential equation   when

Comparing with ,

we have P =  and Q = .

=>  I.F.=

Solution is (I.F.) =

………….. (1)

put value of x and y, we get

=>0cos2π3=1cosπ3+c=>0cos2π3=1cosπ3+c

=> 0=2+c0=2+c

=> c = -2 put value of c in (1), we get

=>ycos2x=1cosx2=>ycos2x=1cosx−2

=>y=cosx2cos2x=>y=cosx−2cos2x

### 14.  when

Ans. Given: Differential equation   when

Comparing with ,

we have P =  and Q = .

=> I.F.=

Solution is (I.F.) =

……….(i)

Now putting

Putting the value of  in eq. (i),

=>

### 15.  when

Ans. Given: Differential equation

Comparing with ,

we have P =  and Q = .

I.F.=

Solution is (I.F.) =

……….(i)

Now putting  in eq. (i),

Putting  in eq. (i),

=>

=> y = 4sin3x – 2sin2x

### 16. Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of coordinates of that point.

Ans. Slope of the tangent to the curve at any point (x,y) = Sum of coordinates of the point

Comparing with ,

we have P =  and Q = .

=>  I.F.=

Solution is

=> y(I.F.) =

Applying Product rule of Integration,

=>

……….(i)

Now, since curve (i) passes through the origin (0, 0),

therefore putting  in eq. (i)

Putting  in eq. (i),

### 17. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

Ans.    According to the question, Sum of the coordinates of any point say  on the curve

= Magnitude of the slope of the tangent to the curve + 5

Comparing with ,

we have P =  and Q = .

=> I.F.=

Solution is

=> y(I.F.) =

Applying Product rule of Integration,

……….(i)

Now, since curve (i) passes through the point (0, 2),

therefore putting  in eq. (i)

0+2 = 4+ce0

Putting  in eq. (i),

### 18. Choose the correct answer:

The integrating factor of the differential equation  is:

(A)

(B)

(C)

(D)

Ans.    Given: Differential equation

Comparing with ,

we have P =  and Q = .

=> I.F.=

Therefore, option (C) is correct.

### 19. Choose the correct answer:

The integrating factor of the differential equation (1y2)dxdy+yx=ay(1<y<1)is(1−y2)dxdy+yx=ay(−1<y<1)is

(A)

(B)

(C)

(D)

Ans. Given: Differential equation  (1y2)dxdy+yx=ay(1<y<1)(1−y2)dxdy+yx=ay(−1<y<1)

Comparing with ,

we have P =  and Q =

=>   I.F. =

Therefore, option (D) is correct.