RBSE Class 12 Maths Chapter 12 – Differential Equations Important questions and solutions are provided here. All the important questions and solutions of this chapter are updated, as the per the new pattern prescribed by the board. The RBSE Class 12 important questions and solutions given at BYJU’S will help the students in covering all the concepts, which are important from the examination point of view.

Chapter 12 of RBSE Class 12 contains nine exercises, all of which have questions on integrating the given equations using different methods. However, the first exercise is about finding the degree and order of given equations. After practising these important questions, students will be able to solve any types of equations which involve differentiation and integration.

### RBSE Maths Chapter 12: Exercise 12.1 Textbook Important Questions and Solutions

**Question 1: Find the order and degree of the following differential equation.**

**d ^{2}y/dx^{2} = sin x + cos x**

**Solution:**

Given differential equation is:

d^{2}y/dx^{2} = sin x + cos x

The highest derivative of y = d2y/dx2 = (d2y/dx2)1

Therefore, the order of the given differential equation is 2 and degree is 1.

**Question 2: Find the order and degree of the following differential equation.**

**x dx + y dy = 0**

**Solution:**

Given differential equation is:

x dx + y dy = 0

⇒ dx[x + (y dy/dx)] = 0

⇒ x + y (dy/dx) = 0

The highest derivative of y = dy/dx = (dy/dx)^{1}

Therefore, the order of the given differential equation is 1 and degree is 1.

### RBSE Maths Chapter 12: Exercise 12.2 Textbook Important Questions and Solutions

**Question 3: Find the differential equation of family of curves for x ^{2} + y^{2} = a^{2}.**

**Solution:**

Given,

x^{2} + y^{2} = a^{2}

Differentiating with respect to x on both sides,

d/dx (x^{2}) + d/dx (y^{2}) = d/dx (a^{2})

2x + 2y (dy/dx) = 0

2[x + y(dy/dx)] = 0

x + y(dy/dx) = 0

**Question 4: Find the differential equation of the family of curves for y = a cos(x + b), where a and b are arbitrary variables.**

**Solution:**

Given,

y = a cos(x + b) ….(i)

Differentiating with respect to x on both sides,

dy/dx = s d/dx cos(x + b)

dy/dx = -a sin(x + b)

Again differentiating with respect to x on both sides,

d^{2}y/dx^{2} = -a d/dx sin(x + b)

d^{2}y/dx^{2} = -a cos(x + b)

d^{2}y/dx^{2} = -y [From (i)]

(d^{2}y/dx^{2}) + y = 0

### RBSE Maths Chapter 12: Exercise 12.3 Textbook Important Questions and Solutions

**Question 5: Prove that y ^{2} = 4a(x + a) is a solution of differential equation y [1 – (dy/dx)^{2}] = 2x(dy/dx).**

**Solution:**

Given,

y^{2} = 4a(x + a) ….(i)

Differentiating with respect to x on both sides,

d/dx (y^{2}) = 4a d/dx (x + a)

2y (dy/dx) = 4a (1)

dy/dx = 4a/2y

dy/dx = 2a/y

⇒ 2a = y (dy/dx)

⇒ a = (y/2) (dy/dx)

Substituting the value of a in (i),

y^{2} = 4 (y/2) (dy/dx) [x + (y/2) (dy/dx)]

y^{2} = 2y(dy/dx) [x + (y/2) (dy/dx)]

y2 = 2xy(dy/dx) + y^{2}(dy/dx)^{2}

⇒ y = 2x(dy/dx) + y(dy/dx)^{2}

⇒ y – y(dy/dx)^{2} = 2x(dy/dx)

⇒ y[1 – (dy/dx)^{2}] = 2x(dy/dx)

Hence proved.

**Question 6: Prove that y = ae ^{-2x} + bex is a solution of differential equation (d^{2}y/dx^{2}) + (dy/dx) – 2y = 0.**

**Solution:**

Given,

y = ae^{-2x} + be^{x} ….(i)

Differentiating with respect to x on both sides,

dy/dx = a d/dx(e^{-2x}) + b d/dx(e^{x})

dy/dx = -2ae^{-2x} + be^{x} ….(ii)

Again differentiating with respect to x on both sides,

d^{2}y/dx^{2} = -2a d/dx(e^{-2x}) + b d/dx(e^{x})

= -2a(-2)e^{-2x} + be^{x}

= 4ae^{-2x} + be^{x}

= 2ae^{-2x} + 2ae^{-2x} + be^{x}

= [be^{x} – (dy/dx)] + 2ae^{-2x} + be^{x} [From (ii)]

= -(dy/dx) + 2(ae^{-2x} + be^{x})

= -(dy/dx) + 2y [From (i)]

⇒ (d^{2}y/dx^{2}) + (dy/dx) – 2y = 0

Hence proved.

### RBSE Maths Chapter 12: Exercise 12.4 Textbook Important Questions and Solutions

**Question 7: Solve the following differential equation.**

**(1 + x ^{2})dy = (1 + y^{2})dx**

**Solution:**

Given,

(1 + x^{2})dy = (1 + y^{2})dx

⇒ dx/(1 + x^{2}) = dy/(1 + y^{2})

Integrating both the sides,

∫dx/(1 + x^{2}) = ∫dy/(1 + y^{2})

⇒ tan^{-1}x = tan^{-1}y + tan^{-1}C

⇒ tan^{-1}y = tan^{-1}x + tan^{-1}C

⇒ tan^{-1}[(y – x)/ (1 + xy)] = tan^{-1}C

(y – x)/(1 + xy) = C

y – x = C(1 + xy)

**Question 8: Solve the following differential equation.**

**dy/dx = e ^{x – y} + x^{2}e^{-y}**

**Solution:**

Given,

dy/dx = e^{x – y} + x^{2}e^{-y}

dy/dx = (e^{x}/e^{y}) + (x^{2}/ey)

dy/dx = (e^{x} + x^{2})/ey

ey dy = (e^{x} + x^{2}) dx

Integrating both the sides,

∫e^{y} dy = ∫(e^{x} + x^{2}) dx

e^{y} = e^{x} + (x^{3}/3) + C

### RBSE Maths Chapter 12: Exercise 12.5 Textbook Important Questions and Solutions

**Question 9: Solve the following differential equation.**

**x + y = sin ^{-1}(dy/dx)**

**Solution:**

Given,

x + y = sin^{-1}(dy/dx)

⇒ sin(x + y) = dy/dx ….(i)

Let x + y = u

Differentiating with respect to x on both sides,

d/dx (x + y) = d/dx(u)

1 + (dy/dx) = du/dx

dy/dx = (du/dx) – 1 ….(ii)

From (i) and (ii),

sin u = (du/dx) – 1

du/dx = 1 + sin u

1/(1 + sin u) du = dx

Integrating both the sides,

∫1/(1 + sin u) du = ∫ dx

∫[(1 – sin u)/ (1 – sin^{2}u)] du = x + C

∫[(1 – sin u)/(cos^{2}u)] du = x + C

∫(1/cos^{2}u) du – ∫sin u/cos^{2}u du = x + C

∫sec^{2}u du – ∫sec u tan u du = x + C

tan u – sec u = x + C

Substituting u = x + y,

tan(x + y) – sec(x + y) = x + C

x = tan(x + y) – sec(x + y) + C (since C is a constant)

**Question 10: Solve the following differential equation.**

**dy/dx = [(x – y) + 3]/ [2(x – y) + 5]**

**Solution:**

Let x – y = u

Differentiating with respect to x,

d/dx (x – y) = d/dx (u)

1 – (dy/dx) = du/dx

⇒ dy/dx = 1 – (du/dx)

⇒ 1 – (du/dx) = (u + 3)/ (2u + 5)

⇒ du/dx = 1 – [(u + 3)/ (2u + 5)]

⇒ du/dx = (2u + 5 – u – 3)/(2u + 5)

⇒ du/dx = (u + 2)/(2u + 5)

⇒ [(2u + 5)/(u + 2)] du = dx

⇒ [2 + (1/u + 2)] du = dx

⇒ 2 du + 1/(u + 2) du = dx

Integrating both the sides,

2∫ du + ∫[1/(u + 2)] du = ∫dx

2u + log (u + 2) = x + C

Substituting u = x – y,

2(x – y) + log (x – y + 2) = x + C

### RBSE Maths Chapter 12: Exercise 12.6 Textbook Important Questions and Solutions

**Question 11: Solve the following differential equation.**

**x(dy/dx) + (y ^{2}/x) = y**

**Solution:**

Given,

x(dy/dx) + (y^{2}/x) = y

x(dy/dx) = y – (y^{2}/x)

x(dy/dx) = (xy – y^{2})/x

dy/dx = (xy – y^{2})/x^{2} ….(i)

This is a homogeneous equation.

Substituting y = vx

Differentiating with respect to x,

dy/dx = d/dx(vx)

dy/dx = v + x(dv/dx) ….(ii)

From (i) and (ii),

v + x(dy/dx) = (x.vx – v^{2}x^{2})/x^{2}

v + x(dy/dx) = [(v – v^{2})x^{2}/x^{2}]

v + x(dy/dx) = v – v^{2}

⇒ x(dv/dx) = -v^{2}

⇒ dv/dx = -v^{2}/x

⇒ (1/v^{2})dv = -(1/x)dx

Integrating both the sides,

∫(1/v^{2})dv = -∫(1/x) dx

-1/v = -log x + C

⇒ -1/(y/x) = -log x + C

⇒ -x/y = -log x + C

⇒ -x = -y log x + Cy

⇒ x + CY = y log x

### RBSE Maths Chapter 12: Exercise 12.7 Textbook Important Questions and Solutions

**Question 12: Solve the following differential equation.**

**dy/dx = (6x – 2y – 7)/ (2x + 3y – 6)**

**Solution:**

Given,

dy/dx = (6x – 2y – 7)/ (2x + 3y – 6)

By comparing with the standard form,

a/a’ ≠ b/b’

Let us substitute x + X + h and y = Y + k

dY/dX = [6(X + h) – 2(Y + k) – 7]/ [2(X + h) + 3(Y + k) – 6]

dY/dX = [6X – 2Y + (6h – 2k – 7)]/ [2X + 3Y + (2h + 3k – 6)] ….(i)

Considering the equation of h and k are:

6h – 2k – 7 = 0

2h + 3k – 6 = 0

By solving these equations,

h = 3/2 and k = 1

Now,

dY/dX = (6X – 2Y)/(2X + 3Y)

This is a homogeneous equation.

Let Y = vX

⇒ v + X(dv/dX) = (6X – 2vX)/ (2X + 3vX)

⇒ X(dv/dX) = [(6 – 2v)/(2 + 3v)] – v

⇒ X(dv/dX) = (6 -2v – 2v – 3v^{2})/ (2 + 3v)

⇒ X(dv/dX) = (6 – 4v – 3v^{2})/ (2 + 3v) = -(3v^{2} + 4v – 6)/(2 + 3v)

⇒ (1/X) dX/dv = -(2 + 3v)/ (3v^{2} + 4v – 6)

⇒ (1/X) dX = -[(3v + 2)/ (3v^{2} + 4v – 6)] dv

= -(1/2) [(6v + 4)/ (3v^{2} + 4v – 6)] dv

⇒ 2 log X = -log (3v^{2} + 4v – 6) + log c

⇒ log X^{2} + log (3v^{2} + 4v – 6) = log c

⇒ log X^{2}(3v^{2} + 4v – 6) = log c

⇒ 3X^{2}v^{2} + 4vX^{2} – 6X^{2} = c

⇒ 3Y^{2} + 4XY – 6X^{2} = c

⇒ 3(y – 1)2 + 4[x – (3/2)](y – 1) – 6[x – (3/2)]^{2} = c

### RBSE Maths Chapter 12: Exercise 12.8 Textbook Important Questions and Solutions

**Question 13: Solve the following differential equation.**

**(dy/dx) + y cot x = sin x**

**Solution:**

Given,

(dy/dx) + y cot x = sin x

Comparing with (dy/dx) + Py = Q

P = cot x and Q = sin x

Integrating factor (IF) = e^{∫cot x dx}

IF = e^{log (sin x)}

IF = sin x

Multiplying the given equation with sin x,

sin x (dy/dx) + sin x. cot x y = sin^{2}x

sin x (dy/dx) + cos x.y = 1 – cos^{2}x

d/dx (sin x.y) = 1 – cos^{2}x

Integrating both the sides,

∫[d/dx (sin x.y)] dx = ∫1 dx – ∫cos^{2}x dx

y sin x = x – ∫(1 + cos 2x)/2 dx

y sin x = x – (1/2) ∫ dx – (1/2) ∫ cos 2x dx

y sin x = x – (1/2)x – (1/2) (sin 2x/2) + C

y sin x = (1/2)x – (1/4) sin 2x + C

### RBSE Maths Chapter 12: Exercise 12.9 Textbook Important Questions and Solutions

**Question 14: Solve the following differential equation.**

**(dy/dx) – y tan x = -y ^{2} sec x**

**Solution:**

Given,

(dy/dx) – y tan x = -y^{2} sec x

⇒ (1/y^{2}) (dy/dx) – (1/y) tan x = -sec x ….(i)

Let -1/y = v

⇒ (1/y^{2}) (dy/dx) = dv/dx ….(ii)

From (i) and (ii),

(dv/dx) + tan x.v = -sec x

This is a linear equation in variable v.

Thus, P = tan x and Q = -sec x

IF = e^{∫P dx} = e^{∫tan x dx}

IF = e^{log (sec x)} = sec x ….(iii)

⇒ -(1/y) sec x = -∫sec^{2}x dx + C

-(sec x)/y = -tan x + C

(1/y) = (tan x – C)/sec x

1/y = (sin x – C cos x)/ cos x sec x

1/y = sin x – C cos x

(1/y) – sin x + C cos x = 0

### RBSE Maths Chapter 12: Additional Important Questions and Solutions

**Question 1: Solution of (dy/dx) + cos x tan y = 0 is**

**(a) log(sin y) + sin x + C**

**(b) log sin x sin y = C**

**(c) sin y + log sin x + C**

**(d) sin x sin y = C**

**Solution:**

Correct answer: (a)

Given,

(dy/dx) + cos x tan y = 0

⇒ dy + (cos x tan y) dx = 0

Dividing by tan y,

⇒ (1/tan y) dy + cos x dx = 0

⇒ cot y dy + cos x dx = 0

Integrating on both sides,

∫cot y dy + ∫cos x dx = 0

log(sin y) + sin x + C = 0