CBSE NCERT Solutions for Class 12 Maths Chapter 09 · 2019. 12. 6. · Class-XII-Maths Differential Equations 5 Practice more on Differential Equations Solution: Given differential

Class-XII-Maths Differential Equations

1 Practice more on Differential Equations www.embibe.com

CBSE NCERT Solutions for Class 12 Maths Chapter 09

Back of Chapter Questions

Exercise 9.1

1. Determine order and degree (if defined) of differential equation 𝑑4𝑦

𝑑𝑥4 + sin(𝑦′′′) = 0 [1 Mark]

Solution:

Given differential equation is 𝑑4𝑦

𝑑𝑥4 + sin(𝑦′′′) = 0

We know that the highest order derivative present in the differential equation is 4. Therefore,

its order is four.

[1/2 Mark]

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree

is not defined. [1/2

Mark]

2. Determine order and degree (if defined) of differential equation 𝑦′ + 5𝑦 = 0 [1

Mark]

Solution:

The given differential equation is 𝑦′ + 5𝑦 = 0

The highest order derivative present in the differential equation is 𝑦′.

Therefore, its order is one. [1/2

Mark]

It is a polynomial equation in its derivatives and the highest power raised to 𝑦′ is 1. [1/2

Mark]

Hence, its degree is one.

3. Determine order and degree (if defined) of differential equation (𝑑𝑠

𝑑𝑡)

4+ 3𝑠

𝑑2𝑠

𝑑𝑡2 = 0 [1

Mark]



Solution:

Given differential equation is (𝑑𝑠

𝑑𝑡)

4+ 3

𝑑2𝑠

𝑑𝑡2 = 0

The highest order derivative present in the given differential equation is 𝑑2𝑠

𝑑𝑡2. Therefore, its order

is two.

[1/2 Mark]

It is a polynomial equation in 𝑑2𝑠

𝑑𝑡2. and 𝑑𝑠

𝑑𝑡 and the power raised to

𝑑2𝑠

𝑑𝑡2 is 1. [1/2

Mark]


4. Determine order and degree (if defined) of differential equation (𝑑2𝑦

𝑑𝑥2)2

+ cos (𝑑𝑦

𝑑𝑥) = 0 [1

Mark]

Solution:

Given differential equation is (𝑑2𝑦

𝑑𝑥2)2

+ cos (𝑑𝑦

𝑑𝑥) = 0

The highest order derivative present in the given differential equation is 𝑑2𝑦

𝑑𝑥2. Therefore, its order

is 2.

[1/2 Mark]

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree

is not defined. [1/2

Mark]

5. Determine order and degree (if defined) of differential equation 𝑑2𝑦

𝑑𝑥2 = cos3𝑥 + sin3𝑥 [1 Mark]

Solution:

Given differential equation is 𝑑2𝑦

𝑑𝑥2 = cos3𝑥 + sin3𝑥

⇒𝑑2𝑦

𝑑𝑥2− cos3𝑥 − sin3𝑥 = 0



The highest order derivative present in the differential equation is 𝑑2𝑦

𝑑𝑥2. Therefore, its order is

two.

[1/2

Mark]

It is a polynomial equation in 𝑑2𝑦

𝑑𝑥2 and the power raised to 𝑑2𝑦

𝑑𝑥2 is 1. [1/2

Mark]


6. Determine order and degree (if defined) of differential equation (y′′′)2 + (𝑦′′)3 + (𝑦′)4 +

𝑦5 = 0 [1 Mark]

Solution:

Given differential equation is (y′′′)2 + (𝑦′′)3 + (𝑦′)4 + 𝑦5 = 0

The highest order derivative present in the differential equation is 𝑦′′′.

Therefore, its order is three. [1/2

Mark]

The given differential equation is a polynomial equation in 𝑦′′′, 𝑦′′ , and 𝑦′.

The highest power raised to 𝑦′′′ is 2. Hence, its degree is 2. [1/2

Mark]

7. Determine order and degree (if defined) of differential equation 𝑦′′′ + 2𝑦′′ + 𝑦′ = 0 [1

Mark]

Solution:

Given differential equation is 𝑦′′′ + 2𝑦′′ + 𝑦′ = 0

The highest order derivative present in the differential equation is 𝑦′′′.

Therefore, its order is three. [1/2

Mark]

It is a polynomial equation in 𝑦′′′, 𝑦′′ , and 𝑦′ and the highest power raised to 𝑦′′′ is 1.



Hence, its degree is 1. [𝟏

𝟐

Mark]

8. Determine order and degree (if defined) of differential equation𝑦′ + 𝑦 = 𝑒𝑥 [1

Mark]

Solution:

Given differential equation is 𝑦′ + 𝑦 = 𝑒𝑥

⇒ 𝑦′ + 𝑦 − 𝑒𝑥 = 0

The highest order derivative present in the differential equation is 𝑦′.

Therefore, its order is one. [1/2

Mark]

The given differential equation is a polynomial equation in 𝑦′ and the highest power raised to 𝑦′

is one. Hence, its degree is one. [1/2

Mark]

9. Determine order and degree (if defined) of differential equation 𝑦′′ + (𝑦′)2 + 2𝑦 = 0 [1

Mark]

Solution:

Given differential equation is 𝑦′′ + (𝑦′)2 + 2𝑦 = 0

The highest order derivative present in the differential equation is 𝑦′′.

Therefore, its order is two. [1/2

Mark]

The given differential equation is a polynomial equation in 𝑦′′ and 𝑦′ the highest power raised

to 𝑦′′ is one.

[1/2 Mark]


10. Determine order and degree (if defined) of differential equation 𝑦′′ + 2𝑦′ + sin𝑦 = 0 [1 Mark]



Solution:

Given differential equation is 𝑦′′ + 2𝑦′ + sin𝑦 = 0

The highest order derivative present in the differential equation is 𝑦′′.

Therefore, its order is two. [1/2

Mark]

This is a polynomial equation in 𝑦′′ and 𝑦′ and the highest power raised to 𝑦′′ is one.

Hence, its degree is one. [1/2

Mark]

11. The degree of the differential equation (𝑑2𝑦

𝑑𝑥2)3

+ (𝑑𝑦

𝑑𝑥)

2+ sin (

𝑑𝑦

𝑑𝑥) + 1 = 0 is [1 Mark]

A) 3

B) 2

C) 1

D) not defined

Solution:

Given differential equation is (𝑑2𝑦

𝑑𝑥2)3

+ (𝑑𝑦

𝑑𝑥)

2+ sin (

𝑑𝑦

𝑑𝑥) + 1 = 0

The given differential equation is not a polynomial equation in its derivatives. Therefore, its

degree is not defined.

[1 Mark]

Hence, the correct answer is D.

12. The order of the differential equation 2𝑥2 𝑑2𝑦

𝑑𝑥2 − 3𝑑𝑦

𝑑𝑥+ 𝑦 = 0 is [1 Mark]

A) 2

B) 1

C) 0

D) not defined



Solution:

Given differential equation is 2𝑥2 𝑑2𝑦


𝑑𝑥+ 𝑦 = 0

The highest order derivative present in the given differential equation is 𝑑2𝑦

𝑑𝑥2.

Therefore, its order is two. [1

Mark]

Hence, the correct answer is A.

Exercise 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of

the corresponding differential equation:

1. 𝑦 = 𝑒𝑥 + 1: 𝑦′′ − 𝑦′ = 0 [2

Marks]

Solution:

Given equation is 𝑦 = 𝑒𝑥 + 1

Differentiating both the sides of the given equation with respect to 𝑥, we get:

𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥(𝑒𝑥 + 1)

⇒ 𝑦′ = 𝑒𝑥 … (1) [1

Mark]

Now, differentiating equation (1) with respect to 𝑥, we get:

𝑑

𝑑𝑥(𝑦′) =

𝑑

𝑑𝑥(𝑒𝑥)

⇒ 𝑦′′ = 𝑒𝑥

Substituting the values of 𝑦′ and 𝑦′′ in the given differential equation, we get the L.H.S. as:

𝑦′′ − 𝑦′ = 𝑒𝑥 − 𝑒𝑥 = 0 = R. H. S.

Thus, the given function is the solution of the corresponding differential equation. [1

Mark]



2. 𝑦 = 𝑥2 + 2𝑥 + 𝐶: 𝑦′ − 2𝑥 − 2−= 0 [2 Marks]

Solution:

Given equation is 𝑦 = 𝑥3 + 2𝑥 + 𝐶

Differentiating both sides of this equation with respect to 𝑥 we get:

𝑦′ =𝑑

𝑑𝑥(𝑥2 + 2𝑥 + 𝐶)

⇒ 𝑦′ = 2𝑥 + 2 [1

Mark]

Substituting the value of 𝑦′ in the given differential equation,

we get L.H.S.= 𝑦′ − 2𝑥 − 2 = 2𝑥 + 2 − 2𝑥 − 2 = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation. [1

Mark]

3. 𝑦 = cos 𝑥 + 𝐶 : 𝑦′ + sin 𝑥 = 0 [2

Marks]

Solution:

Given equation is 𝑦 = cos 𝑥 + 𝐶

Differentiating both sides of this equation with respect to 𝑥, we get:

𝑦′ =𝑑

𝑑𝑥(cos 𝑥 + 𝐶)

⇒ 𝑦′ = −sin𝑥 [1

Mark]

Substituting the value of 𝑦′ in the given differential equation, we get

L. H. S = 𝑦’ + sin 𝑥 = − sin 𝑥 + sin 𝑥 = 0 = R.H.S


Mark]



4. 𝑦 = √1 + 𝑥2: 𝑦′ =𝑥𝑦

1+𝑥2 [2

Marks]

Solution:

Given equation is 𝑦 = √1 + 𝑥2

Differentiating both sides of the equation with respect to 𝑥, we get:

𝑦′ =𝑑

𝑑𝑥(√1 + 𝑥2)

⇒ 𝑦′ =1

2√1 + 𝑥2·

𝑑

𝑑𝑥(1 + 𝑥2)

⇒ 𝑦′ =2𝑥

2√1 + 𝑥2

⇒ 𝑦′ =𝑥

√1 + 𝑥2

⇒ 𝑦′ =𝑥

1 + 𝑥2× √1 + 𝑥2

⇒ 𝑦′ =𝑥

1 + 𝑥2· 𝑦

⇒ 𝑦′ =𝑥𝑦

1 + 𝑥2

∴ L.H.S = R.H.S


Marks]

5. 𝑦 = 𝐴𝑥 ∶ 𝑥𝑦′ = 𝑦(𝑥 ≠ 0) [1

Mark]

Solution:

Given equation is 𝑦 = 𝐴𝑥

Differentiating both sides with respect to 𝑥, we get:

𝑦′ =𝑑

𝑑𝑥(𝐴𝑥)

⇒ 𝑦′ = 𝐴

Substituting the value of 𝑦′ in the given differential equation, we get:



L.H.S.=𝑥𝑦′ = 𝑥 ∙ 𝐴 = 𝐴𝑥 = 𝑦 =R.H.S


Mark]

6. 𝑦 = 𝑥sin𝑥: 𝑥𝑦′ = 𝑦 + 𝑥√𝑥2 − 𝑦2(𝑥 ≠ 0 and 𝑥 > 𝑦 or 𝑥 < −𝑦) [2

Marks]

Solution:

Given equation is 𝑦 = 𝑥 sin 𝑥


𝑦′ =𝑑

𝑑𝑥(𝑥sin𝑥)

⇒ 𝑦′ = sin𝑥 ·𝑑

𝑑𝑥(𝑥) + 𝑥 ·

𝑑

𝑑𝑥(sin𝑥)

⇒ 𝑦′ = sin𝑥 + 𝑥cos𝑥 [1

Mark]


L.H.S = 𝑥𝑦′ = 𝑥(sin𝑥 + 𝑥cos𝑥)

= 𝑥sin𝑥 + 𝑥2cos𝑥

= 𝑦 + 𝑥2 · √1 − sin2𝑥

= 𝑦 + 𝑥2√1 − (𝑦

𝑥)

2

= 𝑦 + 𝑥√𝑦2 − 𝑥2

= R.H.S


Mark]

7. 𝑥𝑦 = log 𝑦 + 𝐶 : 𝑦′ =𝑦2

1−𝑥𝑦(𝑥𝑦 ≠ 1)

[2 Marks]



Solution:

Given equation is 𝑥𝑦 = log 𝑦 + 𝐶

Differentiating both sides of this equation with respect to x, we get:

𝑑

𝑑𝑥(𝑥𝑦) =

𝑑

𝑑𝑥(log𝑦)

⇒ 𝑦 ·𝑑

𝑑𝑥(𝑥) + 𝑥 ·

𝑑𝑦

𝑑𝑥=

1

𝑦

𝑑𝑦

𝑑𝑥

⇒ 𝑦 + 𝑥𝑦′ =1

𝑦𝑦′ [1

Mark]

⇒ 𝑦2 + 𝑥𝑦𝑦′ = 𝑦′

⇒ (𝑥𝑦 − 1)𝑦′ = −𝑦2

⇒ 𝑦′ =𝑦2

1 − 𝑥𝑦

∴ L.H.S = R.H.S


Mark]

8. 𝑦 = cos 𝑦 = 𝑥 : (𝑦 sin 𝑥 + cos 𝑦 + 𝑥)𝑦′ = 𝑦 [2 Marks]

Solution:

Given equation is 𝑦 − cos 𝑦 = 𝑥

Differentiating both sides of the equation with respect to 𝑥, we get:

𝑑𝑦

𝑑𝑥−

𝑑

𝑑𝑥(cos𝑦) =

𝑑

𝑑𝑥(𝑥)

⇒ 𝑦′ + sin𝑦 · 𝑦′ = 1

⇒ 𝑦′(1 + sin𝑦) = 1

⇒ 𝑦′ =1

1+sin𝑦 [1

Mark]

Substituting the value of 𝑦′ in the L.H.S of the given differential equation, we get

= (𝑦sin𝑦 + cos𝑦 + 𝑥)𝑦′



= (𝑦sin𝑦 + cos𝑦 + 𝑦 − cos𝑦) ×1

1 + sin𝑦

= 𝑦(1 + sin 𝑦) ·1

1 + sin𝑦

= 𝑦

= R. H. S


Mark]

9. 𝑥 + 𝑦 = tan−1 𝑦 : 𝑦2𝑦′ + 𝑦2 + 1 = 0 [2 Marks]

Solution:

Given equation is 𝑥 + 𝑦 = tan−1 𝑦


𝑑

𝑑𝑥(𝑥 + 𝑦) =

𝑑

𝑑𝑥(tan−1𝑦)

⇒ 1 + 𝑦′ = [1

1 + 𝑦2] 𝑦′

⇒ 𝑦′ [1

1 + 𝑦2− 1] = 1

⇒ 𝑦′ [1 − (1 + 𝑦2)

1 + 𝑦2] = 1

⇒ 𝑦′ [−𝑦2

1 + 𝑦2] = 1

⇒ 𝑦′ =−(1+𝑦2)

𝑦2 [1

Mark]


L.H.S.=𝑦2𝑦′ + 𝑦2 + 1 = 𝑦2 [−(1+𝑦2)

𝑦2 ] + 𝑦2 + 1

= −1 − 𝑦2 + 𝑦2 + 1

= 0

=R.H.S.




Mark]

10. 𝑦 = √𝑎2 − 𝑥2𝑥 ∈ (−𝑎, 𝑎) : 𝑥 + 𝑦𝑑𝑦

𝑑𝑥= 0(𝑦 ≠ 0) [2

Marks]

Solution:

Given function is 𝑦 = √𝑎2 − 𝑥2


𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥(√𝑎2 − 𝑥2)

⇒𝑑𝑦

𝑑𝑥=

1

2√𝑎2 − 𝑥2·

𝑑

𝑑𝑥(𝑎2 − 𝑥2)

⇒𝑑𝑦

𝑑𝑥=

1

2√𝑎2 − 𝑥2(−2𝑥)

⇒𝑑𝑦

𝑑𝑥=

1

√𝑎2−𝑥2 [1

Mark]

Substituting the value of 𝑑𝑦

𝑑𝑥

L.H.S.=𝑥 + 𝑦𝑑𝑦

𝑑𝑥= 𝑥 + √𝑎2 − 𝑥2 ×

−𝑥

√𝑥2−𝑥2

= 𝑥 − 𝑥

= 0

=R.H.S.


Mark]

11. The numbers of arbitrary constants in the general solution of a differential equation of

fourth order are:

[1 Mark]

(A) 0

(B) 2



(C) 3

(D) 4

Solution:

We know that the number of constants in the general solution of a differential equation of order

𝑛 is equal to its order. [𝟏

𝟐

Mark]

Therefore, the number of constants in the general equation of fourth order differential equation

is four.

Hence, the correct answer is D. [𝟏

𝟐

Mark]

12. The numbers of arbitrary constants in the particular solution of a differential equation of

third order are:

[1 Mark]

(A) 3

(B) 2

(C) 1

(D) 0

Solution:

In a particular solution of a differential equation, there will be no arbitrary constants. [𝟏

𝟐

Mark]

Hence, the correct answer is D. [𝟏

𝟐

Mark]

Exercise 9.3

In each of the Exercises 1 to 5, form a differential equation representing the given family of

curves by eliminating arbitrary constants 𝑎 and 𝑏.



1. 𝑥

𝑎+

𝑦

𝑏= 1 [2

Marks]

Solution:

Given curve is 𝑥

𝑎+

𝑦

𝑏= 1

Differentiating both sides of the given equation with respect to 𝑥, we get

1

𝑎+

1

𝑏

𝑑𝑦

𝑑𝑥= 0

⇒1

𝑎+

1

𝑏𝑦′ = 0

[1 Mark

Again, differentiating both sides with respect to 𝑥, we get

0 +1

𝑏𝑦′′ = 0

⇒1

𝑏𝑦′′ = 0

⇒ 𝑦′′ = 0

Hence, the required differential equation of the given curve is 𝑦′′ = 0 [1

Mark]

2. 𝑦2 = 𝑎(𝑏2 − 𝑥2) [2

Marks]

Solution:

Given curve is 𝑦2 = 𝑎(𝑏2 − 𝑥2)


2𝑦𝑑𝑦

𝑑𝑥= 𝑎(−2𝑥)

⇒ 2𝑦𝑦′ = −2𝑎𝑥

⇒ 𝑦𝑦′ = −𝑎𝑥 … (1)

Again, differentiating both sides with respect to 𝑥, we get:

𝑦′ · 𝑦′ + 𝑦𝑦′′ = −𝑎



⇒ (𝑦′)2 + 𝑦𝑦′′ = −𝑎 … (2) [1

Mark]

Dividing equation (2) by equation (1), we get:

(𝑦′)2 + 𝑦𝑦′′

𝑦𝑦′=

−𝑎

−𝑎𝑥

⇒ 𝑥𝑦𝑦′′ + 𝑥(𝑦′)2 − 𝑦𝑦′′ = 0

Hence, the required differential equation of the given curve is 𝑥𝑦𝑦′′ + 𝑥(𝑦′)2 − 𝑦𝑦′′ = 0[1

Mark]

3. 𝑦 = 𝑎𝑒3𝑥 + 𝑏𝑒−2𝑥 [4

Marks]

Solution:

Given curve is 𝑦 = 𝑎𝑒3𝑥 + 𝑏𝑒−2𝑥 … (1)


𝑦′ = 3𝑎𝑒3𝑥 − 2𝑏𝑒−2𝑥 … (2)


𝑦′′ = 9𝑎𝑒3𝑥 + 4𝑏𝑒−2𝑥 … (3)

Multiplying equation (1) with (2) and then adding it to equation (2), we get

(2𝑎𝑒3𝑥 + 2𝑏𝑒−2𝑥) + (3𝑎𝑒3𝑥 − 2𝑏𝑐−2𝑥) = 2𝑦 + 𝑦′

⇒ 5𝑎𝑒3𝑥 = 2𝑦 + 𝑦′

⇒ 𝑎𝑒3𝑥 =2𝑦+𝑦′

5 [2

Marks]

Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we get:

(3𝑎𝑒3𝑥 + 3𝑏𝑒−2𝑥) − (3𝑎𝑒3𝑥 − 2𝑏𝑒−2𝑥) = 3𝑦 − 𝑦′

⇒ 5𝑏𝑒−2𝑥 = 3𝑦 − 𝑦′

⇒ 𝑏𝑒−2𝑥 =3𝑦−𝑦′

5 [1

Mark]

Substituting the values of 𝑎𝑒3𝑥 and 𝑏𝑒−2𝑥 in equation (3), we get:

𝑦′′ = 9 ·(2𝑦 + 𝑦′)

5+ 4

(3𝑦 − 𝑦′)

5



⇒ 𝑦′′ =18𝑦 + 9𝑦′

5+

12𝑦 − 4𝑦′

5

⇒ 𝑦′′ =30𝑦 + 5𝑦′

5

⇒ 𝑦′′ = 6𝑦 + 𝑦′

⇒ 𝑦′′ − 𝑦′ − 6𝑦 = 0

Hence, the required differential equation of the given curve is 𝑦′′ − 𝑦′ − 6𝑦 = 0. [1

Mark]

4. 𝑦 = 𝑒2𝑥(𝑎 + 𝑏𝑥) [4

Marks]

Solution:

Given curve is 𝑦 = 𝑒2𝑥(𝑎 + 𝑏𝑥) … (1)


𝑦′ = 2𝑒2𝑥(𝑎 + 𝑏𝑥) + 𝑒2𝑥 · 𝑏

⇒ 𝑦′ = 𝑒2𝑥(2𝑎 + 2𝑏𝑥 + 𝑏) … (2) [1

Mark]

Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we get:

𝑦′ − 2𝑦 = 𝑒2𝑥(2𝑎 + 2𝑏𝑥 + 𝑏) − 𝑒2𝑥(2𝑎 + 2𝑏𝑥)

⇒ 𝑦′ − 2 = 𝑏𝑒2𝑥 … (3)


𝑦′′𝑘 − 2𝑦′ = 2𝑏𝑒2𝑥 … (4) [1

Mark]


𝑦′′𝑘 − 2𝑦′

𝑦′ − 2𝑦= 2

⇒ 𝑦′′ − 2𝑦′ = 2𝑦′ − 4𝑦

⇒ 𝑦′′ − 4𝑦′ + 4𝑦 = 0

Hence, the required differential equation of the given curve is 𝑦′′ − 4𝑦′ + 4𝑦 = 0 [2

Marks]



5. 𝑦 − 𝑒𝑥(𝑎 cos 𝑥 + 𝑏 sin 𝑥) [4

Marks]

Solution:

Given curve is 𝑦 − 𝑒𝑥(𝑎 cos 𝑥 + 𝑏 sin 𝑥) … (1)


𝑦′ = 𝑒𝑥(𝑎cos𝑥 + 𝑏sin𝑥) + 𝑒𝑥(−𝑎sin𝑥 + 𝑏cos𝑥)

⇒ 𝑦′ = 𝑒𝑥[(𝑎 + 𝑏)cos𝑥 − (𝑎 − 𝑏)sin𝑥] … (2)

Again, differentiating with respect to 𝑥, we get:

𝑦′′ = 𝑒𝑥[(𝑎 + 𝑏)cos𝑥 − (𝑎 − 𝑏)sin𝑥] + 𝑒𝑥[−(𝑎 + 𝑏)sin𝑥 − (𝑎 − 𝑏)cos𝑥]

𝑦′′ = 𝑒𝑥[2𝑏cos𝑥 − 2𝑎sin𝑥]

𝑦′′ = 2𝑒𝑥(𝑏cos𝑥 − 𝑎sin𝑥)

⇒𝑦′′

2= 𝑒𝑥(𝑏cos𝑥 − 𝑎sin𝑥) … (3)

Adding equations (1) and (3), we get:

𝑦 +𝑦′′

2= 𝑒𝑥[(𝑎 + 𝑏)cos𝑥 − (𝑎 − 𝑏)sin𝑥] [3

Marks]

⇒ 𝑦 +𝑦′′

2= 𝑦′

⇒ 2𝑦 + 𝑦′′ = 2𝑦′

⇒ 𝑦′′ − 2𝑦′ + 2𝑦 = 0

Hence, the required differential equation of the given curve is 𝑦′′ − 2𝑦′ + 2𝑦 = 0 [1

Mark]

6. Form the differential equation of the family of circles touching the 𝑦-axis at the origin.[2

Marks]

Solution:

We know that the centre of the circle touching the y-axis at origin lies on the 𝑥-axis.

Let (𝑎, 0) be the centre of the circle.



Since it touches the y-axis at origin, its radius is 𝑎.

Now, the equation of the circle with centre (𝑎, 0) and radius 𝑎 is (𝑥 − 𝑎)2 + 𝑦2 = 𝑎2

⇒ 𝑥2 + 𝑦2 = 2𝑎𝑥 … (1)

Now, differentiating equation (1) with respect to 𝑥, we get:

2𝑥 + 2𝑦𝑦′ = 2𝑎

⇒ 𝑥 + 𝑦𝑦′ = 𝑎 [1

Mark]

Now, on substituting the value of a in equation (1), we get:

𝑥2 + 𝑦2 = 2(𝑥 + 𝑦𝑦′)𝑥

⇒ 𝑥2 + 𝑦2 = 2𝑥2 + 2𝑥𝑦𝑦′

⇒ 2𝑥𝑦𝑦′ + 𝑥2 = 𝑦2

Hence, the required differential equation is 2𝑥𝑦𝑦′ + 𝑥2 = 𝑦2 [1

Mark]

7. Form the differential equation of the family of parabolas having vertex at origin and axis

along positive y-axis.

[2 Marks]

Solution:

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is

𝑥2 = 4𝑎𝑦 … (1)



Differentiating equation (1) with respect to 𝑥, we get:

2𝑥 = 4𝑎𝑦′ … (2) [1

Mark]


2𝑥

𝑥2=

4𝑎𝑦′

4𝑎𝑦

⇒2

𝑥=

𝑦′

𝑦

⇒ 𝑥𝑦′ = 2𝑦

⇒ 𝑥𝑦′ − 2𝑦 = 0

Hence, the required differential equation is 𝑥𝑦′ − 2𝑦 = 0. [1

Mark]

8. Form the differential equation of the family of ellipses having foci on y-axis and centre at

origin.

[4

Marks]

Solution:

The equation of the family of ellipses having foci on the y-axis and the centre at origin is 𝑥2

𝑏2 +

𝑦2

𝑎2 = 1 … (1)




2𝑥

𝑏2 +2𝑦𝑦′

𝑏2 = 0 [1

Mark]

⇒𝑥

𝑏2+

𝑦𝑦′

𝑎2= 0 … (2)

Again, differentiating with respect to x, we get:

1

𝑏2+

𝑦′ · 𝑦′ + 𝑦 · 𝑦′′

𝑎2= 0

⇒1

𝑏2+

1

𝑎2(𝑦′2 + 𝑦𝑦′′) = 0

⇒1

𝑏2 = −1

𝑎2(𝑦′2 + 𝑦𝑦′′) [1

Mark]

Substituting this value in equation (2), we get:

𝑥 [−1

𝑎2((𝑦′)2 + 𝑦𝑦′′)] +

𝑦𝑦′

𝑎2 = 0 [1

Mark]

⇒ −𝑥(𝑦′)2 − 𝑥𝑦𝑦′′ + 𝑦𝑦′ = 0

⇒ 𝑥𝑦𝑦′′ + 𝑥(𝑦′)2 − 𝑦𝑦′ = 0

Hence, the required differential equation is 𝑥𝑦𝑦′′ + 𝑥(𝑦′)2 − 𝑦𝑦′ = 0 [1

Mark]

9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre

at origin.



[4

Marks]

Solution:

The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is

𝑥2

𝑎2−

𝑦2

𝑏2= 1 … (1)

Differentiating both sides of equation (1) with respect to 𝑥, we get:

2𝑥

𝑎2 −2𝑦𝑦′

𝑏2 = 0 [1

Mark]

⇒𝑥

𝑎2 −𝑦𝑦′

𝑏2 = 0 … (2)


1

𝑎2−

𝑦′ · 𝑦′ + 𝑦𝑦′′

𝑏2= 0

⇒1

𝑎2 =1

𝑏2((𝑦′)2 + 𝑦𝑦′′) [1

Mark]

Substituting the value of 1

𝑎2 in equation (2)

𝑥

𝑏2((𝑦′)2 + 𝑦𝑦′′) −

𝑦𝑦′

𝑏2 = 0 [1

Mark]

⇒ 𝑥(𝑦′)2 + 𝑥𝑦𝑦′′ − 𝑦𝑦′ = 0

⇒ 𝑥𝑦𝑦′′ + 𝑥(𝑦′)2 − 𝑦𝑦′ = 0

Hence, the required differential equation is 𝑥𝑦𝑦′′ + 𝑥(𝑦′)2 − 𝑦𝑦′ = 0. [1

Mark]



10. Form the differential equation of the family of circles having centre on y-axis and radius 3

units.

[2

Marks]

Solution:

Let the centre of the circle on y-axis be (0, b).

The equation of the family of circles with centre at (0, b) and radius 3 is 𝑥2 + (𝑦 − 𝑏)2 = 32

⇒ 𝑥2 + (𝑦 − 𝑏)2 = 9 … (1)


2𝑥 + 2(𝑦 − 𝑏) · 𝑦′ = 0

⇒ (𝑦 − 𝑏) · 𝑦′ = −𝑥

⇒ 𝑦 − 𝑏 =−𝑥

𝑦′ [1

Mark]

Substituting the value of (𝑦– 𝑏) in equation (1), we get:

𝑥2 + (−𝑥

𝑦′)

2

= 9

⇒ 𝑥2 [1 +1

(𝑦′)2] = 9

⇒ 𝑥2((𝑦′)2 + 1) = 9(𝑦′)2

⇒ (𝑥2 − 9)(𝑦′)2 + 𝑥2 = 0



Hence, the required differential equation is (𝑥2 − 9)(𝑦′)2 + 𝑥2 = 0 [1

Mark]

11. Which of the following differential equations has 𝑦 = 𝑐1𝑒𝑥 + 𝑐2𝑒−𝑥

as the general solution? [2

Marks]

(A) 𝑑2𝑦

𝑑𝑥2 + 𝑦 = 0

(B) 𝑑2𝑦

𝑑𝑥2 − 𝑦 = 0

(C) 𝑑2𝑦

𝑑𝑥2 + 1 = 0

(D) 𝑑2𝑦

𝑑𝑥2 − 1 = 0

Solution:

The given equation is 𝑦 = 𝑐1𝑒𝑥 + 𝑐2𝑒−𝑥 ….(𝑖)

Differentiating with respect to 𝑥, we get:

𝑑𝑦

𝑑𝑥= 𝑐1𝑒𝑥 − 𝑐2𝑒−𝑥 [1

Mark]


𝑑2𝑦

𝑑𝑥2= 𝑐1𝑒𝑥 + 𝑐2𝑒−𝑥

⇒𝑑2𝑦

𝑑𝑥2= 𝑦

⇒𝑑2𝑦

𝑑𝑥2− 𝑦 = 0

Hence, the required differential equation of the given equation of curve is 𝑑2𝑦

𝑑𝑥2 − 𝑦 = 0 [1

Mark]

Hence, the correct answer is B.

12. Which of the following differential equation has 𝑦 = 𝑥 as one of its particular solution? [2

Marks]



(A) 𝑑2𝑦

𝑑𝑥2 − 𝑥2 𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 𝑥

(B) 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 𝑥

(C) 𝑑2𝑦

𝑑𝑥2 − 𝑥2 𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 0

(D) 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 0

The given equation of curve is 𝑦 = 𝑥.

Differentiating with respect to 𝑥, we get:

𝑑𝑦

𝑑𝑥= 1 … (1)


𝑑2𝑦

𝑑𝑥2 = 0 … (2) [1

Mark]

Now, on substituting the values of𝑦,𝑑2𝑦

𝑑𝑥2, and 𝑑𝑦

𝑑𝑥 from equation (1) and (2) in each of

the given alternatives, we find that only the differential equation given in alternative C is

correct.

𝑑2𝑦

𝑑𝑥2− 𝑥2

𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 0 − 𝑥2 ∙ 1 + 𝑥 ∙ 𝑥

= −𝑥2 + 𝑥2

= 0

Hence, the correct answer is C [1

Mark]

Exercise 9.4

For each of the differential equations in Exercises 1 to 10, find the general solution: [2 Marks]

1. 𝑑𝑦

𝑑𝑥=

1−cos 𝑥

1+cos 𝑥



Solution:

Given differential equation is 𝑑𝑦

𝑑𝑥=

1−cos𝑥

1+cos𝑥

⇒𝑑𝑦

𝑑𝑥=

2sin2 𝑥2

2cos2 𝑥2

= tan2𝑥

2

⇒𝑑𝑦

𝑑𝑥= (sec2

𝑥

2− 1)

Separating the variables, we get:

𝑑𝑦 = (sec2 𝑥

2− 1) 𝑑𝑥 [1

Mark]

Now, integrating both sides of this equation, we get

∫ 𝑑𝑦 = ∫ (sec2𝑥

2− 1) 𝑑𝑥 = ∫ sec2

𝑥

2𝑑𝑥 − ∫ 𝑑𝑥

⇒ 𝑦 = 2tan𝑥

2− 𝑥 + C

Hence, the required general solution of the given differential equation is 𝑦 = 2tan𝑥

2− 𝑥 + C[1

Mark]

2. 𝑑𝑦

𝑑𝑥= √4 − 𝑦2(−2 < 𝑦 < 2)

[2 Marks]

Solution:


𝑑𝑥= √4 − 𝑦2


⇒𝑑𝑦

√4−𝑦2 = 𝑑𝑥 [1

Mark]

Now, integrating both sides of this equation, we get:

∫𝑑𝑦

√4 − 𝑦2= ∫ 𝑑𝑥

⇒ sin−1𝑦

2= 𝑥 + C



⇒𝑦

2= sin(𝑥 + C)

⇒ 𝑦 = 2 sin(𝑥 + C)

Hence, the required general solution of the given differential equation is 𝑦 = 2 sin(𝑥 + C)[1

Mark]

3. 𝑑𝑦

𝑑𝑥+ 𝑦 = 1(𝑦 ≠ 1) [2

Marks]

Solution:


𝑑𝑥+ 𝑦 = 1

⇒ 𝑑𝑦 + 𝑦 𝑑𝑥 = 𝑑𝑥

⇒ 𝑑𝑦 = (1 − 𝑦)𝑑𝑥


⇒𝑑𝑦

1−𝑦= 𝑑𝑥 [1

Mark]

Now, integrating both sides, we get:

∫𝑑𝑦

1 − 𝑦= ∫ 𝑑 𝑥

⇒ log(1 − 𝑦) = 𝑥 + logC

⇒ −log𝐶 − log(1 − 𝑦) = 𝑥

⇒ log𝐶(1 − 𝑦) = −𝑥

⇒ 𝐶(1 − 𝑦) = 𝑒−𝑥

⇒ 1 − 𝑦 =1

C𝑒−𝑥

⇒ 𝑦 = 1 −1

C𝑒−𝑥

⇒ 𝑦 = 1 + 𝐴𝑒−𝑥 (where 𝐴 = −1

C)

Hence, the required general solution of the given differential equation is 𝑦 = 1 + 𝐴𝑒−𝑥 [1

Mark]



4. sec2 𝑥 tan 𝑦 𝑑𝑥 + sec2 𝑦 tan 𝑥 𝑑𝑦 = 0 [4

Marks]

Solution:

Given differential equation is sec2𝑥tan𝑦𝑑𝑥 + sec2𝑦tan𝑥𝑑𝑦 = 0

⇒sec2𝑥tan𝑦𝑑𝑥 + sec2𝑦tan𝑥𝑑𝑦

tan𝑥tan𝑦= 0

⇒sec2𝑥

tan𝑥𝑑𝑥 +

sec2𝑦

tan𝑦𝑑𝑦 = 0

⇒sec2𝑥

tan𝑥𝑑𝑥 = −

sec2𝑦

tan𝑦𝑑𝑦

Integrating both sides of this equation, we get:

∫ 𝑑𝑥 = − ∫sec2𝑦

tan𝑦𝑑𝑦 … (1) [1

Mark]

Let tan𝑥 = 𝑡

∴𝑑

𝑑𝑥(tan𝑥) =

𝑑𝑡

𝑑𝑥

⇒ sec2𝑥 =𝑑𝑡

𝑑𝑥

⇒ sec2𝑥𝑑𝑥 = 𝑑𝑡

Now, ∫sec2𝑥

tan𝑥𝑑𝑥 = ∫

1

𝑡𝑑𝑡

= log 𝑡

= log(tan 𝑥) [1

Mark]

Similarly, ∫sec2𝑥

tan𝑥𝑑𝑦 = log(tan 𝑦)

Substituting these values in equation (1), we get:

log(tan𝑥) = − log(tan𝑦) + logC

⇒ log(tan𝑥) = log (C

tan𝑦)

⇒ tan𝑥 =C

tan𝑦

⇒ tan𝑥 tan𝑦 = C



Hence, the required general solution of the given differential equation is tan𝑥 tan𝑦 = C [2

Marks]

5. (𝑒𝑥 + 𝑒−𝑥)𝑑𝑦 − (𝑒𝑥 − 𝑒−𝑥)𝑑𝑥 = 0 [2 Marks]

Solution:

Given differential equation is (𝑒𝑥 + 𝑒−𝑥)𝑑𝑦 − (𝑒𝑥 − 𝑒−𝑥)𝑑𝑥 = 0

⇒ (𝑒𝑥 + 𝑒−𝑥)𝑑𝑦 = (𝑒𝑥 − 𝑒−𝑥)𝑑𝑥

⇒ 𝑑𝑦 = [𝑒𝑥 − 𝑒−𝑥

𝑒𝑥 + 𝑒−𝑥] 𝑑𝑥


∫ 𝑑𝑦 = ∫ [𝑒𝑥 − 𝑒−𝑥

𝑒𝑥 + 𝑒−𝑥] 𝑑𝑥 + 𝐶

⇒ 𝑦 = ∫ [𝑒𝑥−𝑒−𝑥

𝑒𝑥+𝑒−𝑥] 𝑑𝑥 + C … (1)

Let 𝑒𝑥 + 𝑒−𝑥) = 𝑡


𝑑

𝑑𝑥(𝑒𝑥 + 𝑒−𝑥) =

𝑑𝑡

𝑑𝑥

⇒ 𝑒𝑥 − 𝑒−𝑥 =𝑑𝑡

𝑑𝑡

⇒ (𝑒𝑥 − 𝑒−𝑥)𝑑𝑥 = 𝑑𝑡 [1

Mark]


𝑦 = ∫1

𝑡

𝑑𝑡 + C

⇒ 𝑦 = log(𝑡) + C

⇒ 𝑦 = log(𝑒𝑥 + 𝑒−𝑥) + C

Hence,

the required general solution of the given differential equation is 𝑦 = log(𝑒𝑥 + 𝑒−𝑥) + C[1

Mark]



6. 𝑑𝑦

𝑑𝑥= (1 + 𝑥2)(1 + 𝑦2) [2

Marks]

Solution:


𝑑𝑥= (1 + 𝑥2)(1 + 𝑦2)

⇒𝑑𝑦

1+𝑦2 = (1 + 𝑥2)𝑑𝑥 [1

Mark]


∫𝑑𝑦

1 + 𝑦2= ∫(1 + 𝑥2) 𝑑𝑥

⇒ tan−1𝑦 = ∫ 𝑑𝑥 + ∫ 𝑥2𝑑𝑥

⇒ tan−1𝑦 = 𝑥 +𝑥3

3+ C

Hence, the required general solution of the given differential equation is tan−1𝑦 = 𝑥 +𝑥3

3+ C[1

Mark]

7. 𝑦log𝑦𝑑𝑥 − 𝑥𝑑𝑦 = 0 [2 Marks]

Solution:

Given differential equation is 𝑦log𝑦𝑑𝑥 − 𝑥𝑑𝑦 = 0

⇒ 𝑦log𝑦𝑑𝑥 = 𝑥𝑑𝑦

⇒𝑑𝑦

𝑦log𝑦=

𝑑𝑥

𝑥

Integrating both sides, we get:

∫𝑑𝑦

𝑦log𝑦= ∫

𝑑𝑥

𝑥… (1)

Let log 𝑦 = 𝑡

∴𝑑

𝑑𝑦(log𝑦) =

𝑑𝑡

𝑑𝑦



⇒1

𝑦=

𝑑𝑡

𝑑𝑦

⇒1

𝑦𝑑𝑦 = 𝑑𝑡 [1

Mark]


∫𝑑𝑡

𝑡= ∫

𝑑𝑥

𝑥

⇒ log𝑡 = log𝑥 + logC

⇒ log(log𝑦) = log𝐶𝑥

⇒ log𝑦 = C𝑥

⇒ 𝑦 = 𝑒𝑐𝑥

Hence, the required general solution of the given differential equation is 𝑦 = 𝑒𝑐𝑥 [1

Mark]

8. 𝑥5 𝑑𝑦

𝑑𝑥= −𝑦5

[2 Marks]

Solution:

Given differential equation is 𝑥5 𝑑𝑦

𝑑𝑥= −𝑦5

⇒𝑑𝑦

𝑦5= −

𝑑𝑥

𝑥5

⇒𝑑𝑥

𝑥5+

𝑑𝑦

𝑦5= 0


∫𝑑𝑥

𝑥5 + ∫𝑑𝑦

𝑦5 = 𝑘 [1

Mark]

⇒ ∫ 𝑥−5 𝑑𝑥 + ∫ 𝑦−5 𝑑𝑦 = 𝑘

⇒𝑥−4

−4+

𝑦−4

−4= 𝑘

⇒ 𝑥−4 + 𝑦−4 = −4𝑘

⇒ 𝑥−4 + 𝑦−4 = 𝐶 (𝐶 = −4𝑘)



Hence, the required general solution of the given differential equation is 𝑥−4 + 𝑦−4 = 𝐶 [1

Mark]

9. 𝑑𝑦

𝑑𝑥= sin−1 𝑥

[4 Marks]

Solution:


𝑑𝑥= sin−1 𝑥

⇒ 𝑑𝑦 = sin−1 𝑥 𝑑𝑥


∫ 𝑑𝑦 = ∫ sin−1 𝑥𝑑𝑥

⇒ 𝑦 = ∫(sin−1𝑥 · 1) 𝑑𝑥

⇒ 𝑦 = sin−1𝑥 · ∫(1) 𝑑𝑥 − ∫ [(𝑑

𝑑𝑥(sin−1𝑥) · ∫(1) 𝑑𝑥)] 𝑑𝑥

⇒ 𝑦 = sin−1𝑥 · 𝑥 − ∫ (1

√1 − 𝑥2· 𝑥) 𝑑𝑥

⇒ 𝑦 = 𝑥sin−1𝑥 + ∫−𝑥

√1−𝑥2𝑑𝑥 … (1) [1

Mark]

Let 1 − 𝑥2 = 𝑡

⇒𝑑

𝑑𝑥(1 − 𝑥2) =

𝑑𝑡

𝑑𝑥

⇒ −2𝑥 =𝑑𝑡

𝑑𝑥

⇒ 𝑥𝑑𝑥 = −1

2𝑑𝑡


𝑦 = 𝑥sin−1𝑥 + ∫1

2√𝑡𝑑𝑡 [1

Mark]

⇒ 𝑦 = 𝑥sin−1𝑥 +1

2· ∫(𝑡)−

12 𝑑𝑡



⇒ 𝑦 = 𝑥sin−1𝑥 +1

2·

𝑡12

12

+ C

⇒ 𝑦 = 𝑥sin−1𝑥 + √𝑡 + C

⇒ 𝑦 = 𝑥sin−1𝑥 + √1 − 𝑥2 + C [2

Marks]

Hence, the required general solution of the given differential equation is 𝑦 = 𝑥sin−1𝑥 +

√1 − 𝑥2 + C

10. 𝑒𝑥tan𝑦𝑑𝑥 + (1 − 𝑒𝑥)sec2𝑦𝑑𝑦 = 0 [4

Marks]

Solution:

Given differential equation is 𝑒𝑥tan𝑦𝑑𝑥 + (1 − 𝑒𝑥)sec2𝑦𝑑𝑦 = 0

(1 − 𝑒𝑥)sec2𝑦𝑑𝑦 = −𝑒𝑥tan𝑦𝑑𝑥


sec2𝑦

tan𝑦𝑑𝑦 =

−𝑒𝑥

1 − 𝑒𝑥𝑑𝑥


∫sec2𝑦

tan𝑦𝑑𝑦 = ∫

−𝑒𝑥

1−𝑒𝑥 𝑑𝑥 … (1) [1

Mark]

Let tan 𝑦 = 𝑢

⇒𝑑

𝑑𝑦(tan𝑦) =

𝑑𝑢

𝑑𝑦

⇒ sec2𝑦 =𝑑𝑢

𝑑𝑦

⇒ sec2𝑦𝑑𝑦 = 𝑑𝑢

∴ ∫sec2𝑦

tan𝑦𝑑𝑦 = ∫

𝑑𝑢

𝑢= log𝑢 = log(tan𝑦) [1

Mark]

Now, Let 1 − 𝑒𝑥 = 𝑡.

∴𝑑

𝑑𝑥(1 − 𝑒𝑥) =

𝑑𝑡

𝑑𝑥



⇒ −𝑒𝑥 =𝑑𝑡

𝑑𝑥

⇒ −𝑒𝑥𝑑𝑥 = 𝑑𝑡

⇒ ∫−𝑒𝑥

1−𝑒𝑥 𝑑𝑥 = ∫𝑑𝑡

𝑡= log𝑡 = log(1 − 𝑒𝑥) [1

Mark]

Substituting the values of ∫sec2𝑦

tan𝑦𝑑𝑦 and ∫

−𝑒𝑥

1−𝑒𝑥 𝑑𝑥

⇒ log(tan𝑦) = log(1 − 𝑒𝑥) + log𝐶

⇒ log(tan𝑦) = log[C(1 − 𝑒𝑥)]

⇒ tan𝑦 = C(1 − 𝑒𝑥)

Hence, the required general solution of the given differential equation is tan𝑦 = C(1 − 𝑒𝑥)[1

Mark]

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying

the given condition:

11. (𝑥3 + 𝑥2 + 𝑥 + 1)𝑑𝑦

𝑑𝑥= 2𝑥2 + 𝑥; 𝑦 = 1 when 𝑥 = 0

[6 Marks]

Solution:

Given differential equation is (𝑥3 + 𝑥2 + 𝑥 + 1)𝑑𝑦

𝑑𝑥= 2𝑥2 + 𝑥

⇒𝑑𝑦

𝑑𝑥=

2𝑥2 + 𝑥

(𝑥3 + 𝑥2 + 𝑥 + 1)

⇒ 𝑑𝑦 =2𝑥2 + 𝑥

(𝑥 + 1)(𝑥2 + 1)𝑑𝑥


∫ 𝑑𝑦 = ∫2𝑥2 + 𝑥

(𝑥 + 1)(𝑥2 + 1)𝑑𝑥 … (1)

Let 2𝑥2+𝑥

(𝑥+1)(𝑥2+1)=

𝐴

𝑥+1+

𝐵𝑥+𝐶

𝑥2+1… (2)

⇒2𝑥2+𝑥

(𝑥+1)(𝑥2+1)=

𝐴𝑥2+𝐴+(𝐵𝑥+𝐶)(𝑥+1)

(𝑥+1)(𝑥2+1) [1

Mark]⇒ 2𝑥2 + 𝑥 = 𝐴𝑥2 + 𝐴 + 𝐵𝑥2 + 𝐵𝑥 + 𝐶𝑥 + 𝐶

⇒ 2𝑥2 + 𝑥 = (𝐴 + 𝐵)𝑥2 + (𝐵 + 𝐶)𝑥 + (𝐴 + 𝐶)



Comparing the coefficients of 𝑥2 and 𝑥, we get:

𝐴 + 𝐵 = 2

𝐵 + 𝐶 = 1

𝐴 + 𝐶 = 0

Solving these equations, we get:

𝐴 =1

2, 𝐵 =

3

2 and 𝐶 =

−1

2 [1

Mark]

Substituting the values of 𝐴, 𝐵, and C in equation (2), we get:

2𝑥2 + 𝑥

(𝑥 + 1)(𝑥2 + 1)=

1

2·

1

(𝑥 + 1)+

1

2

(3𝑥 − 1)

(𝑥2 + 1)

Therefore, equation (1) becomes:

∫ 𝑑𝑦 =1

2∫

1

𝑥+1𝑑𝑥 +

1

2∫

3𝑥−1

𝑥2+1𝑑𝑥 [1

Mark]

⇒ 𝑦 =1

2log(𝑥 + 1) +

3

2∫

𝑥

𝑥2 + 1𝑑𝑥 −

1

2∫

1

𝑥2 + 1𝑑𝑥

⇒ 𝑦 =1

2log(𝑥 + 1) +

3

4· ∫

2𝑥

𝑥2 + 1𝑑𝑥 −

1

2tan−1𝑥 + 𝐶

⇒ 𝑦 =1

2log(𝑥 + 1) +

3

4log(𝑥2 + 1) −

1

2tan−1𝑥 + 𝐶

⇒ 𝑦 =1

4[2 log(𝑥 + 1) + 3 log(𝑥2 + 1)] −

1

2tan−1𝑥 + 𝐶

⇒ 𝑦 =1

4[(𝑥 + 1)2(𝑥2 + 1)3] −

1

2tan−1𝑥 + 𝐶 … (3) [1

Mark]

Now, 𝑦 = 1 when 𝑥 = 0

⇒ I =1

4log(1) −

1

2tan−10 + C

⇒ 1 =1

4× 0 −

1

2× 0 + C

⇒ C = 1 [1

Mark]

Substituting 𝐶 = 1 in equation (3), we get:

𝑦 =1

4[log(𝑥 + 1)2(𝑥2 + 1)3] −

1

2tan−1𝑥 + 1



Hence, the required general solution of the given differential equation is 𝑦 =1

4[log(𝑥 +

1)2(𝑥2 + 1)3] −1

2tan−1𝑥 + 1

[1 Mark]

12. 𝑥(𝑥2 − 1)𝑑𝑦

𝑑𝑥= 1; 𝑦 = 0 when 𝑥 = 2 [6

Marks]

Solution:

Given differential equation is 𝑥(𝑥2 − 1)𝑑𝑦

𝑑𝑥= 1

⇒ 𝑑𝑦 =𝑑𝑥

𝑥(𝑥2 − 1)

⇒ 𝑑𝑦 =1

𝑥(𝑥 − 1)(𝑥 + 1)𝑑𝑥


∫ 𝑑𝑦 = ∫1

𝑥(𝑥−1)(𝑥+1)𝑑𝑥 … (1) [1

Mark]

Let 1

𝑥(𝑥−1)(𝑥+1)=

𝐴

𝑥+

𝐵

𝑥−1+

𝐶

𝑥+1 … (2)

⇒1

𝑥(𝑥−1)(𝑥+1)=

𝐴(𝑥−1)(𝑥+1)+𝐵𝑥(𝑥+1)+𝐶𝑥(𝑥−1)

𝑥(𝑥−1)(𝑥+1) [1

Mark]

=(𝐴 + 𝐵 + 𝐶)𝑥2 + (𝐵 − 𝐶)𝑥 − 𝐴

𝑥(𝑥 − 1)(𝑥 + 1)

Comparing the coefficients of 𝑥2, 𝑥, and constant, we get:

𝐴 = −1

𝐵 − 𝐶 = 0

𝐴 + 𝐵 + 𝐶 = 0

Solving these equations, we get

𝐵 =1

2 and 𝐶 =

1

2 [1

Mark]

Substituting the values of 𝐴, 𝐵, and 𝐶 in equation (2), we get:



1

𝑥(𝑥 − 1)(𝑥 + 1)=

−1

𝑥+

1

2(𝑥 − 1)+

1

2(𝑥 + 1)


∫ 𝑑𝑦 = − ∫1

𝑥𝑑𝑥 +

1

2∫

1

𝑥 − 1𝑑𝑥 +

1

2∫

1

𝑥 + 1𝑑𝑥

⇒ 𝑦 = −log𝑥 +1

2log(𝑥 − 1) +

1

2log(𝑥 + 1) + log𝑘 [1

Mark]

⇒ 𝑦 =1

2log [

𝑘2(𝑥 − 1)(𝑥 + 1)

𝑥2] … (3)


⇒ 0 =1

2log [

𝑘2(2 − 1)(2 + 1)

4]

⇒ log (3𝑘2

4) = 0

⇒3𝑘2

4= 1

⇒ 3𝑘2 = 4

⇒ 𝑘2 =4

3 [1

Mark]

Substituting the value of 𝑘2 in equation (3), we get:

𝑦 =1

2log [

4(𝑥 − 1)(𝑥 + 1)

3𝑥2]

⇒ 𝑦 =1

2log [

4(𝑥2 − 1)

3𝑥2]

Hence, the required general solution of the given differential equation is 𝑦 =1

2log [

4(𝑥2−1)

3𝑥2 ][1

Mark]

13. cos (𝑑𝑦

𝑑𝑥) = 𝑎(𝑎 ∈ 𝑅); 𝑦 = 1 when 𝑥 = 0 [2

Marks]

Solution:

Given differential equation is cos (𝑑𝑦

𝑑𝑥) = 𝑎



⇒𝑑𝑦

𝑑𝑥= cos−1𝑎

⇒ 𝑑𝑦 = cos−1𝑎𝑑𝑥


∫ 𝑑𝑦 = cos−1𝑎 ∫ 𝑑𝑥

⇒ 𝑦 = cos−1𝑎 · 𝑥 + 𝐶

⇒ 𝑦 = 𝑥cos−1𝑎 + C … (1)


⇒ 1 = 0 cos−1 𝑎 + 𝐶

⇒ 𝐶 = 1 [1

Mark]


𝑦 = 𝑥cos−1𝑎 + 1

⇒𝑦 − 1

𝑥= cos−1𝑎

⇒ cos (𝑦 − 1

𝑥) = 𝑎

Hence, the required general solution of the given differential equation is cos (𝑦−1

𝑥) = 𝑎 [1

Mark]

14. 𝑑𝑦

𝑑𝑥= 𝑦tan𝑥; 𝑦 = 1 when 𝑥 = 0 [2

Marks]

Solution:


𝑑𝑥= 𝑦tan𝑥

⇒𝑑𝑦

𝑦= tan𝑥𝑑𝑥


∫𝑑𝑦

𝑦= − ∫ tan𝑥 𝑑𝑥

⇒ log𝑦 = log(sec𝑥) + log𝐶



⇒ log𝑦 = log(C sec 𝑥)

⇒ 𝑦 = Csec𝑥 … (1) [1

Mark]

Now,𝑦 = 1 when 𝑥 = 0

⇒ 1 = 𝐶 × sec0

⇒ 1 = 𝐶 × 1

⇒ 𝐶 = 1


𝑦 = sec 𝑥

⇒ 𝑦 − sec 𝑥 = 0

Hence, the required general solution of the given differential equation is 𝑦 − sec 𝑥 = 0 [1

Mark]

15. Find the equation of a curve passing through the point (0, 0) and whose differential

equation is. 𝑦′ = 𝑒𝑥 sin 𝑥 [4

Marks]

Solution:

Given that the differential equation of the curve is 𝑦′ = 𝑒𝑥sin𝑥

⇒𝑑𝑦

𝑑𝑥= 𝑒𝑥sin𝑥

⇒ 𝑑𝑦 = 𝑒𝑥sin𝑥


∫ 𝑑 𝑦 = ∫ 𝑒𝑥 sin𝑥𝑑𝑥 … (1) [1

Mark]

𝐼 = ∫ 𝑒𝑥 sin𝑥𝑑𝑥

⇒ 𝐼 = sin𝑥 ∫ 𝑒𝑥 𝑑𝑥 − ∫ (𝑑

𝑑𝑥(sin𝑥) · ∫ 𝑒𝑥 𝑑𝑥) 𝑑𝑥

⇒ 𝐼 = sin𝑥 · 𝑒𝑥 − ∫ cos𝑥 · 𝑒𝑥𝑑𝑥

⇒ 𝐼 = sin𝑥 · 𝑒𝑥 − [cos𝑥 · ∫ 𝑒𝑥𝑑𝑥 − ∫ (𝑑

𝑑𝑥(cos𝑥) · ∫ 𝑒𝑥 𝑑𝑥) 𝑑𝑥]



⇒ 𝐼 = sin𝑥 · 𝑒𝑥 − [cos𝑥 · 𝑒𝑥 − ∫(−sin𝑥) · 𝑒𝑥𝑑𝑥]

⇒ 𝐼 = 𝑒𝑥sin𝑥 − 𝑒𝑥cos𝑥 − 𝐼

⇒ 2𝐼 = 𝑒𝑥(sin𝑥 − cos𝑥)

⇒ 𝐼 =𝑒𝑥(sin𝑥−cos𝑥)

2 [1

Mark]


𝑦 =𝑒𝑥(sin𝑥 − cos𝑥)

2+ C … (2)

Now the curve passes through point (0, 0)

∴ 0 =𝑒0(sin0 − cos0)

2+ C

⇒ 0 =1(0 − 1)

2+ C

⇒ 𝐶 =1

2 [1

Mark]

Substituting 𝐶 =1

2 in equation (2), we get:

𝑦 =𝑒𝑥(sin𝑥 − cos𝑥)

2+

1

2

⇒ 2𝑦 = 𝑒𝑥(sin𝑥 − cos𝑥) + 1

⇒ 2𝑦 − 1 = 𝑒𝑥(sin𝑥 − cos𝑥)

Hence, the required equation of the given curve is 2𝑦 − 1 = 𝑒𝑥(sin 𝑥 − cos 𝑥) [1

Mark]

16. For the differential equation find the solution curve passing through the point (1, –1). [4

Marks]

Solution:

Given differential equation of the curve is 𝑥𝑦𝑑𝑦

𝑑𝑥= (𝑥 + 2)(𝑦 + 2)

⇒ (𝑦

𝑦 + 2) 𝑑𝑦 = (

𝑥 + 2

𝑥) 𝑑𝑥

⇒ (1 −2

𝑦+2) 𝑑𝑦 = (1 +

2

𝑥)




∫ (1 −2

𝑦 + 2) 𝑑𝑦 = ∫ (1 +

2

𝑥) 𝑑𝑥

⇒ ∫ 𝑑 𝑦 − 2 ∫1

𝑦 + 2𝑑𝑦 = ∫ 𝑑𝑥 + 2 ∫

1

𝑥𝑑𝑥

⇒ 𝑦 − 2log(𝑦 + 2) = 𝑥 + 2log𝑥 + C

⇒ 𝑦 − 𝑥 − C = log𝑥2 + log(𝑦 + 2)2

⇒ 𝑦 − 𝑥 − C = log[𝑥2(𝑦 + 2)2] … (1) [2

Marks]

Now, the curve passed through point (1, −1)

⇒ −1 − 1 − 𝐶 = log[(1)2(−1 + 2)2]

⇒ −2 − 𝐶 = log1 = 0

⇒ 𝐶 = −2 [1

Mark]

Substituting 𝐶 =– 2 in equation (1), we get:

𝑦 − 𝑥 + 2 = log[𝑥2(𝑦 + 2)2]

Hence, the required solution of the given curve is 𝑦 − 𝑥 + 2 = log[𝑥2(𝑦 + 2)2] [1

Mark]

17. Find the equation of a curve passing through the point (0, – 2) given that at any point

(𝑥, 𝑦) on the curve, the product of the slope of its tangent and 𝑦-coordinate of the point

is equal to the 𝑥-coordinate of the point. [4

Marks]

Solution:

Let 𝑥 and 𝑦 be the x-coordinate and 𝑦-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation,

𝑑𝑦

𝑑𝑥

According to the given information, we get:

𝑦 ∙𝑑𝑦

𝑑𝑥= 𝑥



⇒ 𝑦𝑑𝑦 = 𝑥𝑑𝑥


∫ 𝑦𝑑𝑦 = ∫ 𝑥𝑑𝑥

⇒𝑦2

2=

𝑥2

2+ 𝐶

⇒ 𝑦2 − 𝑥2 = 2𝐶 … (1) [2

Marks]

Now, the curve passes through point (0, – 2).

∴ (– 2)2– 02 = 2𝐶

⇒ 2𝐶 = 4 [1

Mark]

Substituting 2𝐶 = 4 in equation (1), we get:

𝑦2– 𝑥2 = 4

Hence, the required equation of the curve is 𝑦2– 𝑥2 = 4 [1

Mark]

18. At any point (𝑥, 𝑦) of a curve, the slope of the tangent is twice the slope of the line segment

joining the point of contact to the point (– 4, – 3). Find the equation of the curve given that

it passes through (– 2, 1).

[4 Marks]

Solution:

Given that (𝑥, 𝑦) is the point of contact of the curve and its tangent.

Let 𝑚1 be the slope of the line segment joining the given points and 𝑚2 be the slope of the

tangent

We know, slope (𝑚1) of the line segment joining (𝑥, 𝑦) and (– 4, – 3) is 𝑥+3

𝑥+4

And slope (𝑚2) of the tangent = 𝑑𝑦

𝑑𝑥

According to the given information:

𝑚2 = 2𝑚1

⇒𝑑𝑦

𝑑𝑥=

2(𝑦 + 3)

𝑥 + 4



⇒𝑑𝑦

𝑦 + 3=

2𝑑𝑥

𝑥 + 4


∫𝑑𝑦

𝑦 + 3= 2 ∫

𝑑𝑥

𝑥 + 4

⇒ log(𝑦 + 3) = 2 log(𝑥 + 4) + log𝐶

⇒ log(𝑦 + 3) log𝐶(𝑥 + 4)2

⇒ 𝑦 + 3 = 𝐶(𝑥 + 4)2 … (1) [2

Marks]

This is the general equation of the curve.

It is given that it passes through point (– 2, 1).

⇒ 1 + 3 = 𝐶(−2 + 4)2

⇒ 4 = 4𝐶

⇒ 𝐶 = 1 [1

Mark]


𝑦 + 3 = (𝑥 + 4)2

Hence, the required equation of the curve is 𝑦 + 3 = (𝑥 + 4)2 [1

Mark]

19. The volume of spherical balloon being inflated changes at a constant rate. If initially its

radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after 𝑡 seconds.

[4 Marks]

Solution:

Let the rate of change of the volume of the balloon be 𝑘 (where 𝑘 is a constant).

⇒𝑑𝑣

𝑑𝑡= 𝑘

⇒𝑑

𝑑𝑡(

4

3𝜋𝑟3) = 𝑘 [Volume of sphere =

4

3𝜋𝑟3]

⇒4

3𝜋 · 3𝑟2 ·

𝑑𝑟

𝑑𝑡= 𝑘



⇒ 4𝜋𝑟2𝑑𝑟 = 𝑘𝑑𝑡 [1

Mark]


4𝜋 ∫ 𝑟2 𝑑𝑟 = 𝑘 ∫ 𝑑𝑡

⇒ 4𝜋 ·𝑟3

3= 𝑘𝑡 + C

⇒ 4𝜋𝑟3 = 3(𝑘𝑡 + C) … (1)

Now, at 𝑡 = 0, 𝑟 = 3:

⇒ 4𝜋 × 33 = 3(𝑘 × 0 + 𝐶)

⇒ 108𝜋 = 3𝐶

⇒ 𝐶 = 36𝜋 [1

Mark]

At 𝑡 = 3, 𝑟 = 6:

⇒ 4𝜋 × 63 = 3(𝑘 × 3 + 𝐶)

⇒ 864𝜋 = 3(3𝑘 + 36𝜋)

⇒ 3𝑘 =– 288𝜋 – 36𝜋 = 252𝜋

⇒ 𝑘 = 84𝜋 [1

Mark]

Substituting the values of 𝑘 and 𝐶 in equation (1), we get:

4𝜋𝑟3 = 3[84𝜋𝑡 + 36𝜋]

⇒ 4𝜋𝑟3 = 4𝜋(63𝑡 + 27)

⇒ 𝑟3 = 63𝑡 + 27

⇒ 𝑟 = (63𝑡 + 27)13

Thus, the radius of the balloon after 𝑡 seconds is (63𝑡 + 27)1

3 [1

Mark]

20. In a bank, principal increases continuously at the rate of 𝑟% per year. Find the value of 𝑟 if

Rs 100 doubles itself in 10 years (log𝑒 2 = 0.6931).

[4 Marks]

Solution:



Let 𝑝 and 𝑡 represent the principal and time respectively.

It is given that the principal increases continuously at the rate of 𝑟% per year.

⇒𝑑𝑝

𝑑𝑡= (

𝑟

100) 𝑝

⇒𝑑𝑝

𝑝= (

𝑟

100) 𝑑𝑡


∫𝑑𝑝

𝑝=

𝑟

100∫ 𝑑𝑡

⇒ log𝑝 =𝑟𝑡

100+ 𝑘

⇒ 𝑝 = 𝑒𝑟𝑡

100+𝑘 … (1) [1

Mark]

It is given that when 𝑡 = 0, 𝑝 = 100.

⇒ 100 = 𝑒𝑘 … (2) [1

Mark]

Now, if 𝑡 = 10, then 𝑝 = 2 × 100 = 200.


200 = 𝑒𝑟

10+𝑘

⇒ 200 = 𝑒𝑟

10 · 𝑒𝑘

⇒ 200 = 𝑒𝑟

10 · 100 (From (2))

⇒ 𝑒𝑟

10 = 2

⇒𝑟

10= log𝑒2

⇒𝑟

10= 0.6931

⇒ 𝑟 = 0.6931

Hence, the value of 𝑟 is 6.93% [2

Marks]

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs



1000 is deposited with this bank, how much will it worth after 10 years (𝑒0.5 = 1.648) [4

Marks]

Solution:

Let 𝑝 and 𝑡 be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.

⇒𝑑𝑝

𝑑𝑡= (

5

100) 𝑝

⇒𝑑𝑝

𝑑𝑡=

𝑝

20

⇒𝑑𝑝

𝑝=

𝑑𝑡

20 [1

Mark]


∫𝑑𝑝

𝑝=

1

20∫ 𝑑𝑡

⇒ log 𝑝 =𝑡

20+ 𝐶

⇒ 𝑝 = 𝑒𝑡

20+𝐶 … (1) [1

Mark]

Now, when 𝑡 = 0, 𝑝 = 1000.

⇒ 1000 = 𝑒𝐶 … (2) [1

Mark]

At 𝑡 = 10, equation (1) becomes:

𝑝 = 𝑒12

+C

⇒ 𝑝 = 𝑒0.5 × 𝑒C

⇒ 𝑝 = 1.648 × 1000

⇒ 𝑝 = 1648

Hence, after 10 years the amount will worth Rs 1648. [1

Mark]

22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In

how many hours will the count reach 2,00,000, if the rate of growth of bacteria is



proportional to the number present?

[6 Marks]

Solution:

Let 𝑦 be the number of bacteria at any instant 𝑡.

It is given that the rate of growth of the bacteria is proportional to the number present.

∴𝑑𝑦

𝑑𝑡∝ 𝑦

⇒𝑑𝑦

𝑑𝑡= 𝑘𝑦 (where 𝑘 is a constant)

⇒𝑑𝑦

𝑦= 𝑘𝑑𝑡 [1

Mark]


∫𝑑𝑦

𝑦= 𝑘 ∫ 𝑑𝑡

⇒ log𝑦 = 𝑘𝑡 + C (1) [1

Mark]

Let 𝑦0 be the number of bacteria at 𝑡 = 0.

⇒ log 𝑦0 = 𝐶

Substituting the value of 𝐶 in equation (1), we get:

log𝑦 = 𝑘𝑡 + log𝑦0

⇒ log𝑦 − log𝑦0 = 𝑘𝑡

⇒ log (𝑦

𝑦0) = 𝑘𝑡

⇒ 𝑘𝑡 = log (𝑦

𝑦0) … (2) [1

Mark]

Also, it is given that the number of bacteria increases by 10% in 2 hours.

⇒ 𝑦 =110

100𝑦0

⇒𝑦

𝑦0=

11

10… (3) [1

Mark]


𝑘 · 2 = log (11

10)



⇒ 𝑘 =1

2log (

11

10)


1

2log (

11

10) · 𝑡 = log (

𝑦

𝑦0)

⇒ 𝑡 =2 log(

𝑦

𝑦0)

log(11

10)

… (4) [1

Mark]

Now, let the time when the number of bacteria increases from 100000 to 200000 be 𝑡1.

⇒ 𝑦 = 2𝑦0 at 𝑡 = 𝑡1

From equation (4), we get:

𝑡1 =2 log (

𝑦𝑦0

)

log (1110)

=2log2

log (1110)

Hence, in 2log2

log(11

10) hours the number of bacteria increases from 100000 to 200000. [1

Mark]

23. The general solution of the differential equation 𝑑𝑦

𝑑𝑥= 𝑒𝑥+𝑦 is [2

Marks]

(A) 𝑒𝑥 + 𝑒−𝑦 = C

(B) 𝑒𝑥 + 𝑒𝑦 = C

(C) 𝑒−𝑥 + 𝑒𝑦 = C

(D) 𝑒−𝑥 + 𝑒−𝑦 = C

Solution:


𝑑𝑥= 𝑒𝑥+𝑦 = 𝑒𝑥 · 𝑒𝑦

⇒𝑑𝑦

𝑒𝑦= 𝑒𝑥𝑑𝑥

⇒ 𝑒−𝑦𝑑𝑦 = 𝑒𝑥𝑑𝑥




∫ 𝑒−𝑦 𝑑𝑦 = ∫ 𝑒𝑥 𝑑𝑥 [1

Mark]

⇒ −𝑒−𝑦 = 𝑒𝑥 + 𝑘

⇒ 𝑒𝑥 + 𝑒−𝑦 = −𝑘

⇒ 𝑒𝑥 + 𝑒−𝑦 = 𝑐 … (𝑐 = −𝑘)

Therefore, the general solution of the given differential equation is 𝑒𝑥 + 𝑒−𝑦 = 𝑐 … (𝑐 = −𝑘)

Hence, the correct answer is A. [1

Mark]

Exercise 9.5

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous

and solve each of them.

1. (𝑥2 + 𝑥𝑦)𝑑𝑦 = (𝑥2 + 𝑦2)𝑑𝑥

[6 Marks]

Solution:

Given differential equation is (𝑥2 + 𝑥𝑦)𝑑𝑦 = (𝑥2 + 𝑦2)𝑑𝑥

It can be written as 𝑑𝑦

𝑑𝑥=

𝑥2+𝑦2

𝑥2+𝑥𝑦

Let 𝐹(𝑥, 𝑦) =𝑥2+𝑦2

𝑥2+𝑥𝑦

Now, 𝐹(𝜆𝑥, 𝜆𝑦) =(𝜆𝑥)2+(𝜆𝑦)2

(𝜆𝑥)2+(𝜆𝑥)(𝜆𝑦)=

𝑥2+𝑦2

𝑥2+𝑥𝑦= 𝜆0 · 𝐹(𝑥, 𝑦)

This shows that the given equation is a homogeneous equation. [1

Mark]

To solve it, we make the substitution as:

𝑦 = 𝑣𝑥


𝑑𝑦

𝑑𝑥= 𝑣 + 𝑥

𝑑𝑣

𝑑𝑥

Substituting the values of v and 𝑑𝑦

𝑑𝑥 in the given equation, we get:



𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑥2+(𝑣𝑥)2

𝑥2+𝑥(𝑣𝑥) [1

Mark]

⇒ 𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

1 + 𝑣2

1 + 𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

1 + 𝑣2

1 + 𝑣− 𝑣 =

(1 + 𝑣2) − 𝑣(1 + 𝑣)

1 + 𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

1 − 𝑣

1 + 𝑣

⇒ (1 + 𝑣

1 − 𝑣) = 𝑑𝑣 =

𝑑𝑥

𝑥

⇒ (2 − 1 + 𝑣

1 − 𝑣) 𝑑𝑣 =

𝑑𝑥

𝑥

⇒ (2

1−𝑣− 1) 𝑑𝑣 =

𝑑𝑥

𝑥 [1

Mark]


−2 log(1 − 𝑣) − 𝑣 = log𝑥 − log𝑘

⇒ 𝑣 = −2 log(1 − 𝑣) − log𝑥 + log𝑘

⇒ 𝑣 = log [𝑘

𝑥(1 − 𝑣)2]

⇒𝑦

𝑥= log [

𝑘

𝑥 (1 −𝑦𝑥)

2]

⇒𝑦

𝑥= log [

𝑘𝑥

(𝑥 − 𝑦)2]

⇒𝑘𝑥

(𝑥 − 𝑦)2= 𝑒

𝑦𝑥

⇒ (𝑥 − 𝑦)2 = 𝑘𝑥𝑒𝑦𝑥

Hence, the required solution of the given differential equation is (𝑥 − 𝑦)2 = 𝑘𝑥𝑒𝑦

𝑥 [1

Mark]

2. 𝑦′ =𝑥+𝑦

𝑥 [4

Marks]



Solution:

The given differential equation is 𝑦′ =𝑥+𝑦

𝑥

⇒𝑑𝑦

𝑑𝑥=

𝑥 + 𝑦

𝑥… (1)

Let 𝐹(𝑥, 𝑦) =𝑥+𝑦

𝑥.

Now, 𝐹(𝜆𝑥, 𝜆𝑦) =𝜆𝑥+𝜆𝑦

𝜆𝑥=

𝑥+𝑦

𝑥= 𝜆0𝐹(𝑥, 𝑦)

Thus, the given equation is a homogeneous equation. [1

Mark]


𝑦 = 𝑣𝑥


𝑑𝑦


𝑑𝑣

𝑑𝑥

Substituting the values of 𝑦 and 𝑑𝑦

𝑑𝑥 in equation (1), we get:

𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑥 + 𝑣𝑥

𝑥


𝑑𝑥= 1 + 𝑣

𝑥𝑑𝑣

𝑑𝑥= 1

⇒ 𝑑𝑣 =𝑑𝑥

𝑥 [2

Marks]


𝑣 = log𝑥 + C

⇒𝑦

𝑥= log𝑥 + C

⇒ 𝑦 = 𝑥log𝑥 + C𝑥

Hence, the required solution of the given differential equation is 𝑦 = 𝑥log𝑥 + C𝑥 [1

Mark]

3. (𝑥 − 𝑦)𝑑𝑦 − (𝑥 + 𝑦)𝑑𝑥 = 0

[4 Marks]



Solution:

Given differential equation is (𝑥 − 𝑦)𝑑𝑦 − (𝑥 + 𝑦)𝑑𝑥 = 0

⇒𝑑𝑦

𝑑𝑥=

𝑥 + 𝑦

𝑥 − 𝑦… (1)

Let 𝐹(𝑥, 𝑦) =𝑥+𝑦

𝑥−𝑦

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =𝜆𝑥 + 𝜆𝑦

𝜆𝑥 − 𝜆𝑦=

𝑥 + 𝑦

𝑥 − 𝑦= 𝜆0 · 𝐹(𝑥, 𝑦)

Thus, the given differential equation is a homogeneous equation. [1

Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥




𝑑𝑥

𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑥 + 𝑣𝑥

𝑥 − 𝑣𝑥=

1 + 𝑣

1 − 𝑣

𝑥𝑑𝑣

𝑑𝑥=

1+𝑣

1−𝑣− 𝑣 =

1+𝑣−𝑣(1−𝑣)

1−𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

1 + 𝑣2

1 − 𝑣

⇒1 − 𝑣

(1 + 𝑣2)𝑑𝑣 =

𝑑𝑥

𝑥

⇒ (1

1+𝑣2 −𝑣

1−𝑣2) 𝑑𝑣 =𝑑𝑥

𝑥 [1

Mark]


tan−1𝑣 −1

2log(1 + 𝑣2) = log𝑥 + 𝐶

⇒ tan−1 (𝑦

𝑥) −

1

2log [1 + (

𝑦

𝑥)

2

] = log𝑥 + 𝐶

⇒ tan−1 (𝑦

𝑥) −

1

2log (

𝑥2 + 𝑦2

𝑥2 ) = log𝑥 + 𝐶

⇒ tan−1 (𝑦

𝑥) −

1

2[log(𝑥2 + 𝑦2) − log𝑥2] = log𝑥 + 𝐶

⇒ tan−1 (𝑦

𝑥) =

1

2log(𝑥2 + 𝑦2) + C

Hence, the required solution of the given differential equation is

tan−1 (𝑦

𝑥) =

1

2log(𝑥2 + 𝑦2) + C [2

Marks]

4. (𝑥2 − 𝑦2)𝑑𝑥 + 2𝑥𝑦𝑑𝑦 = 0 [4

Marks]

Solution:

The given differential equation is (𝑥2 − 𝑦2)𝑑𝑥 + 2𝑥𝑦𝑑𝑦 = 0

⇒𝑑𝑦

𝑑𝑥=

−(𝑥2 − 𝑦2)

2𝑥𝑦… (1)

Let 𝐹(𝑥, 𝑦) =−(𝑥2−𝑦2)

2𝑥𝑦



∴ 𝐹(𝜆𝑥, 𝜆𝑦) = [(𝜆𝑥)2 − (𝜆𝑦)2

2(𝜆𝑥)(𝜆𝑦)] =

−(𝑥2 − 𝑦2)

2𝑥𝑦= 𝜆0 · 𝐹(𝑥, 𝑦)

Therefore, the given differential equation is a homogeneous equation. [1

Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥 [1

Mark]


𝑑𝑥

𝑣 + 𝑥𝑑𝑣

𝑑𝑥= − [

𝑥2 − (𝑣𝑥)2

2𝑥 · (𝑣𝑥)]


𝑑𝑥=

𝑣2 − 1

2𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣2 − 1

2𝑣− 𝑣 =

𝑣2 − 1 − 2𝑣2

2𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥= −

(1 + 𝑣2)

2𝑣

⇒2𝑣

1+𝑣2 𝑑𝑣 = −𝑑𝑥

𝑥 [1

Mark]


log(1 + 𝑣2) = −log𝑥 + log𝐶 = log𝐶

𝑥

⇒ 1 + 𝑣2 =𝐶

𝑥

⇒ [1 +𝑦2

𝑥2] =

𝐶

𝑥

⇒ 𝑥2 + 𝑦2 = 𝐶𝑥

Hence, the required solution of the given differential equation is 𝑥2 + 𝑦2 = 𝐶𝑥 [1

Mark]

5. 𝑥2 𝑑𝑦

𝑑𝑥− 𝑥2 − 2𝑦2 + 𝑥𝑦 [4

Marks]



Solution:

Given differential equation is 𝑥2 𝑑𝑦

𝑑𝑥− 𝑥2 − 2𝑦2 + 𝑥𝑦

𝑑𝑦

𝑑𝑥=

𝑥2 − 2𝑦2 + 𝑥𝑦

𝑥2… (1)

Let 𝐹(𝑥, 𝑦) =𝑥2−2𝑦2+𝑥𝑦

𝑥2

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =(𝜆𝑥)2−2(𝜆𝑦)2+(𝜆𝑥)(𝜆𝑦)

(𝜆𝑥)2 =𝑥2−2𝑦2+𝑥𝑦

𝑥2 = 𝜆0 · 𝐹(𝑥, 𝑦) [1

Mark]

Therefore, the given differential equation is a homogeneous equation.


𝑦 = 𝑣𝑥

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥 [1

Mark]


𝑑𝑥

𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑥2 − 2(𝑣𝑥)2 + 𝑥 · (𝑣𝑥)

𝑥2


𝑑𝑥= 1 − 2𝑣2 + 𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥= 1 − 2𝑣2

⇒𝑑𝑣

1 − 2𝑣2=

𝑑𝑥

𝑥

⇒1

2·

𝑑𝑣

12 − 𝑣2

=𝑑𝑥

𝑥

⇒1

2· [

𝑑𝑣

(1

√2)

2−𝑣2

] =𝑑𝑥

𝑥 [1

Mark]


1

2·

1

2 ×1

√2

log |

1

√2+ 𝑣

√2 − 𝑣| = log|𝑥| + 𝐶

⇒1

2√2log |

1

√2+

𝑦𝑥

1

√2−

𝑦𝑥

| = log|𝑥| + C



⇒1

2√2log |

𝑥+√2𝑦

𝑥−√2𝑦| = log|𝑥| + C [1

Mark]

Hence, the required solution for the given differential equation is 1

2√2log |

𝑥+√2𝑦

𝑥−√2𝑦| = log|𝑥| +

C

6. 𝑥𝑑𝑦 − 𝑦𝑑𝑥 = √𝑥2 + 𝑦2𝑑𝑥 [4

Marks]

Solution:

Given differential equation is 𝑥𝑑𝑦 − 𝑦𝑑𝑥 = √𝑥2 + 𝑦2𝑑𝑥

⇒ 𝑥𝑑𝑦 = [𝑦 + √𝑥2 + 𝑦2] 𝑑𝑥

𝑑𝑦

𝑑𝑥=

𝑦 + √𝑥2 + 𝑦2

𝑥2… (1)

Let 𝐹(𝑥, 𝑦) =𝑦+√𝑥2+𝑦2

𝑥2 .

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =𝜆𝑥 + √(𝜆𝑥)2 + (𝜆𝑦)2

𝜆𝑥=

𝑦 + √𝑥2 + 𝑦2

𝑥= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥 [1

Mark]

Substituting the values of 𝑣 and 𝑑𝑦


𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑣𝑥 + √𝑥2 + (𝑣𝑥)2

𝑥


𝑑𝑥= 𝑣 + √1 + 𝑣2

⇒𝑑𝑣

√1+𝑣2=

𝑑𝑥

𝑥 [1

Mark]




log |𝑣 + √1 + 𝑣2| = log|𝑥| + logC

⇒ log |𝑦

𝑥+ √1 +

𝑦2

𝑥2| = log|C𝑥|

⇒ log |𝑦 + √𝑥2 + 𝑦2

𝑥| = log|C𝑥|

⇒ 𝑦 + √𝑥2 + 𝑦2 = C𝑥2

Hence, the required solution of the given differential equation is 𝑦 + √𝑥2 + 𝑦2 = C𝑥2 [1

Mark]

7. {𝑥cos (𝑦

𝑥) + 𝑦sin (

𝑦

𝑥)} 𝑦𝑑𝑥 = {𝑦sin (

𝑦

𝑥) − 𝑥cos (

𝑦

𝑥)} 𝑥𝑑𝑦 [4

Marks]

Solution:

Given differential equation is {𝑥cos (𝑦

𝑥) + 𝑦sin (

𝑦

𝑥)} 𝑦𝑑𝑥 = {𝑦sin (

𝑦

𝑥) − 𝑥cos (

𝑦

𝑥)} 𝑥𝑑𝑦

𝑑𝑦

𝑑𝑥=

{𝑥cos (𝑦𝑥) + 𝑦sin (

𝑦𝑥)} 𝑦

{𝑦sin (𝑦𝑥) − 𝑥cos (

𝑦𝑥)} 𝑥

… (1)

Let 𝐹(𝑥, 𝑦) ={𝑥cos(

𝑦

𝑥)+𝑦sin(

𝑦

𝑥)}𝑦

{𝑦sin(𝑦

𝑥)−𝑥cos(

𝑦

𝑥)}𝑥

∴ 𝐹(𝜆𝑥, 𝜆𝑦) ={𝜆𝑥cos (

𝜆𝑦𝜆𝑥

) + 𝜆𝑦sin (𝜆𝑦𝜆𝑥

)} 𝜆𝑦

{𝜆𝑦sin (𝜆𝑦𝜆𝑥

) − 𝜆𝑥sin (𝜆𝑦𝜆𝑥

)} 𝜆𝑥

={𝑥cos (

𝑦𝑥) + 𝑦sin (

𝑦𝑥)} 𝑦

{𝑦sin (𝑦𝑥) − 𝑥cos (

𝑦𝑥)} 𝑥

= 𝜆0 ⋅ 𝐹(𝐹, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑𝑦

𝑑𝑥= 𝑣 + 𝑥 =

𝑑𝑣

𝑑𝑥



Substituting the values of y and 𝑑𝑦


𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

(𝑥cos𝑣+𝑣𝑥sin𝑣)·𝑣𝑥

(𝑣𝑥sin𝑣−𝑥cos𝑣)·𝑥 [1

Mark]


𝑑𝑥=

𝑣cos𝑣 + 𝑣2sin𝑣

𝑣sin𝑣 − cos𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣cos𝑣 + 𝑣2sin𝑣

𝑣sin𝑣 − cos𝑣− 𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣cos𝑣 + 𝑣2sin𝑣 − 𝑣2sin𝑣 + 𝑣cos𝑣


⇒ 𝑥𝑑𝑣

𝑑𝑥=

2𝑣cos𝑣


⇒ [𝑣sin𝑣 − cos𝑣

𝑣cos𝑣] 𝑑𝑣 =

2𝑑𝑥

𝑥

⇒ (tan 𝑣 −1

𝑣) 𝑑𝑣 =

2𝑑𝑥

𝑥 [1

Mark]


log(sec𝑣) − log𝑣 = 2log𝑥 + log𝐶

⇒ log (sec𝑣

𝑣) = log(C𝑥2)

⇒ (sec𝑣

𝑣) = C𝑥2

⇒ sec𝑣 = C𝑥2𝑣

⇒ sec (𝑦

𝑥) = C · 𝑥2 ·

𝑦

𝑥

⇒ sec (𝑦

𝑥) = C𝑥𝑦

⇒ cos (𝑦

𝑥) =

1

C𝑥𝑦=

1

C·

1

𝑥𝑦

⇒ 𝑥𝑦cos (𝑦

𝑥) = 𝑘 (𝑘 =

1

𝐶)

Hence, the required solution of the given differential equation is 𝑥𝑦cos (𝑦

𝑥) = 𝑘. [1

Mark]

8. 𝑥𝑑𝑦

𝑑𝑥− 𝑦 + 𝑥sin (

𝑦

𝑥) = 0 [4

Marks]



Solution:

Give differential equation is 𝑥𝑑𝑦

𝑑𝑥− 𝑦 + 𝑥sin (

𝑦

𝑥) = 0

⇒ 𝑥𝑑𝑦

𝑑𝑥= 𝑦 − 𝑥sin (

𝑦

𝑥)

⇒𝑑𝑦

𝑑𝑥=

𝑦 − 𝑥sin (𝑦𝑥)

𝑥. . (1)

Let 𝐹(𝑥, 𝑦) =𝑦−𝑥sin(

𝑦

𝑥)

𝑥

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =𝜆𝑦 − 𝜆𝑥sin (

𝜆𝑦𝜆𝑥

)

𝜆𝑥=

𝑦 − 𝑥sin (𝑦𝑥)

𝑥= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥

Substituting the values of y and 𝑑𝑦


𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑣𝑥−𝑥sin𝑣

𝑥 [1

Mark]


𝑑𝑥= 𝑣 − sin𝑣

⇒ −𝑑𝑣

sin𝑣=

𝑑𝑥

𝑥

⇒ cosec 𝑣 𝑑𝑣 = −𝑑𝑥

𝑥 [1

Mark]


log|cosec𝑣 − cot𝑣| = −log𝑥 + log𝐶 = log𝐶

𝑥

⇒ cosec (𝑦

𝑥) − cot (

𝑦

𝑥) =

𝐶

𝑥

⇒1

sin (𝑦𝑥)

−cos (

𝑦𝑥)

sin (𝑦𝑥)

=𝐶

𝑥



⇒ 𝑥 [1 − cos (𝑦

𝑥)] = 𝐶sin (

𝑦

𝑥)

Hence, the required solution of the given differential equation is 𝑥 [1 − cos (𝑦

𝑥)] =

𝐶sin (𝑦

𝑥)[1 Mark]

9. 𝑦𝑑𝑥 + 𝑥log (𝑦

𝑥) 𝑑𝑦 − 2𝑥𝑑𝑦 = 0 [6

Marks]

Solution:

Given differential equation is 𝑦𝑑𝑥 + 𝑥log (𝑦

𝑥) 𝑑𝑦 − 2𝑥𝑑𝑦 = 0

⇒ 𝑦𝑑𝑥 = [2𝑥 − 𝑥log (𝑦

𝑥)] 𝑑𝑦

⇒𝑑𝑦

𝑑𝑥=

𝑦

2𝑥 − 𝑥log (𝑦𝑥)

… (1)

Let 𝐹(𝑥, 𝑦) =𝑦

2𝑥−𝑥log(𝑦

𝑥)

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =𝜆𝑦

2(𝜆𝑥) − (𝜆𝑥) log (𝜆𝑦𝜆𝑥

)=

𝑦

2𝑥 − log (𝑦𝑥)

= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥


𝑑𝑥

𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑣𝑥

2𝑥−𝑥 log 𝑦 [1

Mark]


𝑑𝑥=

𝑣

2 − log𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣

2 − log𝑣− 𝑣



⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣 − 2𝑣 + 𝑣log𝑣

2 − log𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣log𝑣 − 𝑣

2 − log𝑣

⇒2 − log𝑣

𝑣(log𝑣 − 1)𝑑𝑣 =

𝑑𝑥

𝑥

⇒ [1 + (1 − log𝑣)

𝑣(log𝑣 − 1)] 𝑑𝑣 =

𝑑𝑥

𝑥

⇒ [1

𝑣(log𝑣−1)−

1

𝑣] 𝑑𝑣 =

𝑑𝑥

𝑥 [1

Mark]


∫1

𝑣(log𝑣 − 1)𝑑𝑣 − ∫

1

𝑣𝑑𝑣 = ∫

1

𝑥𝑑𝑥

⇒ ∫𝑑𝑣

𝑣(log𝑣 − 1)− log𝑣 = log𝑥 + log𝐶 … (2)

⇒ Let log𝑣 − 1 = 𝑡

⇒𝑑

𝑑𝑣(log𝑣 − 1) =

𝑑𝑡

𝑑𝑣

⇒1

𝑣=

𝑑𝑡

𝑑𝑣

⇒𝑑𝑣

𝑣= 𝑑𝑡 [1

Mark]


⇒ ∫𝑑𝑡

𝑡− log𝑣 = log𝑥 + logC

⇒ log𝑡 − log (𝑦

𝑥) = log(C𝑥)

⇒ log [log (𝑦

𝑥) − 1] − log (

𝑦

𝑥) = log(C𝑥)

⇒ log [log (

𝑦𝑥) − 1

𝑦𝑥

] = log(C𝑥)

⇒𝑥

𝑦[log (

𝑦

𝑥) − 1] = C𝑥

⇒ log (𝑦

𝑥) − 1 = C𝑦

Hence, the required solution of the given differential equation is log (𝑦

𝑥) − 1 = C𝑦 [2

Marks]



10. (1 + 𝑒𝑥

𝑦) 𝑑𝑥 + 𝑒𝑥

𝑦 (1 −𝑥

𝑦) 𝑑𝑦 = 0 [4

Marks]

Solution:

Given differential equation is (1 + 𝑒𝑥

𝑦) 𝑑𝑥 + 𝑒𝑥

𝑦 (1 −𝑥

𝑦) 𝑑𝑦 = 0

⇒ (1 + 𝑒𝑥𝑦) 𝑑𝑥 = −𝑒

𝑥𝑦 (1 −

𝑥

𝑦) 𝑑𝑦

⇒𝑑𝑥

𝑑𝑦=

−𝑒𝑥𝑦 (1 −

𝑥𝑦

)

1 + 𝑒𝑥𝑦

… (1)

Let 𝐹(𝑥, 𝑦) =−𝑒

𝑥𝑦(1−

𝑥

𝑦)

1+𝑒𝑥𝑦

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =−𝑒

𝜆𝑥𝜆𝑦 (1 −

𝜆𝑥𝜆𝑦

)

1 + 𝑒𝜆𝑥𝜆𝑦

=−𝑒

𝑥𝑦 (1 −

𝑥𝑦

)

1 + 𝑒𝑥𝑦

= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑥 = 𝑣𝑦

⇒𝑑

𝑑𝑦(𝑥) =

𝑑

𝑑𝑦(𝑣𝑦)

⇒𝑑𝑥

𝑑𝑦= 𝑣 + 𝑦

𝑑𝑣

𝑑𝑦

Substituting the values of 𝑥 and 𝑑𝑥

𝑑𝑦

𝑣 + 𝑦𝑑𝑣

𝑑𝑦=

−𝑒𝑣(1−𝑣)

1+𝑒𝑣 [1

Mark]

⇒ 𝑦𝑑𝑣

𝑑𝑦=

−𝑒𝑣 + 𝑣𝑒𝑣

1 + 𝑒𝑣− 𝑣

⇒ 𝑦𝑑𝑣

𝑑𝑦=

−𝑒𝑣 + 𝑣𝑒𝑣 − 𝑣 − 𝑣𝑒𝑣

1 + 𝑒𝑣

⇒ 𝑦𝑑𝑣

𝑑𝑦= − [

𝑣 + 𝑒𝑣

1 + 𝑒𝑣]



⇒ [1+𝑒𝑣

𝑣+𝑒𝑣] 𝑑𝑣 = −𝑑𝑦

𝑦 [1

Mark]


⇒ log(𝑣 + 𝑒𝑣) = −log𝑦 + log𝐶 = log (𝐶

𝑦)

⇒ [𝑥

𝑦+ 𝑒

𝑥𝑦] =

𝐶

𝑦

⇒ 𝑥 + 𝑦𝑒𝑥𝑦 = 𝐶

Hence, the required solution of the given differential equation is 𝑥 + 𝑦𝑒𝑥

𝑦 = 𝐶 [1

Mark]

For each of the differential equations in Exercises from 11 to 15, find the particular solution

satisfying the given condition:

11. (𝑥 + 𝑦)𝑑𝑦 + (𝑥 − 𝑦)𝑑𝑦 = 0; 𝑦 = 1 when 𝑥 = 1 [6

Marks]

Solution:

Given differential equation is (𝑥 + 𝑦)𝑑𝑦 + (𝑥 − 𝑦)𝑑𝑦 = 0

⇒ (𝑥 + 𝑦)𝑑𝑦 = −(𝑥 − 𝑦)𝑑𝑥

⇒𝑑𝑦

𝑑𝑥=

−(𝑥 − 𝑦)

𝑥 + 𝑦… (1)

Let 𝐹(𝑥, 𝑦) =−(𝑥−𝑦)

𝑥+𝑦

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =−(𝜆𝑥 − 𝜆𝑦)

𝜆𝑥 − 𝜆𝑦=

−(𝑥 − 𝑦)

𝑥 + 𝑦= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥





𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

−(𝑥−𝑣𝑥)

𝑥+𝑣𝑥 [1

Mark]


𝑑𝑥=

𝑣 − 1

𝑣 + 1

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣 − 1

𝑣 + 1− 𝑣 =

𝑣 − 1 − 𝑣(𝑣 + 1)

𝑣 + 1

⇒ 𝑥𝑑𝑣

𝑑𝑥=

𝑣 − 1 − 𝑣2 − 𝑣

𝑣 + 1=

−(1 + 𝑣2)

𝑣 + 1

⇒(𝑣 + 1)

1 + 𝑣2𝑑𝑣 = −

𝑑𝑥

𝑥

⇒ [𝑣

1+𝑣2 +1

1+𝑣2] 𝑑𝑣 = −𝑑𝑥

𝑥 [1

Mark]


1

2log(1 + 𝑣2) + tan−1𝑣 = −log𝑥 + 𝑘

⇒ log(1 + 𝑣2) + 2tan−1𝑣 = −2log𝑥 + 2𝑘

⇒ log[(1 + 𝑣2) · 𝑥2] + 2tan−1𝑣 = 2𝑘

⇒ log [(1 +𝑦2

𝑥2) · 𝑥2] + 2tan−1𝑦

𝑥= 2𝑘

⇒ log(𝑥2 + 𝑦2) + 2tan−1 𝑦

𝑥= 2𝑘 … (2) [1

Mark]

Now, 𝑦 = 1 at 𝑥 = 1

⇒ log2 + 2tan−11 = 2𝑘

⇒ log2 + 2 ×𝜋

4= 2𝑘

⇒𝜋

2+ log2 = 2𝑘 [1

Mark]

Substituting the value of 2𝑘 in equation (2), we get:

log(𝑥2 + 𝑦2) + 2tan−1 (𝑦

𝑥) =

𝜋

2+ log2

Hence, the required solution of the given differential equation is log(𝑥2 + 𝑦2) +

2tan−1 (𝑦

𝑥) =

𝜋

2+ log2

[1 Mark]



12. 𝑥2𝑑𝑦 + (𝑥𝑦 + 𝑦2)𝑑𝑥 = 0; 𝑦 = 1 when 𝑥 = 1 [6

Marks]

Solution:

𝑥2𝑑𝑦 + (𝑥𝑦 + 𝑦2)𝑑𝑥 = 0

⇒ 𝑥2𝑑𝑦 = −(𝑥𝑦 + 𝑦2)𝑑𝑥

⇒𝑑𝑦

𝑑𝑥=

−(𝑥𝑦 + 𝑦2)

𝑥2… (1)

Let 𝐹(𝑥, 𝑦) =−(𝑥𝑦+𝑦2)

𝑥2

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =[𝜆𝑥 · 𝜆𝑦 + (𝜆𝑦)2]

(𝜆𝑥)2=

−(𝑥𝑦 + 𝑦2)

𝑥2= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥


𝑑𝑥

𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

−[𝑥·𝑣𝑥+(𝑣𝑥)2]

𝑥2 = −𝑣 − 𝑣2 [1

Mark]

⇒ 𝑥𝑑𝑣

𝑑𝑥= −𝑣2 − 2𝑣 = −𝑣(𝑣 + 2)

⇒𝑑𝑣

𝑣(𝑣 + 2)= −

𝑑𝑥

𝑥

⇒1

2[(𝑣 + 2) − 𝑣

𝑣(𝑣 + 2)] 𝑑𝑣 = −

𝑑𝑥

𝑥

⇒1

2[

1

𝑣−

1

𝑣+2]𝑑𝑣 = −

𝑑𝑥

𝑥 [1

Mark]


1

2[log𝑣 − log(𝑣 + 2)] = −log𝑥 + logC



⇒1

2log (

𝑣

𝑣 + 2) = log

C

𝑥

⇒𝑣

𝑣 + 2= (

C

𝑥)

2

⇒

𝑦𝑥

𝑦𝑥

+ 2= (

C

𝑥)

2

⇒𝑦

𝑦 + 2𝑥=

C2

𝑥2

⇒𝑥2𝑦

𝑦+2𝑥= C2 … (2) [1

Mark]

Now, 𝑦 = 1 at 𝑥 = 1

⇒1

1 + 2= C2

⇒ C2 =1

3 [1

Mark]

Substituting 𝐶2 =1

3

𝑥2𝑦

𝑦 + 2𝑥=

1

3

⇒ 𝑦 + 2𝑥 = 3𝑥2𝑦

Hence, the required solution of the given differential equation is 𝑦 + 2𝑥 = 3𝑥2𝑦 [1

Mark]

13. [𝑥sin2 (𝑥

𝑦− 𝑦)] 𝑑𝑥 + 𝑥𝑑𝑦 = 0; 𝑦

𝜋

4 when 𝑥 = 1 [6

Marks]

Solution:

Given differential equation is [𝑥sin2 (𝑦

𝑥) − 𝑦] 𝑑𝑥 + 𝑥𝑑𝑦 = 0

⇒𝑑𝑦

𝑑𝑥=

− [𝑥sin2 (𝑦𝑥) − 𝑦]

𝑥… (1)

Let 𝐹(𝑥, 𝑦) =−[𝑥sin2(

𝑦

𝑥)−𝑦]

𝑥



∴ 𝐹(𝜆𝑥, 𝜆𝑦) =− [𝜆𝑥 · sin2 (

𝜆𝑥𝜆𝑦

) − 𝜆𝑦]

𝜆𝑥=

− [𝑥sin2 (𝑦𝑥

) − 𝑦]

𝑥= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]

To solve this differential equation, we make the substitution as:

𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦

𝑑𝑥= 𝑣 + 𝑥 =

𝑑𝑣

𝑑𝑥



𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

−[𝑥sin2𝑣−𝑣𝑥]

𝑥 [1

Mark]


𝑑𝑥= −[sin2𝑣 − 𝑣] = 𝑣 − sin2𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥= −sin2𝑣

⇒𝑑𝑣

sin2𝑣= −sin2𝑣

⇒ cosec2𝑣𝑑𝑣 = −𝑑𝑥

𝑥 [1

Mark]


−cot𝑣 = − log|𝑥| − 𝐶

⇒ cot𝑣 = log|𝑥| + 𝐶

⇒ cot (𝑦

𝑥) = log|𝑥| + log𝐶

⇒ cot (𝑦

𝑥) = log|C𝑥| … (2)

Now, 𝑦 =𝜋

4 at 𝑥 = 1

⇒ cot (𝜋

4) = log|C|

⇒ 1 = logC

⇒ C = 𝑒1 = 𝑒 [2

Marks]

Substituting 𝐶 = 𝑒 in equation (2), we get:



cot (𝑦

𝑥) = log|𝑒𝑥|

Hence, the required solution of the given differential equation is cot (𝑦

𝑥) = log|𝑒𝑥| [1

Mark]

14. 𝑑𝑦

𝑑𝑥−

𝑦

𝑥+ cosec (

𝑦

𝑥) = 0; 𝑦 = 0 when 𝑥 = 1 [6

Marks]

Solution:


𝑑𝑥−

𝑦

𝑥+ cosec (

𝑦

𝑥) = 0

⇒𝑑𝑦

𝑑𝑥=

𝑦

𝑥− cosec (

𝑦

𝑥) … (1)

Let 𝐹(𝑥, 𝑦) =𝑦

𝑥− cosec (

𝑦

𝑥)

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =𝜆𝑦

𝜆𝑥− cosec (

𝜆𝑦

𝜆𝑥)

⇒ 𝐹(𝜆𝑥, 𝜆𝑦) =𝑦

𝑥− cosec (

𝑦

𝑥) = 𝐹(𝑥, 𝑦) = 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥



𝑣 + 𝑥𝑑𝑣

𝑑𝑥= 𝑣 − cosec𝑣 [1

Mark]

⇒ −𝑑𝑣

cosec𝑣= −

𝑑𝑥

𝑥

⇒ −sin𝑣𝑑𝑣 =𝑑𝑥

𝑥 [1

Mark]




cos𝑣 = log𝑥 + logC = log|C𝑥|

⇒ cos (𝑦

𝑥) = log|C𝑥| … (2) [1

Mark]

This is the required solution of the given differential equation.

Now, 𝑦 = 0 at 𝑥 = 1.

⇒ cos(0) = logC

⇒ 1 = logC

⇒ C = 𝑒1 = 𝑒 [1

Mark]

Substituting 𝐶 = 𝑒 in equation (2), we get:

cos (𝑦

𝑥) = log|(𝑒𝑥)|

Hence, the required solution of the given differential equation is cos (𝑦

𝑥) = log|(𝑒𝑥)| [1

Mark]

15. 2𝑥𝑦 + 𝑦2 − 2𝑥2 𝑑𝑦

𝑑𝑥= 0; 𝑦 = 2 when 𝑥 = 1 [6

Marks]

Solution:

Given differential equation is 2𝑥𝑦 + 𝑦2 − 2𝑥2 𝑑𝑦

𝑑𝑥= 0

⇒ 2𝑥2𝑑𝑦

𝑑𝑥= 2𝑥𝑦 + 𝑦2

⇒𝑑𝑦

𝑑𝑥=

2𝑥𝑦 + 𝑦2

2𝑥2… (1)

Let 𝐹(𝑥, 𝑦) =2𝑥𝑦+𝑦2

2𝑥2

∴ 𝐹(𝜆𝑥, 𝜆𝑦) =2(𝜆𝑥)(𝜆𝑦) + (𝜆𝑦)2

2(𝜆𝑥)2=

2𝑥𝑦 + 𝑦2

2𝑥2= 𝜆0 · 𝐹(𝑥, 𝑦)


Mark]


𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)



⇒𝑑𝑦


𝑑𝑣

𝑑𝑥

Substituting the value of 𝑦 and 𝑑𝑦


𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

2𝑥(𝑣𝑥)+(𝑣𝑥)2

2𝑥2 [1

Mark]


𝑑𝑥=

2𝑣 + 𝑣2

2


𝑑𝑥= 𝑣 +

𝑣2

2

⇒2

𝑣2 𝑑𝑣 =𝑑𝑥

𝑥 [1

Mark]


2 ·𝑣−2+1

−2 + 1= log|𝑥| + C

⇒ −2

𝑣= log|𝑥| + C

⇒ −2𝑦𝑥

= log|𝑥| + C

⇒ −2𝑥

𝑦= log|𝑥| + C …(2) [1

Mark]

Now, 𝑦 = 2 at 𝑥 = 1

⇒ −1 = log(1) + 𝐶

⇒ 𝐶 = −1 [1

Mark]

Substituting 𝐶 = −1 in equation (2), we get:

−2𝑥

𝑦= log|𝑥| − 1

⇒2𝑥

𝑦= 1 − log|𝑥|

⇒ 𝑦 =2𝑥

1 − log|𝑥|, (𝑥 ≠ 0, 𝑥 ≠ 𝑒)

Hence, the required solution of the given differential equation is 𝑦 =2𝑥

1−log|𝑥| [1

Mark]



16. A homogeneous differential equation of the form 𝑑𝑥

𝑑𝑦= ℎ (

𝑥

𝑦) can be solved by making

the substitution.

[1 Mark]

(A) 𝑦 = 𝑣𝑥

(B) 𝑣 = 𝑦𝑥

(C) 𝑥 = 𝑣𝑦

(D) 𝑥 = 𝑣

Solution:

For solving the homogeneous equation of the form 𝑑𝑥

𝑑𝑦= ℎ (

𝑥

𝑦), we need to make the

substitution as

𝑥 = 𝑣𝑦.

Hence, the correct answer is C. [1

Mark]

17. Which of the following is a homogeneous differential equation? [2

Marks]

(A) (4𝑥 + 6𝑦 + 5)𝑑𝑦 − (3𝑦 + 2𝑥 + 4)𝑑𝑥 = 0

(B) (𝑥𝑦)𝑑𝑥 − (𝑥3 + 𝑦3)𝑑𝑦 = 0

(C) (𝑥3 + 2𝑦2)𝑑𝑥 + 2𝑥𝑦𝑑𝑦 = 0

(D) 𝑦2𝑑𝑥 + (𝑥2 − 𝑥𝑦2 − 𝑦2)𝑑𝑦 = 0

Solution:

Function 𝐹(𝑥, 𝑦) is said to be the homogenous function of degree n, if 𝐹(𝜆𝑥, 𝜆𝑦) =

𝜆𝑛 𝐹(𝑥, 𝑦) for any non-zero constant (𝜆).

Consider the equation given in alternative D:

𝑦2𝑑𝑥 + (𝑥2 − 𝑥𝑦 − 𝑦2)𝑑𝑦 = 0

⇒𝑑𝑦

𝑑𝑥=

−𝑦2

𝑥2−𝑥𝑦−𝑦2 =𝑦2

𝑦2+𝑥𝑦−𝑥2 [1

Mark]

Let 𝐹(𝑥, 𝑦) =𝑦2

𝑦2+𝑥𝑦−𝑥2



⇒ 𝐹(𝜆𝑥, 𝜆𝑦) =(𝜆𝑦)2

(𝜆𝑦)2 + (𝜆𝑥)(𝜆𝑦) − (𝜆𝑥)2

=𝜆2𝑦2

𝜆2(𝑦2 + 𝑥𝑦 − 𝑥2)

== 𝜆0 (𝑦2

𝑦2 + 𝑥𝑦 − 𝑥2)

= 𝜆0 · 𝐹(𝑥, 𝑦)

Hence, the differential equation given in alternative D is a homogenous equation. [1

Mark]

Exercise 9.6

For each of the differential equations given in Exercises 1 to 12, find the general solution:

1. 𝑑𝑦

𝑑𝑥+ 2𝑦 = sin𝑥 [4

Marks]

Solution:


𝑑𝑥+ 2𝑦 = sin 𝑥

This is in the form of 𝑑𝑦

𝑑𝑥+ 𝑃𝑦 = 𝑄 (where 𝑃 = 2 and 𝑄 = sin 𝑥)

Now,=𝑒∫ 𝑃𝑑𝑥 = 𝑒∫ 2𝑑𝑥 = 𝑒2𝑥

The solution of the given differential equation is given by the relation,

𝑦(𝐼. 𝐹. ) = ∫(𝑄 × 𝐼. 𝐹. ) 𝑑𝑥 + 𝐶

⇒ 𝑦𝑒2𝑥 = ∫ sin 𝑥 · 𝑒2𝑥𝑑𝑥 + C … (1) [1

Mark]

Let 𝐼 = ∫ sin 𝑥 ∙ 𝑒2𝑥.

⇒ 𝐼 = sin𝑥 · ∫ 𝑒2𝑥 𝑑𝑥 − ∫ (𝑑

𝑑𝑥(sin𝑥) · ∫ 𝑒2𝑥 𝑑𝑥) 𝑑𝑥

⇒ 𝐼 = sin𝑥 ·𝑒2𝑥

2− ∫ (cos𝑥 ·

𝑒2𝑥

2) 𝑑𝑥

⇒ 𝐼 =𝑒2𝑥sin𝑥

2−

1

2[cos𝑥 · ∫ 𝑒2𝑥 − ∫ (

𝑑

𝑑𝑥(cos𝑥) · ∫ 𝑒2𝑥 𝑑𝑥) 𝑑𝑥]




2−

1

2[cos𝑥 ·

𝑒2𝑥

2− ∫ [(−sin𝑥) ·

𝑒2𝑥

2] 𝑑𝑥]


2−

𝑒2𝑥cos𝑥

4−

1

4∫(sin𝑥 · 𝑒2𝑥) 𝑑𝑥

⇒ 𝐼 =𝑒2𝑥

4(2sin𝑥 − cos𝑥) −

1

4𝐼

⇒5

4𝐼 =

𝑒2𝑥

4(2sin𝑥 − cos𝑥)

⇒ 𝐼 =𝑒2𝑥

5(2sin𝑥 − cos𝑥) [2

Marks]


𝑦𝑒2𝑥 =𝑒2𝑥

5(2sin𝑥 − cos𝑥) + C

⇒ 𝑦 =1

5(2sin𝑥 − cos𝑥) + C𝑒−2𝑥

Hence, the required general solution of the given differential equation is

𝑦 =1

5(2sin𝑥 − cos𝑥) + C𝑒−2𝑥 [1

Mark]

2. 𝑑𝑦

𝑑𝑥+ 3𝑦 = 𝑒−2𝑥 [2

Marks]

Solution:

The given differential equation is 𝑑𝑦

𝑑𝑥+ 3𝑦 = 𝑒−2𝑥


𝑑𝑥+ 𝑃𝑦 = 𝑄 (where 𝑃 = 3 and 𝑄 = 𝑒−2𝑥)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑃𝑑𝑥 = 𝑒∫ 3𝑑𝑥 = 𝑒3𝑥 [1

Mark]


𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦𝑒3𝑥 = ∫(𝑒−2𝑥 × 𝑒3𝑥) + C

⇒ 𝑦𝑒3𝑥 = ∫ 𝑒𝑥 𝑑𝑥 + C



⇒ 𝑦𝑒3𝑥 = 𝑒𝑥 + C

⇒ 𝑦 = 𝑒−2𝑥 + Ce−3𝑥


𝑦 = 𝑒−2𝑥 + Ce−3𝑥 [1

Mark]

3. 𝑑𝑦

𝑑𝑥+

𝑦

𝑥= 𝑥2

[2 Marks]

Solution:


𝑑𝑥+

𝑦

𝑥= 𝑥2


𝑑𝑥+ 𝑃𝑦 = 𝑄 (where 𝑃 =

1

𝑥and 𝑄 = 𝑥2)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑃𝑑𝑥 = 𝑒∫1

𝑥𝑑𝑥 = 𝑒log 𝑥 = 𝑥 [1

Mark]


𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦(𝑥) = ∫(𝑥2 · 𝑥) 𝑑𝑥 + C

⇒ 𝑥𝑦 = ∫ 𝑥3 𝑑𝑥 + C

⇒ 𝑥𝑦 =𝑥4

4+ C

Hence, the required general solution of the given differential equation is 𝑥𝑦 =𝑥4

4+ C [1

Mark]

4. 𝑑𝑦

𝑑𝑥+ sec 𝑥𝑦 = tan 𝑥 (0 ≤ 𝑥 <

𝜋

2) [2

Marks]

Solution:


𝑑𝑥+ sec 𝑥𝑦 = tan 𝑥 (0 ≤ 𝑥 <

𝜋

2)




𝑑𝑥+ 𝑃𝑦 = 𝑄 (where 𝑃 = sec 𝑥 and 𝑄 = tan 𝑥)

Now 𝐼. 𝐹 = 𝑒∫ 𝑃𝑑𝑥 = 𝑒∫ sec𝑥𝑑𝑥 = 𝑒log(sec𝑥+tan𝑥) = sec𝑥 + tan𝑥 [1

Mark]

The general solution of the given differential equation is given by the relation,

𝑦(IF) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦(sec𝑥 + tan𝑥) = ∫ tan 𝑥(sec𝑥 + tan𝑥)𝑑𝑥 + C

⇒ 𝑦(sec𝑥 + tan𝑥) = ∫ sec 𝑥tan𝑥𝑑𝑥 + ∫ tan2 𝑥𝑑𝑥 + C

⇒ 𝑦(sec𝑥 + tan𝑥) = sec𝑥 + ∫(sec2𝑥 − 1) 𝑑𝑥 + C

⇒ 𝑦(sec𝑥 + tan𝑥) = sec𝑥 + tan𝑥 − 𝑥 + C

Hence, the required general solution of the given differential equation is 𝑦(sec𝑥 + tan𝑥) =

sec𝑥 + tan𝑥 − 𝑥 + C

[1 Mark]

5. cos2 𝑥 .𝑑𝑦

𝑑𝑥+ 𝑦 = tan 𝑥 (0 ≤ 𝑥 <

𝜋

2)

[4 Marks]

Solution:

Given differential equation is cos2 𝑥 .𝑑𝑦

𝑑𝑥+ 𝑦 = tan 𝑥

⇒𝑑𝑦

𝑑𝑥+

𝑦

cos2 𝑥=

tan 𝑥

cos2 𝑥

⇒𝑑𝑦

𝑑𝑥+ 𝑦 sec2 𝑥 = tan 𝑥 sec2 𝑥


𝑑𝑥+ 𝑃𝑦 = 𝑄 (where 𝑃 = sec2 𝑥 and 𝑄 = tan 𝑥 sec2 𝑥)

Now, I. F = e∫ 𝑝𝑑𝑥

I.F = e∫ sec2𝑥·𝑑𝑥

I.F. = etan𝑥 [1

Mark]

Solution of the given equation is 𝑦 × I. F. = ∫ 𝑄 × I. F × 𝑑𝑥 + 𝐶

𝑦 · 𝑒tan𝑥 = ∫ sec2𝑥 · tan𝑥 · 𝑒tan𝑥 · 𝑑𝑥 + 𝐶 ….. (1)



Let 𝐼 = ∫ sec2𝑥 · tan𝑥 · 𝑒tan𝑥 · 𝑑𝑥

Putting 𝑡 = tan𝑥

⇒ sec2𝑥 · 𝑑𝑥 = 𝑑𝑡 [1

Mark]

I = ∫ tan𝑥 · etan𝑥 · (sec2𝑥 · 𝑑𝑥)

⇒ I = ∫ 𝑡 · 𝑒𝑡 · 𝑑𝑡

⇒ I = t∫ 𝑒𝑡𝑑𝑡 − ∫ [𝑑𝑡

𝑑𝑡∫ 𝑒𝑡𝑑𝑡] 𝑑𝑡

⇒ I = t · et − ∫ et𝑑𝑡

⇒ I = 𝑡et − et

[1

Mark]

Putting 𝑡 = tan 𝑥, we get

𝐼 = tan𝑥 · 𝑒tan𝑥 − 𝑒tan𝑥

⇒ 𝐼 = 𝑒tan𝑥(tan𝑥 − 1)

Substituting value of 𝐼 in (1), we get

𝑦𝑒tan𝑥 = 𝑒tan𝑥(tan𝑥 − 1) + 𝐶

⇒ 𝑦 = (tan𝑥 − 1) + 𝐶. 𝑒−tan𝑥 [1

Mark]

Hence, the required general solution of the given differential equation is 𝑦 = (tan𝑥 − 1) +

𝐶. 𝑒−tan𝑥

6. 𝑥𝑑𝑦

𝑑𝑥+ 2𝑦 = 𝑥2log𝑥

[4 Marks]

Solution:

The given differential equation is 𝑥𝑑𝑦

𝑑𝑥+ 2𝑦 = 𝑥2log𝑥

⇒𝑑𝑦

𝑑𝑥+

2

𝑥𝑦 = 𝑥log𝑥

This equation is in the form of a linear differential equation as:

𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄 (where 𝑝 =

2

𝑥 and 𝑥 log 𝑥)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫ 𝑑𝑥2

𝑥 = 𝑒2 log 𝑥 = 𝑒log𝑥2= 𝑥2 [1

Mark]




𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦 · 𝑥2 = ∫(𝑥log𝑥 · 𝑥2) 𝑑𝑥 + C

⇒ 𝑥2𝑦 = ∫(𝑥3log𝑥) 𝑑𝑥 + C

⇒ 𝑥2𝑦 = log𝑥 · ∫ 𝑥3 𝑑𝑥 − ∫ [𝑑

𝑑𝑥(log𝑥) · ∫ 𝑥3 𝑑𝑥] 𝑑𝑥 + C [1

Mark]

⇒ 𝑥2𝑦 = log𝑥 ·𝑥4

4− ∫ (

1

𝑥·

𝑥4

4) 𝑑𝑥 + C

⇒ 𝑥2𝑦 =𝑥4log𝑥

4−

1

4∫ 𝑥3 𝑑𝑥 + C

⇒ 𝑥2𝑦 =𝑥4log𝑥

4−

1

4·

𝑥4

4+ C [1

Mark]

⇒ 𝑥2𝑦 =1

16𝑥4(4log𝑥 − 1) + C

⇒ 𝑦 =1

16𝑥2(4log𝑥 − 1) + C𝑥−2


𝑦 =1

16𝑥2(4log𝑥 − 1) + C𝑥−2 [1

Mark]

7. 𝑥log𝑥𝑑𝑦

𝑑𝑥+ 𝑦 =

2

𝑥log𝑥 [4

Marks]

Solution:

The given differential equation is 𝑥log𝑥𝑑𝑦

𝑑𝑥+ 𝑦 =

2

𝑥log𝑥

⇒𝑑𝑦

𝑑𝑥+

𝑦

𝑥log𝑥=

2

𝑥2

This equation is the form of a linear differential equation as:

𝑑𝑦


1

𝑥log𝑥and 𝑄 =

2

𝑥2)

Now, 𝐼. 𝐹. = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫

1

𝑥log𝑑𝑥

= 𝑒log(log𝑥) = log𝑥 [1

Mark]




𝑦(𝐼. 𝐹. ) = ∫ (Q × I.F.)𝑑𝑥 + C

⇒ 𝑦log𝑥 = ∫ (2

𝑥2log𝑥) 𝑑𝑥 + C … (1)

Now, ∫ (2

𝑥2 log𝑥) 𝑑𝑥 = 2 ∫ (log𝑥 ·1

𝑥2) 𝑑𝑥

= 2 [log𝑥 · ∫1

𝑥2 𝑑𝑥 − ∫ {𝑑

𝑑𝑥(log𝑥) · ∫

1

𝑥2 𝑑𝑥} 𝑑𝑥] [1

Mark]

= 2 [log𝑥 (−1

𝑥) − ∫ (

1

𝑥· (−

1

𝑥)) 𝑑𝑥]

= 2 [−log𝑥

𝑥+ ∫

1

𝑥2𝑑𝑥]

= 2 [−log𝑥

𝑥−

1

𝑥]

= −2

𝑥(1 + log𝑥) [1

Mark]

Substituting the value of ∫ (2

𝑥2 log 𝑥) 𝑑𝑥 in equation (1), we get:

𝑦log𝑥 = −2

𝑥(1 + log𝑥) + 𝐶


𝑦log𝑥 = −2

𝑥(1 + log𝑥) + 𝐶 [1

Mark]

8. (1 + 𝑥2)𝑑𝑦 + 2𝑥𝑦𝑑𝑥 = cot𝑥𝑑𝑥(𝑥 ≠ 0) [4

Marks]

Solution:

Given differential equation is (1 + 𝑥2)𝑑𝑦 + 2𝑥𝑦𝑑𝑥 = cot𝑥𝑑𝑥

⇒𝑑𝑦

𝑑𝑥+

2𝑥𝑦

1+𝑥2 =cot𝑥

1+𝑥2 [1

Mark]

This equation is a linear differential equation of the form:

𝑑𝑦


2𝑥

1 + 𝑥2and 𝑄 =

cot 𝑥

1 + 𝑥2)



Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫ 𝑑𝑥2𝑥

1+𝑥2 = 𝑒log(1+𝑥2) = 1 + 𝑥2 [1

Mark]


𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = ∫ [cot𝑥

1 + 𝑥2× (1 + 𝑥2)] 𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = ∫ cot 𝑥𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = log|sin𝑥| + C


𝑦(1 + 𝑥2) = log|sin𝑥| + C [2

Marks]

9. 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 − 𝑥 + 𝑥𝑦cot𝑥 = 0(𝑥 ≠ 0) [4

Marks]

Solution:

Given differential equation is 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 − 𝑥 + 𝑥𝑦cot𝑥 = 0

⇒ 𝑥𝑑𝑦

𝑑𝑥+ 𝑦(1 + 𝑥cot𝑥) = 𝑥

⇒𝑑𝑦

𝑑𝑥+ (

1

𝑥+ cot𝑥) 𝑦 = 1 [1

Mark]

This equation is a linear differential equation of the form:

𝑑𝑦


1

𝑥+ cot𝑥 and 𝑄 =

cot 𝑥

1 + 𝑥2)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫2𝑥

1+𝑥2𝑑𝑥= 𝑒∫ log(1+𝑥2) = 1 + 𝑥2 [1

Mark]


𝑦(𝐼. 𝐹. ) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = ∫ [cot𝑥

1 + 𝑥2× (1 + 𝑥2)] 𝑑𝑥 + C



⇒ 𝑦(1 + 𝑥2) = ∫ cot 𝑥𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = log|sin𝑥| + 𝐶


𝑦(1 + 𝑥2) = log|sin𝑥| + 𝐶 [2

Marks]

10. 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 − 𝑥 + 𝑥𝑦cot𝑥 = 0(𝑥 ≠ 0) [4

Marks]

Solution:

Given differential equation is 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 − 𝑥 + 𝑥𝑦cot𝑥 = 0

⇒ 𝑥𝑑𝑦

𝑑𝑥+ 𝑦(1 + 𝑥cot𝑥) = 𝑥

⇒𝑑𝑦

𝑑𝑥+ (

1

𝑥+ cot𝑥) 𝑦 = 1 [1

Mark]

The equation is a linear differential equation of the form:

𝑑𝑦


1

𝑥cot 𝑥 and 𝑄 = 1)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫(1

𝑥+cot 𝑥)𝑑𝑥

= 𝑒log 𝑥+log(sin 𝑥) = 𝑒∫ log(𝑥 sin 𝑥) = 𝑥 sin 𝑥 [1

Mark]


𝑦(𝐼. 𝐹. ) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦(𝑥sin𝑥) = ∫(1 × 𝑥sin𝑥) 𝑑𝑥 + C

⇒ 𝑦(𝑥sin𝑥) = ∫(𝑥sin𝑥) 𝑑𝑥 + C

⇒ 𝑦(𝑥sin𝑥) = 𝑥 ∫ sin 𝑥𝑑𝑥 − ∫ [𝑑

𝑑𝑥(𝑥) · ∫ sin𝑥 𝑑𝑥] + C

⇒ 𝑦(𝑥sin𝑥) = 𝑥(−cos𝑥) − ∫ 1 · (−cos𝑥)𝑑𝑥 + C

⇒ 𝑦(𝑥sin𝑥) = −𝑥cos𝑥 + sin𝑥 + C

⇒ 𝑦 =−𝑥cos𝑥

𝑥sin𝑥+

sin𝑥

𝑥sin𝑥+

C

𝑥sin𝑥



⇒ 𝑦 = −cot · 𝑥 +1

𝑥+

C

𝑥sin𝑥 [2

Marks]

Hence, the required general solution of the given differential equation is 𝑦 = −cot · 𝑥 +1

𝑥+

C

𝑥sin𝑥

11. (𝑥 + 𝑦)𝑑𝑦

𝑑𝑥= 1 [4

Marks]

Solution:

Given differential equation is (𝑥 + 𝑦)𝑑𝑦

𝑑𝑥= 1

⇒𝑑𝑦

𝑑𝑥=

1

𝑥 + 𝑦

⇒𝑑𝑥

𝑑𝑦= 𝑥 + 𝑦

⇒𝑑𝑥

𝑑𝑦− 𝑥 = 𝑦

This is a linear differential equation of the form:

𝑑𝑦

𝑑𝑥+ 𝑝𝑥 = 𝑄 (where 𝑝 = −1 and 𝑄 = 𝑦)

Now, 𝐼. 𝐹 = 𝑒∫ 𝜌𝑑𝑦 = 𝑒∫ −𝑑𝑦 = 𝑒−𝑦. [2

Marks]


𝑥(𝐼. 𝐹. ) = ∫(𝑄 × I.F.) 𝑑𝑦 + 𝐶

⇒ 𝑥𝑒−𝑦 = ∫(𝑦 · 𝑒−𝑦) 𝑑𝑦 + 𝐶

⇒ 𝑥𝑒−𝑦 = 𝑦 · ∫ 𝑒−𝑦 𝑑𝑦 − ∫ [𝑑

𝑑𝑦(𝑦) ∫ 𝑒−𝑦 𝑑𝑦] 𝑑𝑦 + 𝐶

⇒ 𝑥𝑒−𝑦 = 𝑦(−𝑒−𝑦) − ∫(−𝑒−𝑦) 𝑑𝑦 + 𝐶

⇒ 𝑥𝑒−𝑦 = −𝑦𝑒−𝑦 + ∫ 𝑒−𝑦 𝑑𝑦 + 𝐶

⇒ 𝑥𝑒−𝑦 = −𝑦𝑒−𝑦 − 𝑒−𝑦 + 𝐶

⇒ 𝑥 = −𝑦 − 1 + C𝑒𝑦



⇒ 𝑥 + 𝑦 + 1 = C𝑒𝑦 [2

Marks]

Hence, the required general solution of the given differential equation is 𝑥 + 𝑦 + 1 = C𝑒𝑦

12. 𝑦𝑑𝑥 + (𝑥 − 𝑦2)𝑑𝑦 = 0 [4

Marks]

Solution:

Given differential equation is 𝑦𝑑𝑥 + (𝑥 − 𝑦2)𝑑𝑦 = 0

⇒ 𝑦𝑑𝑥 = (𝑦2 − 𝑥)𝑑𝑦

⇒𝑑𝑥

𝑑𝑦=

𝑦2 − 𝑥

𝑦= 𝑦 −

𝑥

𝑦

⇒𝑑𝑥

𝑑𝑦+

𝑥

𝑦= 𝑦 [1

Mark]


𝑑𝑦

𝑑𝑥+ 𝑝𝑥 = 𝑄 (where 𝑝 =

1

𝑦 and 𝑄 = 𝑦)

Now 𝐼. 𝐹 = 𝑒∫ 𝜌𝑑𝑦 = 𝑒∫

1

𝑦𝑑𝑦

= 𝑒log𝑦 = 𝑦. [1

Mark]


𝑥(I.F.) = ∫(Q × I.F.) 𝑑𝑦 + C

⇒ 𝑥𝑦 = ∫ (𝑦 · 𝑦)𝑑𝑦 + C

⇒ 𝑥𝑦 = ∫ 𝑦2𝑑𝑦 + C

⇒ 𝑥𝑦 =𝑦3

3+ C

⇒ 𝑥 =𝑦2

3+

C

𝑦

Hence, the required general solution of the given differential equation is 𝑥 =𝑦2

3+

C

𝑦 [2

Marks]



13. (𝑥 + 3𝑦2)𝑑𝑦

𝑑𝑥= 𝑦(𝑦 > 0) [4

Marks]

Solution:

Given differential equation is (𝑥 + 3𝑦2)𝑑𝑦

𝑑𝑥= 𝑦

⇒𝑑𝑦

𝑑𝑥=

𝑦

𝑥 + 3𝑦2

⇒𝑑𝑥

𝑑𝑦=

𝑥 + 3𝑦2

𝑦=

𝑥

𝑦+ 3𝑦

⇒𝑑𝑥

𝑑𝑦−

𝑥

𝑦= 3𝑦 [1

Mark]


𝑑𝑥

𝑑𝑦+ 𝑝𝑥 = 𝑄 (where 𝑝 = −

1

𝑦and 𝑄 = 3𝑦)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑦 = 𝑒−∫

𝑑𝑣

𝑦 = 𝑒− log 𝑦 = 𝑒log(

1

𝑦)

=1

𝑦. [1

Mark]


𝑥(I.F.) = ∫(Q × IF) 𝑑𝑦 + C

⇒ 𝑥 ×1

𝑦= ∫ (3𝑦 ×

1

𝑦) 𝑑𝑦 + C

⇒𝑥

𝑦= 3𝑦 + C

⇒ 𝑥 = 3𝑦2 + Cy

Hence, the required general solution of the given differential equation is 𝑥 = 3𝑦2 + Cy [2

Marks]

14. 𝑑𝑦

𝑑𝑥+ 2𝑦 tan 𝑥 = sin 𝑥; 𝑦 = 0 when 𝑥 =

𝜋

3 [4

Marks]

Solution:


𝑑𝑥+ 2𝑦 tan 𝑥 = sin 𝑥



This is a linear equation of the form:

𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄(where 𝑝 = 2 tan 𝑥 and 𝑄 = sin 𝑥

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫ 2tan𝑥𝑑𝑥 = 𝑒2log|sec𝑥| = 𝑒log(sec2𝑥) = sec2𝑥. [1

Mark]


𝑦(I.F.) = ∫(Q × I.F, ) 𝑑𝑥 + C

⇒ 𝑦(sec2𝑥) = ∫(sin𝑥 · sec2𝑥) 𝑑𝑥 + C

⇒ 𝑦sec2𝑥 = ∫(sec𝑥 · tan𝑥) 𝑑𝑥 + C

⇒ 𝑦sec2𝑥 = sec𝑥 + C … (1) [1

Mark]

Now, 𝑦 = 0 at 𝑥 =𝜋

3

Therefore,

0 × sec2𝜋

3= sec

𝜋

3+ C

⇒ 0 = 2 + C

⇒ C = −2 [1

Mark]


𝑦sec2𝑥 = sec𝑥 − 2

⇒ 𝑦 = cos𝑥 − 2cos2𝑥

Hence, the required solution of the given differential equation is 𝑦 = cos 𝑥 − 2 cos2 𝑥. [1

Mark]

15. (1 + 𝑥2)𝑑𝑦

𝑑𝑥+ 2𝑥𝑦 =

1

1+𝑥2 ; 𝑦 = 0 when 𝑥 = 1 [4

Marks]

Solution:

Given differential equation is (1 + 𝑥2)𝑑𝑦

𝑑𝑥+ 2𝑥𝑦 =

1

1+𝑥2

⇒𝑑𝑦

𝑑𝑥+

2𝑥𝑦

1 + 𝑥2=

1

(1 + 𝑥2)2




𝑑𝑦


2𝑥

1 + 𝑥2 and 𝑄 =

1

(1 + 𝑥2)2)

Now, 𝐼. 𝐹 = 𝑒∫ 𝜌𝑑𝑥 = 𝑒∫2𝑥𝑑𝑥

1+𝑥2 = 𝑒log(1+𝑥2) = 1 + 𝑥2 [1

Mark]


𝑦(I.F) = ∫(Q × I.F) 𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = ∫ [1

(1 + 𝑥2)2· (1 + 𝑥2)] 𝑑𝑥 + C

⇒ 𝑦(1 + 𝑥2) = ∫1

1 + 𝑥2𝑑𝑥 + 𝐶

⇒ 𝑦(1 + 𝑥2) = tan−1𝑥 + 𝐶 … (1) [1

Mark]

Now, 𝑦 = 0 at 𝑥 = 1.

Therefore,

0 = tan−11 + C

⇒ C = −𝜋

4 [1

Mark]

Substituting 𝐶 = −𝜋


𝑦(1 + 𝑥2) = tan−1𝑥 −𝜋

4


𝑦(1 + 𝑥2) = tan−1𝑥 −𝜋

4 [1

Mark]

16. 𝑑𝑦

𝑑𝑥− 3𝑦cot𝑥 = sin2𝑥; 𝑦 = 2 when 𝑥 =

𝜋

2 [4

Marks]

Solution:


𝑑𝑥− 3𝑦cot𝑥 = sin2𝑥




𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄 (where 𝑝 = −3cot𝑥 and 𝑄 = sin 𝑥)

Now, 𝐼. 𝐹 = 𝑒∫ 𝜌𝑑𝑥 = 𝑒−3 ∫ cot𝑥𝑑𝑥 = 𝑒−3log|sin𝑥| = 𝑒log|

1

|sin3𝑥|

=1

sin3𝑥 [1

Mark]


𝑦(I.F.) = ∫ (Q × I.F.)𝑑𝑥 + C

⇒ 𝑦 ·1

sin3𝑥= ∫ [sin2𝑥 ·

1

sin3𝑥] 𝑑𝑥 + C [1

Mark]

⇒ 𝑦cosec3𝑥 = 2 ∫(cot𝑥cosec𝑥) 𝑑𝑥 + C

⇒ 𝑦cosec3𝑥 = 2cosec𝑥 + 𝐶

⇒ 𝑦 = −2

cosec2𝑥+

3

cosec3𝑥

⇒ 𝑦 = −2sin2𝑥 + Csin3𝑥 … (1)


2

Therefore, we get:

2 = −2 + C

⇒ C = 4 [1

Mark]


𝑦 = −2sin2𝑥 + 4sin3𝑥

⇒ 𝑦 = 4sin3𝑥 − 2sin2𝑥

Hence, the required particular solution of the given differential equation is

𝑦 = 4sin3𝑥 − 2sin2𝑥 [1

Mark]

17. Find the equation of a 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 the coordinates of the

point. [4 Marks]

Solution:

Let 𝐹(𝑥, 𝑦) be the curve passing through the origin.

At point (𝑥, 𝑦), the slope of the curve will be 𝑑𝑦

𝑑𝑥




𝑑𝑦

𝑑𝑥= 𝑥 + 𝑦

⇒𝑑𝑦

𝑑𝑥− 𝑦 = 𝑥


𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄 (where 𝑝 = −1 and 𝑄 = 𝑥)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫(−1)𝑑𝑥 = 𝑒−𝑥. [1

Mark]


𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦𝑒−𝑥 = ∫ 𝑥 𝑒−𝑥𝑑𝑥 + C … (1)

Now, ∫ 𝑥𝑒−𝑥 𝑑𝑥 = 𝑥 ∫ 𝑒−𝑥 𝑑𝑥 − ∫ [𝑑

𝑑𝑥(𝑥) · ∫ 𝑒−𝑥 𝑑𝑥] 𝑑𝑥

= −𝑥𝑒−𝑥 − ∫ −𝑒−𝑥 𝑑𝑥 [1

Mark]

= −𝑥𝑒−𝑥 + (−𝑒−𝑥)

= −𝑒−𝑥(𝑥 + 1)

Substituting in equation (1), we get:

𝑦𝑒−𝑥 = −𝑒−𝑥(𝑥 + 1) + 𝐶

⇒ 𝑦 = −(𝑥 + 1) + 𝐶𝑒𝑥

⇒ 𝑥 + 𝑦 + 1 = 𝐶𝑒𝑥 … (2) [1

Mark]

The curve passes through the origin.


1 = 𝐶

⇒ 𝐶 = 1


𝑥 + 𝑦 + 1 = 𝑒𝑥

Hence, the required equation of curve passing through the origin is 𝑥 + 𝑦 + 1 = 𝑒𝑥 [1

Mark]



18. Find the equation of a 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 tangent

to the curve at that point by 5.

[4 Marks]

Solution:

Let 𝐹(𝑥, 𝑦) be the curve and let (𝑥, 𝑦) be a point on the curve. The slope of the tangent

to the curve at (𝑥, 𝑦) is 𝑑𝑦

𝑑𝑥.


𝑑𝑦

𝑑𝑥+ 5 = 𝑥 + 𝑦

⇒𝑑𝑦

𝑑𝑥− 𝑦 = 𝑥 − 5


𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄(where 𝑝 = −1 and 𝑄 = 𝑥 − 5)

Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑘𝑥 = 𝑒∫(−1)𝑑𝑥 = 𝑒−𝑥. [1

Mark]

The general equation of the curve is given by the relation,

𝑦(I.F.) = ∫(Q × IF) 𝑑𝑥 + C

⇒ 𝑦 · 𝑒−𝑥 = ∫(𝑥 − 5) 𝑒−𝑥𝑑𝑥 + C

Now, ∫(𝑥 − 5) 𝑒−𝑥𝑑𝑥 = (𝑥 − 5) ∫ 𝑒−𝑥 𝑑𝑥 − ∫ [𝑑

𝑑𝑥(𝑥 − 5) · ∫ 𝑒−𝑥 𝑑𝑥] 𝑑𝑥

= (𝑥 − 5)(−𝑒−𝑥) − ∫(−𝑒−𝑥) 𝑑𝑥 [1

Mark]

= (5 − 𝑥)𝑒−𝑥 + (−𝑒−𝑥)

= (4 − 𝑥)𝑒−𝑥


𝑦𝑒−𝑥 = (4 − 𝑥)𝑒−𝑥 + 𝐶

⇒ 𝑦 = 4 − 𝑥 + 𝐶𝑒𝑥

⇒ 𝑥 + 𝑦 − 4 = 𝐶𝑒𝑥 … (2) [1

Mark]

The curve passes through point (0, 2).




0 + 2– 4 = 𝐶𝑒0

⇒– 2 = 𝐶

⇒ 𝐶 =– 2


𝑥 + 𝑦 − 4 = −2𝑒𝑥

⇒ 𝑦 = 4 − 𝑥 − 2𝑒𝑥

Hence, the required equation of the curve is 𝑦 = 4 − 𝑥 − 2𝑒𝑥 [1

Mark]

19. The integrating factor of the differential equation 𝑥𝑑𝑦

𝑑𝑥− 𝑦 = 2𝑥2 is [2

Marks]

(A) 𝑒−𝑥

(B) 𝑒−𝑦

(C) 1

𝑥

(D) 𝑥

Solution:

The given differential equation is 𝑥𝑑𝑦

𝑑𝑥− 𝑦 = 2𝑥2

⇒𝑑𝑦

𝑑𝑥−

𝑦

𝑥= 2𝑥


𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄 (where 𝑝 = −

1

𝑥and 𝑄 = 2𝑥) [1

Mark]

The integrating factor (𝐼. 𝐹) is given by the relation,

𝑒∫ 𝑝𝑑𝑥

∴ 𝐼. 𝐹 = 𝑒∫1𝑥

𝑑𝑥 = 𝑒−log𝑥 = 𝑒log(𝑥−1) = 𝑥−1 =1

𝑥

Hence, the correct answer is C. [1

Mark]



20. The integrating factor of the differential equation.

(1 − 𝑦2)𝑑𝑥

𝑑𝑦+ 𝑦𝑥 = 𝑎𝑦(−1 < 𝑦 < 1) [2

Marks]

(A) 1

𝑦2−1

(B) 1

√𝑦2−1

(C) 1

1−𝑦2

(D) 1

√1−𝑦2

Solution:

The given differential equation is (1 − 𝑦2)𝑑𝑥

𝑑𝑦+ 𝑦𝑥 = 𝑎𝑦

⇒𝑑𝑦

𝑑𝑥+

𝑦𝑥

1 − 𝑦2=

𝑎𝑦

1 − 𝑦2


𝑑𝑥

𝑑𝑦+ 𝑝𝑦 = 𝑄 (where 𝑝 =

𝑦

1−𝑦2 and 𝑄 =𝑎𝑦

1−𝑦2) [1

Mark]

The integrating factor (I.F) is given by the relation,

𝑒∫ 𝑝𝑑𝑡

∴ I.F = 𝑒∫ 𝑝𝑑𝑦 = 𝑒∫

𝑦1−𝑦2𝑑𝑦

= 𝑒−12

log(1−𝑦2) = 𝑒log[

1

√1−𝑦2]

=1

√1 − 𝑦2

Hence, the correct answer is D. [1

Mark]

Miscellaneous exercise 9

1. For each of the differential equations given below, indicate its order and degree (if

defined).

(I) 𝑑2𝑦

𝑑𝑥2 + 5𝑥 (𝑑𝑦

𝑑𝑥)

2− 6𝑦 = log𝑥 [1

Mark]

(II) (𝑑𝑦

𝑑𝑥)

3− 4 (

𝑑𝑦

𝑑𝑥)

2+ 7𝑦 = sin 𝑥 [1

Mark]



(III) 𝑑4𝑦

𝑑𝑥4 − sin (𝑑3𝑦

𝑑𝑥3) = 0 [1

Mark]

Solution:

(I)The differential equation is given as 𝑑2𝑦

𝑑𝑥2 + 5𝑥 (𝑑𝑦

𝑑𝑥)

2− 6𝑦 = log𝑥

⇒𝑑2𝑦

𝑑𝑥2+ 5𝑥 (

𝑑𝑦

𝑑𝑥)

2

− 6𝑦 − log𝑥 = 0


𝑑𝑥2. Thus, its order is two.

The highest power raised to 𝑑2𝑦

𝑑𝑥2 is one. Hence, its degree is one.

[1 Mark]

(II) The differential equation is given as:

(𝑑𝑦

𝑑𝑥)

3

− 4 (𝑑𝑦

𝑑𝑥)

2

+ 7𝑦 = sin𝑥

⇒ (𝑑𝑦

𝑑𝑥)

3

− 4 (𝑑𝑦

𝑑𝑥)

2

+ 7𝑦 − sin𝑥 = 0

The highest order derivative present in the differential equation is 𝑑𝑦

𝑑𝑥. Thus, its order is one.

The highest power raised to is three. Hence, its degree is three.

[1 Mark]

(iii) The differential equation is given as:

𝑑4𝑦

𝑑𝑥4− sin (

𝑑3𝑦

𝑑𝑥3) = 0


𝑑𝑥4. Thus, its order is four.

However, the given differential equation is not a polynomial equation.

Hence, its degree is not defined. [1

Mark]

2. For each of the exercises given below, verify that the given function (implicit or explicit)

is a solution of the corresponding differential equation.

(I) 𝑦 = 𝑎𝑒𝑥 + 𝑏𝑒−𝑥 + 𝑥2: 𝑥𝑑2𝑦

𝑑𝑥2 + 2𝑑𝑦

𝑑𝑥− 𝑥𝑦 + 𝑥2 − 2 = 0 [2

MarkS]



(II) 𝑦 = 𝑒𝑥(𝑎 cos 𝑥 + 𝑏 sin 𝑥):𝑑2𝑦


𝑑𝑥+ 2𝑦 = 0 [4

MarkS]

(III) 𝑦 = 𝑥sin3𝑥:𝑑2𝑦

𝑑𝑥2 + 9𝑦 − 6cos3𝑥 = 0 [2

MarkS]

(IV) 𝑥2 = 2𝑦2log𝑦: (𝑥2 + 𝑦2)𝑑𝑦

𝑑𝑥− 𝑥𝑦 = 0 [2

MarkS]

Solution:

(I) 𝑦 = 𝑎𝑒𝑥 + 𝑏𝑒−𝑥 + 𝑥2


𝑑𝑦

𝑑𝑥= 𝑎

𝑑

𝑑𝑥(𝑒𝑥) + 𝑏

𝑑

𝑑𝑥(𝑒−𝑥) +

𝑑

𝑑𝑥(𝑥2)

⇒𝑑𝑦

𝑑𝑥= 𝑎𝑒𝑥 − 𝑏𝑒−𝑥 + 2𝑥

Again, differentiating both sides with respect to x, we get:

𝑑2𝑦

𝑑𝑥2 = 𝑎𝑒𝑥 + 𝑏𝑒−𝑥 + 2 [1

Mark]

Now, on substituting the values of 𝑑𝑦

𝑑𝑥 and

𝑑2𝑦

𝑑𝑥2 in the differential equation, we get:

L.H.S.

𝑥𝑑2𝑦

𝑑𝑥2+ 2

𝑑𝑦

𝑑𝑥− 𝑥𝑦 + 𝑥2 − 2

= 𝑥(𝑎𝑒𝑥 + 𝑏𝑒−𝑥 + 2) + 2(𝑎𝑒𝑥 − 𝑏𝑒−𝑥 + 2𝑥) − 𝑥(𝑎𝑒𝑥 + 𝑏𝑒−𝑥 + 𝑥2) + 𝑥2 − 2

= (𝑎𝑥𝑒𝑥 + 𝑏𝑥𝑒−𝑥 + 2𝑥) + (2𝑎𝑒𝑥 − 2𝑏𝑒−𝑥 + 4𝑥) − (𝑎𝑥𝑒𝑥 + 𝑏𝑥𝑒−𝑥 + 𝑥3) + 𝑥2 − 2

= 2𝑎𝑒𝑥 − 2𝑏𝑒−𝑥 + 𝑥2 + 6𝑥 − 2

≠ 0

⇒ L. H. S. ≠ R. H. S. [1

Mark]

Hence, the given function is not a solution of the corresponding differential equation.

(II) 𝑦 = 𝑒𝑥(𝑎 cos 𝑥 + 𝑏 sin 𝑥) = 𝑎𝑒𝑥 cos 𝑥 + 𝑏𝑒𝑥sin𝑥


𝑑𝑦

𝑑𝑥= 𝑎 ·

𝑑

𝑑𝑥(𝑒𝑥cos𝑥) + 𝑏 ·

𝑑

𝑑𝑥(𝑒𝑥sin𝑥)



⇒𝑑𝑦

𝑑𝑥= 𝑎(𝑒𝑥cos𝑥 − 𝑒𝑥sin𝑥) + 𝑏 · (𝑒𝑥sin𝑥 + 𝑒𝑥cos𝑥)

⇒𝑑𝑦

𝑑𝑥= (𝑎 + 𝑏)𝑒𝑥cos𝑥 + (𝑏 − 𝑎)𝑒𝑥sin𝑥 [1

Mark]


𝑑2𝑦

𝑑𝑥2 = (𝑎 + 𝑏) ·𝑑

𝑑𝑥(𝑒𝑥cos𝑥) + (𝑏 − 𝑎)

𝑑

𝑑𝑥(𝑒𝑥sin𝑥)

⇒𝑑2𝑦

𝑑𝑥2= (𝑎 + 𝑏) · [𝑒𝑥cos𝑥 − 𝑒𝑥sin𝑥] + (𝑏 − 𝑎)[𝑒𝑥sin𝑥 + 𝑒𝑥cos𝑥]

⇒𝑑2𝑦

𝑑𝑥2= 𝑒𝑥[(𝑎 + 𝑏)(cos𝑥 − sin𝑥) + (𝑏 − 𝑎)(sin𝑥 + cos𝑥)]

⇒𝑑2𝑦

𝑑𝑥2= 𝑒𝑥[𝑎cos𝑥 − 𝑎sin𝑥 + 𝑏cos𝑥 − 𝑏sin𝑥 + 𝑏sin𝑥 + 𝑏cos𝑥 − 𝑎sin𝑥 − 𝑎cos𝑥]

⇒𝑑2𝑦

𝑑𝑥2 = [2𝑒𝑥(𝑏cos𝑥 − 𝑎sin𝑥)] [1

Mark]

Now, on substituting the values of 𝑑2𝑦

𝑑𝑥2 and 𝑑𝑦

𝑑𝑥 in the L.H.S. of the given differential equation,

we get:

𝑑2𝑦

𝑑𝑥2+ 2

𝑑𝑦

𝑑𝑥+ 2𝑦

= 2𝑒𝑥(𝑏cos𝑥 − 𝑎sin𝑥) − 2𝑒𝑥[(𝑎 + 𝑏)cos𝑥 + (𝑏 − 𝑎)sin𝑥] + 2𝑒𝑥(𝑎cos𝑥 + 𝑏sin𝑥)

= 𝑒𝑥 [(2𝑏cos𝑥 − 2𝑎sin𝑥) − (2𝑎cos𝑥 + 2𝑏cos𝑥)

−(2𝑏sin𝑥 − 2𝑎sin𝑥) + (2𝑎cos𝑥 + 2𝑏sin𝑥)]

= 𝑒𝑥[(2𝑏 − 2𝑎 − 2𝑏 + 2𝑎)cos𝑥] + 𝑒𝑥[(−2𝑎 − 2𝑏 + 2𝑎 + 2𝑏)sin𝑥]

= 0

Hence, the given function is a solution of the corresponding differential equation. [2

MarkS]

(III) 𝑦 = 𝑥 sin 3𝑥


𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥(𝑥sin3𝑥) = sin3𝑥 + 𝑥 · cos3𝑥 · 3

⇒𝑑𝑦

𝑑𝑥= sin3𝑥 + 3𝑥cos3𝑥


𝑑2𝑦

𝑑𝑥2=

𝑑

𝑑𝑥(sin3𝑥) + 3

𝑑

𝑑𝑥(𝑥cos3𝑥)



⇒𝑑2𝑦

𝑑𝑥2= 3cos3𝑥 + 3[cos3𝑥 + 𝑥(−sin3𝑥) · 3]

⇒𝑑2𝑦

𝑑𝑥2 = 6cos3𝑥 − 9𝑥sin3𝑥 [1

Mark]

Substituting the value of 𝑑2𝑦

𝑑𝑥2 in the L.H.S. of the given differential equation, we get:

𝑑2𝑦

𝑑𝑥2+ 9𝑦 − 6cos3𝑥

= (6 · cos3𝑥 − 9𝑥sin3𝑥) + 9𝑥sin3𝑥 − 6cos3𝑥

= 0


Mark]

(IV) 𝑥2 = 2𝑦2 log 𝑦


2𝑥 = 2 ·𝑑

𝑑𝑥= [𝑦2log𝑦]

⇒ 𝑥 = [2𝑦 · log𝑦 ·𝑑𝑦

𝑑𝑥+ 𝑦2 ·

1

𝑦·

𝑑𝑦

𝑑𝑥]

⇒ 𝑥 =𝑑𝑦

𝑑𝑥(2𝑦log𝑦 + 𝑦)

⇒𝑑𝑦

𝑑𝑥=

𝑥

𝑦(1+2log𝑦) [1

Mark]

Substituting the value of 𝑑𝑦

𝑑𝑥 in the L.H.S. of the given differential equation, we get:

(𝑥2 + 𝑦2)𝑑𝑦

𝑑𝑥− 𝑥𝑦

= (2𝑦2log𝑦 + 𝑦2) ·𝑥

𝑦(1 + 2log𝑦)− 𝑥𝑦

= 𝑦2(1 + 2log𝑦) ·𝑥

𝑦(1 + 2log𝑦)− 𝑥𝑦

= 𝑥𝑦 − 𝑥𝑦

= 0


Mark]



3. Form the differential equation representing the family of curves given by

(𝑥 − 𝑎)2 + 2𝑦2 = 𝑎2

where a is an arbitrary constant. [2

Marks]

Solution:

(𝑥 − 𝑎)2 + 2𝑦2 = 𝑎2

⇒ 𝑥2 + 𝑎2 − 2𝑎𝑥 + 2𝑦2 = 𝑎2

⇒ 2𝑦2 = 2𝑎𝑥 − 𝑥2 … (1)

Differentiating with respect to x, we get:

2𝑦𝑑𝑦

𝑑𝑥=

2𝑎 − 2𝑥

2

⇒𝑑𝑦

𝑑𝑥=

𝑎 − 𝑥

2𝑦

⇒𝑑𝑦

𝑑𝑥=

2𝑎𝑥−2𝑥2

4𝑥𝑦… (2) [1

Mark]

From equation (1), we get:

2𝑎𝑥 = 2𝑦2 + 𝑥2

On substituting this value in equation (3), we get:

𝑑𝑦

𝑑𝑥=

2𝑦2 + 𝑥2 − 2𝑥2

4𝑥𝑦

⇒𝑑𝑦

𝑑𝑥=

2𝑦2 − 𝑥2

4𝑥𝑦

Hence, the differential equation of the family of curves is given as 𝑑𝑦

𝑑𝑥=

2𝑦2−𝑥2

4𝑥𝑦 [1

Mark]

4. Prove that 𝑥2 − 𝑦2 = 𝑐(𝑥2 + 𝑦2)2 is the general solution of differential equation,

(𝑥3 − 3𝑥𝑦2)𝑑𝑥 = (𝑦3 − 3𝑥𝑦2)𝑑𝑥 = (𝑦3 − 3𝑥2𝑦)𝑑𝑦 where 𝑐 is a parameter.

[6 Marks]

Solution:

(𝑥3 − 3𝑥𝑦2)𝑑𝑥 = (𝑦3 − 3𝑥2𝑦)𝑑𝑦



⇒𝑑𝑦

𝑑𝑥=

𝑥3 − 3𝑥𝑦2

𝑦3 − 3𝑥2𝑦… (1)

This is a homogeneous equation. To simplify it, we need to make the substitution as:

𝑦 = 𝑣𝑥

⇒𝑑

𝑑𝑥(𝑦) =

𝑑

𝑑𝑥(𝑣𝑥)

⇒𝑑𝑦


𝑑𝑣

𝑑𝑥 [1

Mark]

Substituting the values of 𝑦 and 𝑑𝑣

𝑑𝑥

𝑣 + 𝑥𝑑𝑣

𝑑𝑥=

𝑥3 − 3𝑥(𝑣𝑥)2

(𝑣𝑥)3 − 3𝑥2(𝑣𝑥)


𝑑𝑥=

1 − 3𝑣2

𝑣3 − 3𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

1 − 3𝑣2

𝑣3 − 3𝑣− 𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

1 − 3𝑣2 − 𝑣(𝑣3 − 3𝑣)

𝑣3 − 3𝑣

⇒ 𝑥𝑑𝑣

𝑑𝑥=

1 − 𝑣4

𝑣3 − 3𝑣

⇒ (𝑣3−3𝑣

1−𝑣4 ) 𝑑𝑣 =𝑑𝑥

𝑥 [1

Mark]


∫ (𝑣3−3𝑣

1−𝑣4 ) 𝑑𝑣 = log𝑥 + logC′ … (2)

Now, ∫ (𝑣3−3𝑣

1−𝑣4 ) 𝑑𝑣 = ∫𝑣3𝑑𝑣

1−𝑣4 − 3 ∫𝑣𝑑𝑣

1−𝑣4

⇒ ∫ (𝑣3−3𝑣

1−𝑣4 ) 𝑑𝑣 = 𝐼1 − 3𝐼2, where 𝐼1 = ∫𝑣3𝑑𝑣

1−𝑣4 and 𝐼2 = ∫𝑣𝑑𝑣

1−𝑣4 … (3) [1

Mark]

Let 1 − 𝑣4 = 𝑡

∴𝑑

𝑑𝑣(1 − 𝑣4) =

𝑑𝑡

𝑑𝑣

⇒ −4𝑣3 =𝑑𝑡

𝑑𝑣

⇒ 𝑣3𝑑𝑣 = −𝑑𝑡

4𝑡= −

1

4log𝑡 = −

1

4log(1 − 𝑣4)

And, 𝐼2 = ∫𝑣𝑑𝑣

1−𝑣4 = ∫𝑣𝑑𝑣

1−(𝑣2)2 [1

Mark]



Let 𝑣2 = 𝑝

∴𝑑

𝑑𝑣(𝑣2) =

𝑑𝑝

𝑑𝑣

⇒ 2𝑣 =𝑑𝑝

𝑑𝑣

⇒ 𝑣𝑑𝑣 =𝑑𝑝

2

⇒ 𝐼2 =1

2∫

𝑑𝑝

1−𝑝2 =1

2×2log |

1+𝑝

1−𝑝| =

1

4log |

1+𝑣2

1−𝑣2| [1

Mark]

Substituting the values of 𝐼1 and 𝐼2 in equation (3), we get:

∫ (𝑣3 − 3𝑣

1 − 𝑣4 ) 𝑑𝑣 = −1

4log(1 − 𝑣4) −

3

4log |

1 − 𝑣2

1 + 𝑣2|


1

4log(1 − 𝑣4) −

3

4log |

1 + 𝑣2

1 − 𝑣2| = log𝑥 + log𝐶′

⇒ −1

4log [(1 − 𝑣4) (

1 + 𝑣2

1 − 𝑣2)

3

] = log𝐶′𝑥

⇒(1 + 𝑣2)4

(1 − 𝑣2)2= (𝐶′𝑥)−4

⇒(1 +

𝑦2

𝑥2)4

(1 −𝑦2

𝑥2)2 =

1

C'4𝑥4

⇒(𝑥2 + 𝑦2)4

𝑥4(𝑥2 − 𝑦2)2=

1

C4𝑥4

⇒ (𝑥2 − 𝑦2)2 = C4(𝑥2 + 𝑦2)4

⇒ 𝑥2 − 𝑦2 = C(𝑥2 + 𝑦2)2, where 𝐶 = 𝐶′2

Hence proved. [1

Mark]

5. Form the differential equation of the family of circles in the first quadrant which touch

the coordinate axes.

[4 Marks]



Solution:

The equation of a circle in the first quadrant with centre (𝑎, 𝑎) and radius (𝑎) which touches

the coordinate axes is:

(𝑥 − 𝑎)2 + (𝑦 − 𝑎)2 = 𝑎2 … (1)

Differentiating equation (1) with respect to x, we get:

2(𝑥 − 𝑎) + 2(𝑦 − 𝑎)𝑑𝑦

𝑑𝑥= 0

⇒ (𝑥 − 𝑎) + (𝑦 − 𝑎)𝑦′ = 0

⇒ 𝑥 − 𝑎 + 𝑦𝑦′ − 𝑎𝑦′ = 0

⇒ 𝑥 + 𝑦𝑦′ − 𝑎(1 + 𝑦′) = 0

⇒ 𝑎 =𝑥+𝑦𝑦′

1+𝑦′ [2

Marks]

Substituting the value of a in equation (1), we get:

[𝑥 − (𝑥 + 𝑦𝑦′

1 + 𝑦′ )]

2

+ [𝑦 − (𝑥 + 𝑦𝑦′

1 + 𝑦′ )]

2

= (𝑥 + 𝑦𝑦′

1 + 𝑦′ )

2

⇒ [(𝑥 − 𝑦)𝑦′

(1 + 𝑦′)]

2

+ [𝑦 − 𝑥

1 + 𝑦′]2

= [𝑥 + 𝑦𝑦′

1 + 𝑦′]

2

⇒ (𝑥 − 𝑦)2 · 𝑦′2 + (𝑥 − 𝑦)2 = (𝑥 + 𝑦𝑦′)2

⇒ (𝑥 − 𝑦)2[1 + (𝑦′)2] = (𝑥 + 𝑦𝑦′)2

Hence, the required differential equation of the family of circles is

(𝑥 − 𝑦)2[1 + (𝑦′)2] = (𝑥 + 𝑦𝑦′)2. [2

Marks]

6. Find the general solution of the differential equation 𝑑𝑦

𝑑𝑥+ √

1−𝑦2

1−𝑥2 = 0

[2 Marks]



Solution:


𝑑𝑥+ √

1−𝑦2

1−𝑥2 = 0

⇒𝑑𝑦

𝑑𝑥= −

√1 − 𝑦2

√1 − 𝑥2

⇒𝑑𝑦

√1−𝑦2=

−𝑑𝑥

√1−𝑥2 [1

Mark]


sin−1𝑦 = −sin−1𝑥 + 𝐶

⇒ sin−1𝑥 + sin−1𝑦 = 𝐶

Hence, the general solution of given differential equation is sin−1𝑥 + sin−1𝑦 = 𝐶 [1

Mark]

7. Show that the general solution of the differential equation 𝑑𝑦

𝑑𝑥+

𝑦2+𝑦+1

𝑥2+𝑥+1= 0 is given by

(𝑥 + 𝑦 + 1) = 𝐴(1 − 𝑥 − 𝑦 − 2𝑥𝑦), where 𝐴 is parameter [4

Marks]

Solution:


𝑑𝑥+

𝑦2+𝑦+1

𝑥2+𝑥+1= 0

⇒𝑑𝑦

𝑑𝑥= −

(𝑦2 + 𝑦 + 1)

𝑥2 + 𝑥 + 1

⇒𝑑𝑦

𝑦2 + 𝑦 + 1=

−𝑑𝑥

𝑥2 + 𝑥 + 1

⇒𝑑𝑦

𝑦2+𝑦+1+

𝑑𝑥

𝑥2+𝑥+1= 0 [1

Mark]


∫𝑑𝑦

𝑦2 + 𝑦 + 1+ ∫

𝑑𝑥

𝑥2 + 𝑥 + 1= C

⇒ ∫𝑑𝑦

(𝑦 +12)

2

+ (√32 )

2 + ∫𝑑𝑥

(𝑥 +12)

2

+ (√32 )

2 = C



⇒2

√3tan−1 [

𝑦+1

2

√3

2

] +2

√3tan−1 [

𝑥+1

2

√3

2

] = C [1

Mark]

⇒ tan−1 [2𝑦 + 1

√3] + tan−1 [

2𝑥 + 1

√3] =

√3C

2

⇒ tan−1 [

2𝑦 + 1

√3+

2𝑥 + 1

√3

1 −(2𝑦 + 1)

√3∙

(2𝑥 + 1)

√3

] =√3C

2

⇒ tan−1 [

2𝑥 + 2𝑦 + 2

√3

1 − (4𝑥𝑦 + 2𝑥 + 2𝑦 + 1

3 )] =

√3C

2

⇒ tan−1 [2√3(𝑥 + 𝑦 + 1)

3 − 4𝑥𝑦 − 2𝑥 − 2𝑦 − 1] =

√3C

2

⇒ tan−1 [√3(𝑥+𝑦+1)

2(1−𝑥−𝑦−2𝑥𝑦)] =

√3C

2 [1

Mark]

⇒√3(𝑥+𝑦+1)

2(1−𝑥−𝑦−2𝑥𝑦)= tan (

√3C

2) = 𝐵, where 𝐵 = tan (

√3C

2)

⇒ 𝑥 + 𝑦 + 1 =2𝐵

√3(1 − 𝑥𝑦 − 2𝑥𝑦)

⇒ 𝑥 + 𝑦 + 1 = 𝐴(1 − 𝑥 − 𝑦 − 2𝑥𝑦), where 𝐴 =2𝐵

√3

Hence proved [1

Mark]

8. Find the equation of the curve passing through the point (0,𝜋

4) whose differential

equation is, sin 𝑥 cos 𝑦𝑑𝑥 + cos 𝑥 sin 𝑦𝑑𝑦 = 0

[4 Marks]

Solution:

The differential equation of the given curve is sin 𝑥 cos 𝑦𝑑𝑥 + cos 𝑥 sin 𝑦𝑑𝑦 = 0

⇒sin𝑥cos𝑦𝑑𝑥 + cos𝑥sin𝑦𝑑𝑦

cos𝑥cos𝑦= 0

⇒ tan𝑥𝑑𝑥 + tan𝑦𝑑𝑦 = 0 [1

Mark]




log(sec𝑥) + log(sec𝑦) = logC

log(sec𝑥 · sec𝑦) = log𝐶

⇒ sec 𝑥 ∙ sec 𝑦 = 𝐶 … (1) [1

Mark]

The curve passes through point (0,𝜋

4)

∴ 1 × √2 = 𝐶

⇒ 𝐶 = √2 [1

Mark]

On substituting 𝐶 = √2

sec𝑥 · sec𝑦 = √2

⇒ sec𝑥 ·1

cos𝑦= √2

⇒ cos𝑦 =sec𝑥

√2

Hence, the required equation of the curve is cos 𝑦 =sec 𝑥

√2 [1

Mark]

9. Find the particular solution of the different equation\

(1 + 𝑒2𝑥)𝑑𝑦 + (1 + 𝑦2)𝑒𝑥𝑑𝑥 = 0, given that 𝑦 = 1 when 𝑥 = 0 [4

Marks]

Solution:

Given differential equation is (1 + 𝑒2𝑥)𝑑𝑦 + (1 + 𝑦2)𝑒𝑥𝑑𝑥 = 0

⇒𝑑𝑦

1 + 𝑦2+

𝑒𝑥𝑑𝑥

1 + 𝑒2𝑥= 0


tan−1 𝑦 + ∫𝑒𝑥𝑑𝑥

1+𝑒2𝑥 = 𝐶 … (1) [1

Mark]

Let 𝑒𝑥 = 𝑡 ⇒ 𝑒2𝑥 = 𝑡2.

⇒𝑑

𝑑𝑥(𝑒𝑥) =

𝑑𝑡

𝑑𝑥

⇒ 𝑒𝑥 =𝑑𝑡

𝑑𝑥



⇒ 𝑒𝑥𝑑𝑥 = 𝑑𝑡 [1

Mark]

Substituting these values in equation (1), we get:

tan−1𝑦 + ∫𝑑𝑡

1 + 𝑡2= 𝐶

⇒ tan−1𝑦 + tan−1𝑡 = C

⇒ tan−1𝑦 + tan−1(𝑒𝑥) = C … (2) [1

Mark]

Now, 𝑦 = 1 at 𝑥 = 0.


tan−11 + tan−11 = 𝐶

⇒𝜋

4+

𝜋

4= C

⇒ C =𝜋

2

Substituting 𝐶 =𝜋


tan−1𝑦 + tan−1(𝑒𝑥) =𝜋

2


tan−1𝑦 + tan−1(𝑒𝑥) =𝜋

2 [1

Mark]

10. Solve the differential equation 𝑦𝑒𝑥

𝑦𝑑𝑥 = (𝑥𝑒𝑥

𝑦 + 𝑦2) 𝑑𝑦(𝑦 ≠ 0) [2

Marks]

Solution:

Given differential equation is 𝑦𝑒𝑥

𝑦𝑑𝑥 = (𝑥𝑒𝑥

𝑦 + 𝑦2) 𝑑𝑦

⇒ 𝑦𝑒𝑥𝑦

𝑑𝑥

𝑑𝑦= 𝑥𝑒

𝑥𝑦 + 𝑦2

⇒ 𝑒𝑥𝑦 [𝑦 ·

𝑑𝑥

𝑑𝑦− 𝑥] = 𝑦2

⇒ 𝑒𝑥

𝑦 ·[𝑦·

𝑑𝑥

𝑑𝑦−𝑥]

𝑦2 = 1 … (1) [𝟏

𝟐

Mark]



Let 𝑒𝑥

𝑦 = 𝑧

Differentiating it with respect to y, we get:

𝑑

𝑑𝑦(𝑒

𝑥𝑦) =

𝑑𝑧

𝑑𝑦

⇒ 𝑒𝑥𝑦 ·

𝑑

𝑑𝑦(

𝑥

𝑦) =

𝑑𝑧

𝑑𝑦

⇒ 𝑒𝑥

𝑦 · [𝑦·

𝑑𝑥

𝑑𝑦−𝑥

𝑦2 ] =𝑑𝑧

𝑑𝑦… (2) [

𝟏

𝟐

Mark]

From equation (1) and equation (2), we get:

𝑑𝑧

𝑑𝑦= 1

⇒ 𝑑𝑧 = 𝑑𝑦


𝑧 = 𝑦 + C

⇒ 𝑒𝑥

𝑦 = 𝑦 + C

Hence, the required particular solution of the given differential equation is [1

Mark]

11. Find particular solution of the differential equation.

(𝑥 − 𝑦)(𝑑𝑥 + 𝑑𝑦) = 𝑑𝑥 − 𝑑𝑦

Given that 𝑦 = −1, when 𝑥 = 0 (Hint: put 𝑥 − 𝑦 = 𝑡) [4

Marks]

Solution:

Given differential equation is (𝑥 − 𝑦)(𝑑𝑥 + 𝑑𝑦) = 𝑑𝑥 − 𝑑𝑦

⇒ (𝑥 − 𝑦 + 1)𝑑𝑦 = (1 − 𝑥 + 𝑦)𝑑𝑥

⇒𝑑𝑦

𝑑𝑥=

1 − 𝑥 + 𝑦

𝑥 − 𝑦 + 1

⇒𝑑𝑦

𝑑𝑥=

1−(𝑥−𝑦)

1+(𝑥−𝑦)… (1) [

𝟏

𝟐

Mark]

Let 𝑥 − 𝑦 = 𝑡



⇒𝑑

𝑑𝑥(𝑥 − 𝑦) =

𝑑𝑡

𝑑𝑥

⇒ 1 −𝑑𝑦

𝑑𝑥=

𝑑𝑡

𝑑𝑥

⇒ 1 −𝑑𝑡

𝑑𝑥=

𝑑𝑦

𝑑𝑥 [

𝟏

𝟐

Mark]

Substituting the values of 𝑥 − 𝑦 and 𝑑𝑦


1 −𝑑𝑡

𝑑𝑥=

1 − 𝑡

1 + 𝑡

⇒𝑑𝑡

𝑑𝑥= 1 − (

1 − 𝑡

1 + 𝑡)

⇒𝑑𝑡

𝑑𝑥=

(1 + 𝑡) − (1 − 𝑡)

1 + 𝑡

⇒𝑑𝑡

𝑑𝑥=

2𝑡

1 + 𝑡

⇒ (1 + 𝑡

𝑡) 𝑑𝑡 = 2𝑑𝑥

⇒ (1 +1

𝑡) 𝑑𝑡 = 2𝑑𝑥 … (2) [1

Mark]


𝑡 + log|𝑡| = 2𝑥 + C

⇒ (𝑥 − 𝑦) + log|𝑥 − 𝑦| = 2𝑥 + C

⇒ log|𝑥 − 𝑦| = 𝑥 + 𝑦 + C … (3)

Now, 𝑦 =– 1 at 𝑥 = 0.

Therefore, equation (3) becomes: log 1 = 0– 1 + 𝐶

⇒ 𝐶 = 1 [1

Mark]

Substituting 𝐶 = 1 in equation (3) we get:

log|𝑥 − 𝑦| = 𝑥 + 𝑦 + 1


log|𝑥 − 𝑦| = 𝑥 + 𝑦 + 1 [1

Mark]



12. Solve the differential equation [𝑒−2√𝑥

√𝑥−

𝑦

√𝑥]

𝑑𝑥

𝑑𝑦= 1(𝑥 ≠ 0) [2

Marks]

Solution:

Given differential equation is [𝑒−2√𝑥

√𝑥−

𝑦

√𝑥]

𝑑𝑥

𝑑𝑦= 1

⇒𝑑𝑦

𝑑𝑥=

𝑒−2√𝑥

√𝑥−

𝑦

√𝑥

⇒𝑑𝑦

𝑑𝑥+

𝑦

√𝑥=

𝑒−2√𝑥

√𝑥

The equation is a liner differential equation of the form

𝑑𝑦

𝑑𝑥+ 𝑃𝑦 = 𝑄,where 𝑃 =

1

√𝑥 and 𝑄 =

𝑒−2√𝑥

√𝑥

Now, I. F = 𝑒∫(𝑄×I.F.) = 𝑒∫

1

√𝑥𝑑𝑥

= 𝑒2√𝑥 [1 Mark]

The general solution of the given differential equation is given by,

𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦𝑒2√x = ∫ (𝑒−2√𝑥

√𝑥× 𝑒2√𝑥) 𝑑𝑥 + C

⇒ 𝑦𝑒2√𝑥 = ∫1

√𝑥𝑑𝑥 + C

⇒ 𝑦𝑒2√𝑥 = 2√𝑥 + C [1 Mark]

13. Find a particular solution of the differential equation

𝑑𝑦

𝑑𝑥+ 𝑦cot𝑥 = 4𝑥cosec𝑥(𝑥 ≠ 0)

Given that 𝑦 = 0 when 𝑥 =𝜋

2 [2 Marks]

Solution:


𝑑𝑥+ 𝑦 cot 𝑥 = 4𝑥 cosec 𝑥

This equation is a linear differential equation of the form

𝑑𝑦

𝑑𝑥+ 𝑝𝑦 = 𝑄, where 𝑝 = cot 𝑥 and 𝑄 = 4𝑥 cosec.



Now, 𝐼. 𝐹 = 𝑒∫ 𝑝𝑑𝑥 = 𝑒∫ cot𝑥𝑑𝑥 = 𝑒log|sin𝑥| = sin𝑥

[𝟏

𝟐 Mark]


𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦sin𝑥 = ∫(4𝑥cosec𝑥 · sin𝑥) 𝑑𝑥 + C [𝟏

𝟐 Mark]

⇒ 𝑦sin𝑥 = 4 ∫ 𝑥 𝑑𝑥 + C

⇒ 𝑦sin𝑥 = 4 ·𝑥2

2+ C

⇒ 𝑦sin𝑥 = 2𝑥2 + C … (1)


2.


0 = 2 ×𝜋2

4+ C

⇒ 𝐶 = −𝜋2

2 [

𝟏

𝟐 Mark]

Substituting 𝐶 = −𝜋2

2

𝑦sin𝑥 = 2𝑥2 −𝜋2

2


𝑦sin𝑥 = 2𝑥2 −𝜋2

2 [

𝟏

𝟐 Mark]

14. Find a particular solution of the differential equation

(𝑥 + 1)𝑑𝑦

𝑑𝑥= 2𝑒−𝑦 − 1

Given that 𝑦 = 0 when 𝑥 = 0 [4 Marks]

Solution:

Given differential equation is (𝑥 + 1)𝑑𝑦

𝑑𝑥= 2𝑒−𝑦 − 1

⇒𝑑𝑦

2𝑒−𝑦 − 1=

𝑑𝑥

𝑥 + 1



⇒𝑒𝑦𝑑𝑦

2 − 𝑒𝑦=

𝑑𝑥

𝑥 + 1


∫𝑒𝑦𝑑𝑦

2−𝑒𝑦 = log|𝑥 + 1| + log𝐶 … (1) [1 Mark]

Let 2 − 𝑒𝑦 = 𝑡

∴𝑑

𝑑𝑦(2 − 𝑒𝑦) =

𝑑𝑡

𝑑𝑦

⇒ −𝑒𝑦 =𝑑𝑡

𝑑𝑦

⇒ −𝑒𝑦𝑑𝑡 = 𝑑𝑡


∫−𝑑𝑡

𝑡= log|𝑥 + 1| + logC [1 Mark]

⇒ − log|𝑡| = log|C(𝑥 + 1)|

⇒ − log|2 − 𝑒𝑦| = log|C(𝑥 + 1)|

⇒1

2 − 𝑒𝑦= C(𝑥 + 1)

⇒ 2 − 𝑒𝑦 =1

C(𝑥 + 1)… (2)

Now, at 𝑥 = 0 and 𝑦 = 0, equation (2) becomes:

⇒ 2 − 1 =1

C

⇒ C = 1 [1 Mark]


2 − 𝑒𝑦 =1

𝑥 + 1

⇒ 𝑒𝑦 = 2 −1

𝑥 + 1

⇒ 𝑒𝑦 =2𝑥 + 2 − 1

𝑥 + 1

⇒ 𝑒𝑦 =2𝑥 + 1

𝑥 + 1

⇒ 𝑦 = log |2𝑥 + 1

𝑥 + 1| , (𝑥 ≠ −1)


𝑦 = log |2𝑥+1

𝑥+1| , (𝑥 ≠ −1) [1 Mark]



15. The population of a village increases continuously at the rate proportional to the

number of its inhabitants present at any time. If the population of the village was 20000

in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

[4 Marks]

Solution:

Let the population at any instant (𝑡) be y.

It is given that the rate of increase of population is proportional to the number of

inhabitants at any instant.

∴𝑑𝑦

𝑑𝑡0

⇒𝑑𝑦

𝑑𝑡= 𝑘𝑦 (𝑘 is a constant)

⇒𝑑𝑦

𝑦= 𝑘𝑑𝑡


log 𝑦 = 𝑘𝑡 + 𝐶 … (1)

In the year 1999, 𝑡 = 0 and 𝑦 = 20000.

Therefore, we get:

log 20000 = 𝐶 … (2)

In the year 2004, 𝑡 = 5 and 𝑦 = 25000.

Therefore, we get:

log25000 = 𝑘 · 5 + C [2 Marks]

⇒ log25000 = 5𝑘 + log20000

⇒ 5𝑘 = log (25000

20000) = log (

5

4)

⇒ 𝑘 =1

5log (

5

4) … (3) [1 Mark]

In the year 2009, 𝑡 = 10 years.

Now, on substituting the values of 𝑡, 𝑘, and 𝐶 in equation (1), we get:

log𝑦 = 10 ×1

5log (

5

4) + log(20000)

⇒ log𝑦 = log [20000 × (5

4)

2

]



⇒ 𝑦 = 20000 ×5

4×

5

4

⇒ 𝑦 = 31250

Hence, the population of the village in 2009 will be 31250. [1

Mark]

16. The general solution of the differential equation𝑦𝑑𝑥−𝑥𝑑𝑦

𝑦= 0 is [2

Marks]

A. 𝑥𝑦 = 𝐶

B. 𝑥 = 𝐶𝑦2

C. 𝑦 = 𝐶𝑥

D. 𝑦 = 𝐶𝑥2

Solution:

The given differential equation is 𝑦𝑑𝑥−𝑥𝑑𝑦

𝑦= 0

⇒𝑦𝑑𝑥 − 𝑥𝑑𝑦

𝑥𝑦= 0

⇒1

𝑥𝑑𝑥 −

1

𝑦𝑑𝑦 = 0 [1 Mark]


log|𝑥| − log|𝑦| = log𝑘

⇒ log |𝑥

𝑦| = log𝑘

⇒𝑥

𝑦= 𝑘

⇒ 𝑦 =1

𝑘𝑥

⇒ 𝑦 = C𝑥 where C =1

𝑘

Hence, the correct answer is C. [1 Mark]

17. The general solution of a differential equation of the type 𝑑𝑥

𝑑𝑦+ P1𝑥 = Q1 is [1

Mark]



(A) 𝑦𝑒∫ P1𝑑𝑦 = ∫(Q1𝑒∫ 𝑃𝑑𝑦) 𝑑𝑦 + C

(B) 𝑦 ∙ 𝑒∫ P1𝑑𝑥 = ∫(Q1𝑒∫ 𝑃1𝑑𝑥) 𝑑𝑦 + C

(C) 𝑥𝑒∫ P1𝑑𝑦 = ∫(Q1𝑒∫ 𝑃1𝑑𝑦) 𝑑𝑦 + C

(D) 𝑥𝑒∫ P1𝑑𝑥 = ∫(Q1𝑒∫ 𝑃1𝑑𝑥) 𝑑𝑦 + C

Solution:

The integrating factor of the given differential equation 𝑑𝑥

𝑑𝑦+ P1𝑥 = Q1 is 𝑒∫ 𝑃1𝑑𝑦

The general solution of the differential equation is given by,

𝑥(I.F.) = ∫(Q × I.F.) 𝑑𝑦 + C [𝟏

𝟐 Mark]

⇒ 𝑥 ∙ 𝑒 ∫(𝑄𝑒∫ 𝑃1𝑑𝑦)𝑑𝑦 + 𝐶

Hence, the correct answer is C. [𝟏

𝟐 Mark]

18. The general solution of the differential equation 𝑒𝑥𝑑𝑦 + (𝑦𝑒𝑥 + 2𝑥)𝑑𝑥 = 0 is [2

Marks]

(A) 𝑥𝑒𝑦 + 𝑥2 = 𝐶

(B) 𝑥𝑒𝑦 + 𝑦2 = 𝐶

(C) 𝑦𝑒𝑥 + 𝑥2 = 𝐶

(D) 𝑦𝑒𝑦 + 𝑥2 = 𝐶

Solution:

The given differential equation is:

𝑒𝑥𝑑𝑦 + (𝑦𝑒𝑥 + 2𝑥)𝑑𝑥 = 0

⇒ 𝑒𝑥𝑑𝑦

𝑑𝑥+ 𝑦𝑒𝑥 + 2𝑥 = 0

⇒𝑑𝑦

𝑑𝑥+ 𝑦 = −2𝑥𝑒−𝑥 [

𝟏

𝟐 Mark]

This is a linear differential equation of the form

𝑑𝑦

𝑑𝑥+ 𝑃𝑦 = 𝑄, where 𝑃 = 1 and 𝑄 = −2𝑥𝑒−𝑥

Now, 𝐼. 𝐹 = 𝑒∫ 𝑃𝑑𝑥 = 𝑒∫ 𝑑𝑥 = 𝑒𝑥 [𝟏

𝟐 Mark]




𝑦(I.F.) = ∫(Q × I.F.) 𝑑𝑥 + C

⇒ 𝑦𝑒𝑥 = ∫(−2𝑥𝑒−𝑥 · 𝑒𝑥) 𝑑𝑥 + C

⇒ 𝑦𝑒𝑥 = − ∫ 2𝑥 𝑑𝑥 + C

⇒ 𝑦𝑒𝑥 = −𝑥2 + C

⇒ 𝑦𝑒𝑥 + 𝑥2 = C

Hence, the correct answer is C. [1 Mark]

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CBSE NCERT Solutions for Class 12 Maths Chapter 09 · 2019. 12. 6. · Class-XII-Maths Differential Equations 5 Practice more on Differential Equations Solution: Given differential