NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Maths
Chapter : 9
Chapter Name : Determinants

Q1 Determine order and degree(if de ned) of differential equation
4
d y
′′′
4
+ sin(y ) = 0
dx

4
d y ′′′
+ sin(y ) = 0
dx4
′′′′ m
⇒ y + sin(y ) = 0

The highest order derivative present in the differential equation is y . Therefore, its order is four. ′′′′

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

Page : 382 , Block Name : Exercise 9.1

Q2 Determine order and degree(if de ned) of differential equation
′
y + 5y = 0

Answer. y + 5y = 0 ′

The highest order derivative present in the differential
′
equation is y . Therefore, its order is one.

′
It is a polynomial equation in y . The highest power raised to

′
y is 1. Hence, its degree is one.

Page : 382 , Block Name : Exercise 9.1

Q3 Determine order and degree(if de ned) of differential equation ( (
2
ds 4 d s
) + 3s 2
= 0).
dt dt

Answer. ( (
2
ds 4 d s
) + 3s 2
= 0).
dt dt

2
d s
The highest order derivative present in the given differential equation is . Therefore,
2
dt

its order is two.
2 2
d s ds d s
It is a polynomial equation in and . The power raised to is 1
2 2
dt dt dt

Hence, its degree is one.

Page : 382 , Block Name : Exercise 9.1

Q4
Determine order and degree(if defined) of differential

2 2
d y dy
( ) + cos( ) = 0
dx2 dx

2 2
d y dy
( ) + cos( ) = 0
2
dx dx

The highest order derivative present in the given differential

2
d y
equation is . Therefore, its order is 2 .
2
dx

The given differential equation is not a polynomial equation

in its derivatives. Hence, its degree is not defined.

Page : 382 , Block Name : Exercise 9.1

Q5
Determine order and degree(if defined) of differential equation

2
d y
= cos 3x + sin 3x
2
dx

2
d y
= cos 3x + sin 3x
2
dx

2
d y
⇒ − cos 3x − sin 3x = 0
2
dx

The highest order derivative present in the differential.

2
d y
equation is . Therefore, its order is two.
2
dx

2
d y
2
It is a polynomial equation in dx and the power raised to
2
dx

Hence, its degree is one.

Page : 382 , Block Name : Exercise 9.1

Q6
Determine order and degree(if defined) of differential

′′ 2 ′′ 3 ′ 4 5
equation (y ) + (y ) + (y ) + y = 0

′′ 2 ′ 3 ′ 5
(y ) + (y ) + (y ) + y = 0

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

order is three.

′′ ′ ′
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 .

Page : 382 , Block Name : Exercise 9.1

Q7
Determine order and degree(if defined) of differential

′′′ ′′ ′
equation y + 2y + y = 0

′′ ′′ ′
y + 2y + y = 0

The highest order derivative present in the differential

′′′
equation is y . Therefore, its order is three.
′′′ ′′ ′
It is a polynomial equation in y ,y and y . The highest power

′′′
raised to y is 1. Hence, its degree is 1 .

Page : 382 , Block Name : Exercise 9.1

Q8
Determine order and degree(if defined) of differential

′ x
equation y + y = e

′ x
y + y = e

′ x
⇒ y + y − e = 0

The highest order derivative present in the differential

′
equation is y . Therefore, its order is one.

′
The given differential equation is a polynomial equation in y

′
and the highest power raised to y is one. Hence, its degree is

one.

Page : 383 , Block Name : Exercise 9.1

Q9
Determine order and degree(if defined) of differential

′′ ′ 2
equation y + (y ) + 2y = 0

′′ ′ 2
y + (y ) + 2y = 0

The highest order derivative present in the differential

′′
equation is y . Therefore, its order is two.

′′
The given differential equation is a polynomial equation in y

Hence, its degree is one.

Page : 383 , Block Name : Exercise 9.1

Q10
Determine order and degree(if defined) of differential

′′ ′
equation y + 2y + sin y = 0

′′ ′
y + 2y + sin y = 0

The highest order derivative present in the differential

′′
equation is y . Therefore, its order is two.

′′ ′
This is a polynomial equation in y and y and the highest

′′
power raised to y is one. Hence, its degree is one.

Page : 383 , Block Name : Exercise 9.1

Q11
The degree of the differential equation

2 3 2
d y dy dy
( ) + ( ) + sin( ) + 1 = 0
2
dx dx dx

( A) 3(B)2(C)1(D) not defined

2 3 2
d y dy dy
( ) + ( ) + sin( ) + 1 = 0
2
dx dx dx

The given differential equation is not a polynomial equation

in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

Page : 383 , Block Name : Exercise 9.1

Q12
The order of the differential equation
2
d y dy
2
2x − 3 + y = 0
2
dx dx

(A)2(B)1(C)0(D) not defined

2
d y dy
2
2x − 3 + y = 0
2
dx dx

The highest order derivative present in the given differential

2
d y
equation is . Therefore, its order is two.
2
dx

Hence, the correct answer is A.

Page : 383 , Block Name : Exercise 9.1

Q1 y = e x
+ 1 : y
′′
− y
′
= 0

Differentiating both sides of this equation with respect to x ,

we get:

dy d x
= (e + 1)
dx dx

′ x
⇒ y = e . . . (i)

Now, differentiating equation (i) with respect to x, we get:
d ′ d x
(y ) = (e )
dx dx

′ x
⇒ y = e

′ ′′
Substituting the values of y and y in the given differential equation, we get the L.H.S. as
′ ′ x x
y − y = e − e = 0 = R. H. S

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

Page : 385 , Block Name : Exercise 9.2

Q2 y = x 2
+ 2x + C : y
′
− 2x − 2 = 0

2
y = x + 2x + C

Differentiating both sides of this equation with respect to x ,

we get:

′ d 2
y = (x + 2x + C)
dx

′
⇒ y = 2x + 2
′
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.

Page : 385 , Block Name : Exercise 9.2

Q3 y = cos x + C : y
′
+ sin x = 0

y = cos x + C

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

′ d
y = (cos x + C)
dx

′
⇒ y = − sin x
′
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.

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

Page : 385 , Block Name : Exercise 9.2

Q4 y = √1 + x
xy
2 ′
: y =
1+x2

2
y = √1 + x

Differentiating both sides of the equation with respect to x ,

we get:

′ d 2
y = ( √1 + x )
dx

′ 1 d 2
y = ⋅ (1 + x )
2√1+x2 dx

′ 1
y =
2
2√1+x

′ x
y =
√1+x2

x
′ 2
⇒ y = × √1 + x
2
1 + x
x
′
⇒ y = ⋅ y
2
1 + x
xy
′
⇒ y =
2
1 + x

∴ L. H. S. = R. H. S.

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

Page : 385 , Block Name : Exercise 9.2

Q5 y = Ax : xy
′
= y(x ≠ 0)

y = Ax

Differentiating both sides with respect to x, we get:

′ d
y = (Ax)
dx

′
⇒ y = A

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

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

Page : 385 , Block Name : Exercise 9.2

Q6 y = x sin x : xy
′
= y + x√x
2 2
− y (x ≠ 0 and x > y or x < −y)

y = x sin x

Differentiating both sides of this equation with respect to x1

we get:

′ d
y = (x sin x)
dx

′ d d
⇒ y = sin x ⋅ (x) + x ⋅ (sin x)
dx dx

′
⇒ y = sin x + x cos x
′
Substituting the value of y in the given differential equation, we get:

L.H.S.

= x sin x + x cos x)

2 2
= y + x ⋅ √1 − sin x

y 2
2
= y + x √1 − ( )
x

2 2
= y + x√y − x

= R. H. S.

Page : 385 , Block Name : Exercise 9.2

2

Q7 xy = log y + C
y
′
: y = (xy ≠ 1)
1−xy

xy = log y + C

Differentiating both sides of this equation with respect to x ,

we get:

d d
(xy) = (log y)
dx dx

d 1 1 dy
⇒ y ⋅ (x) + x ⋅ =
dx dx y dx

′ 1 ′
⇒ y + xy = y
y

2 ′ ′
⇒ y + xyy = y

′ 2
⇒ (xy − 1)y = −y

2
y
′
⇒ y =
1−xy

∴ Hence, the given function is the solution of the

corresponding differential equation.

Page : 385 , Block Name : Exercise 9.2

Q8 y − cos y = x : (y sin y + cos y + x)y
′
= y

y − cos y = x . . . (i)

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

dy d d
− (cos y) = (x)
dx dx dx

′ ′
⇒ y + sin y ⋅ y = 1

′
⇒ y (1 + sin y) = 1

′ 1
⇒ y =
1+sin y

′
Substituting the value of y in equation (i), we get:
′
L.H.S. = (y sin y + cos y + x)y

1
= (y sin y + cos y + y − cos y) ×
1 + sin y

1
= y(1 + sin y) ⋅
1 + sin y

= y

= R. H. S.

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

Page : 385 , Block Name : Exercise 9.2

Q9 x + y = tan −1
y :
2
y y
′
+ y
2
+ 1 = 0

−1
x + y = tan y

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

d d −1
(x + y) = (tan y)
dx dx

′ 1 ′
⇒ 1 + y = [ 2
]y
1+y

1
′
⇒ y [ − 1] = 1
2
1 + y

2
1 − (1 + y )
′
⇒ y [ ] = 1
2
1 + y

2
−y
′
⇒ y [ ] = 1
2
1 + y

2
− (1 + y )
′
⇒ y =
2
y

′
substituting the value of y in the given differential equation, we get:
2
− (1 + y )
2 ′ 2 2 2
L.H.S. : y y + y + 1 = y [ ] + y + 1
2
y

2 2
= −1 − y + y + 1

= 0

= R. H. S.

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

Page : 385 , Block Name : Exercise 9.2

dy
Q10 y = √a 2 2
− x x ∈ (−a, a) : x + y
dx
= 0(y ≠ 0)

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

2 2
y = √a − x

dy d
2 2
= (√a − x )
dx dx

dy 1 d
2 2
⇒ = ⋅ (a − x )
dx 2 √a
2
− x
2 dx

1
= (−2x)
2 2
2 √a − x

−x
=
√a 2 − x2
dy
Substituting the value of dx
in the given differential equation, we get:
dy −x
2 2
L.H.S. = x + y = x + √a − x ×
dx √a 2 − x2

= x − x

= 0

= R. H. S

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

Page : 385 , Block Name : Exercise 9.2

Q11 The numbers of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4

Answer. We know that the number of constants in the general solution of a differential equation of order n is
equal to its order.
Therefore, the number of constants in the general equation of fourth order differential equation is four.
Hence, the correct answer is D.

Page : 385 , Block Name : Exercise 9.2

Q12 The numbers of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3 (B) 2 (C) 1 (D) 0

Answer.In a particular solution of a differential equation, there are no arbitrary constants.
Hence, the correct answer is D.

Page : 385 , Block Name : Exercise 9.2

Q1 x y
+ = 1
a b

x y
+ = 1
a b

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

dy
1 1
+ = 0
a b dx

1 1 ′
⇒ + y = 0
a b

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

1 ′′
0 + y = 0
b

1 ′′
⇒ y = 0
b

1 ′′
⇒ y = 0
b

′′
⇒ y = 0

′′
y = 0

Hence, the required differential equation of the given curve is

′′
y = 0

Page : 391 , Block Name : Exercise 9.3

Q2 y 2
= a (b
2
− x )
2

2 2 2
y = a (b − x )

Differentiating both sides with respect to x, we get:

dy
2y = a(−2x)
dx

′
⇒ 2yy = −2ax

′
⇒ yy = −ax . . . (i)

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

′ ′ ′′
y ⋅ y + yy = −a

′ 2 ′
⇒ (y ) + yy = −a . . . (ii)

Dividing equation (ii) by equation (i), we get:

′ 2 ∗
(y ) +yy −a
′
=
yy −ax

′′ ′ 2 ′
⇒ xyy + x(y ) − yy = 0

This is the required differential equation of the given curve.

Page : 391 , Block Name : Exercise 9.3

Q3 y = ae 3x
+ be
−2x

3x −2x
y = ae + be . . . (i)

Differentiating both sides with respect to x, we get:

′ 3x −2x
y = 3ae − 2be . . . (ii)

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

′ 3x −2x
y = 9ae + 4be . . . (iii)

Multiplying equation (i) with (ii) and then adding it to equation (ii), we get:
3x −2x 3x −2x ′
(2ae + 2be ) + (3ae − 2bc ) = 2y + y

3x ′
⇒ 5ae = 2y + y
′
2y+y
3x
⇒ ae =
5

Now, multipling equation (i) with (iii) and subtracting equation

(ii) from it, we get:
3x −2x 3x −2x ′
(3ae + 3be ) − (3ae − 2be ) = 3y − y

−2x ′
⇒ 5be = 3y − y
′
3y−y
−2x
⇒ be =
5

2x −2x
Substituting the values of ae and be in equation (iii), we get:
′ ′
(2y+y ) (3y−y )
′′
y = 9 ⋅ + 4
5 5

′ ′
18y+9y 12y−4y
′′
⇒ y = + 4
5 5

′
30y+5y
′′
⇒ y =
5

′
5y+y
′′
⇒ y =
5

′′ ′
⇒ y − y − 6y = 0

This is the required differential equation of the given curve.

Page : 391 , Block Name : Exercise 9.3

Q4 y = e 2x
(a + bx)

2x
y = e (a + bx) . . . (i)

Differentiating both sides with respect to x, we get:

′ 2x 2x
y = 2e (a + bx) + e ⋅ b

′ 2x
⇒ y = e (2a + 2bx + b) . . . (ii)

Multiplying equation (i) with equation (ii) and then subtracting it from equation (ii), we

get:

′ 2x 2x
y − 2y = e (2a + 2bx + b) − e (2a + 2bx)

′ 2x
⇒ y − 2 = be . . . (iii)

Differentiating both sides with respect to x, we get:

′ ′ 2x
y k − 2y = 2be . . . (iv)

Dividing equation (iv) by equation (iii), we get:
′ ′
y −2y

′
= 2
y −2y

′ ′ ′
⇒ y − 2y = 2y − 4y

′′ ′
⇒ y − 4y + 4y = 0

This is the required differential equation of the given curve.

Page : 391 , Block Name : Exercise 9.3

Q5 y = e (a cos x + b sin x) x

x
y = e (a cos x + b sin x) . . . (i)

Differentiating both sides with respect to x, we get:

′ x x
y = e (a cos x + b sin x) + e (−a sin x + b cos x)

′ x
⇒ y = e [(a + b) cos x − (a − b) sin x] . . . (ii)

Again, differentiating with respect to x, we get:
′′ r x
y = e [(a + b) cos x − (a − b) sin x] + e [−(a + b) sin x − (a − b) cos x]

′′ x
y = e [2b cos x − 2a sin x]

′′ x
y = 2e (b cos x − a sin x)
′
y
x
⇒ = e (b cos x − a sin x) . . . (iii)
2

Adding equations (i) and (iii), we get:
′
y
x
y + = e [(a + b) cos x − (a − b) sin x]
2

′′
y
′
⇒ y + = y
2

′′
⇒ 2y + y

′′ ′
⇒ y − 2y + 2y = 0

′′ ′
⇒ y − 2y + 2y = 0

This is the required differential equation of the given curve.

Page : 391 , Block Name : Exercise 9.3

Q6 Form the differential equation of the family of circles touching the y-axis at the origin.

Answer. 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
2 2 2
(x − a) + y = a

2 2
⇒ x + y = 2ax . . . (i)

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

′
2x + 2yy = 2a

′
⇒ x + yy = a

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

2 2 ′
x + y = 2 (x + yy ) x

2 2 2 ′
⇒ x + y = 2x + 2xyy

′ 2 2
⇒ 2xyy + x = y

This is the required differential equation.

Page : 391 , Block Name : Exercise 9.3

Q7 Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-
axis.

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

y -axis is:

2
x = 4ay . . . (i)

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

′
2x = 4ay . . . (ii)

Dividing equation (ii) by equation (i), we get:

′
2x 4ay
=
2 4ay
x

′
2 y
⇒ =
x y

′
⇒ xy = 2y

′
⇒ xy − 2y = 0

This is the required differential equation.

Page : 391 , Block Name : Exercise 9.3

Q8 Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

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

as follows:

2 2
x y
+ = 1 . . . (i)
2 2
b a

Differentiating equation (i) with respect to x, we get:
′
2x 2yy
+ = 0
2 2
b b

′
x yy
⇒ + = 0 . . . (ii)
2 2
b a

Again, differentiating with respect to x, we get:
′ ′ ∗
1 y y +y.y
+ = 0
2 2
b a

1 1 ′2 ′′
⇒ 2
+ 2
(y + yy ) = 0
b a

1 1 2 ′′
⇒ 2
= − 2
(y + yy )
b a

Substituting this value in equation (ii), we get:
′
1 2 yy
′ ′
x [− ((y ) + yy )] + = 0
2 2
a a

′ 2 ′′ ′
⇒ −x(y ) − xyy + yy = 0

′′ ′ 2 ′
⇒ xyy + x(y ) − yy = 0

This is the required differential equation.

Page : 391 , Block Name : Exercise 9.3

Q9 Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

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

axis is:

2 2
x y
− = 1 . . . (i)
2 2
a b

Differentiating both sides of equation (i) with respect to x, we get:
′
2x 2yy
− = 0
2 2
a b

′
x yy
⇒ − = 0 . . . (ii)
2 2
a b

Again, differentiating both sides with respect to x, we get:
′ ′ ′′
1 y ⋅y +yy
− = 0
2 2
a b

1 1 ′ 2 ′′
⇒ = ((y ) + yy )
a2 b2

2
Substituting the value of a in equation (ii), we get:
′
x 2 yy
′ ′
((y ) + yy ) − = 0
2 2
b b
′ 2 ′′ ′
⇒ x(y ) + xyy − yy = 0

′′ ′ 2 ′
⇒ xyy + x(y ) − yy = 0

This is the required differential equation.

Page : 391 , Block Name : Exercise 9.3

Q10 Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer. Let the centre of the circle on y-axis be (0, b).
The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:
2 2 2
x + (y − b) = 3

2 2
⇒ x + (y − b) = 9 . . . (i)

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

′
2x + 2(y − b) ⋅ y = 0

′
⇒ (y − b) ⋅ y = −x

−x
⇒ y − b = ′
y

Substituting the value of (y − b) in equation (i), we get:

2
2 −x
x + ( ′
) = 9
y

2 1
⇒ x [1 + ] = 9
′ 2
(y )

2 ′ 2 ′ 2
⇒ x ((y ) + 1) = 9(y )

2 ′ 2 2
⇒ (x − 9) (y ) + x = 0

This is the required differential equation.

Page : 391 , Block Name : Exercise 9.3

Q11
Which of the following differential equations has

x −x
y = c1 e + c2 e as the general solution?
2
d y
(A) + y = 0
2
dx

2
d y
(B) − y = 0
2
dx
2
d y
(iii) + 1 = 0
2
dx

2
d y
(iv) − 1 = 0
2
dx

The given equation is:

x −x
y = c1 e + c2 e . . . (i)

Differentiating with respect to x, we get:

dy
x −x
= c1 e − c2 e
dx

Again, differentiating with respect to x, we get:
2
d y
x −x
= c1 e + c2 e
2
dx

2
d y
⇒ = y
2
dx

2
d y
⇒ − y = 0
2
dx

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

Page : 391 , Block Name : Exercise 9.3

Q12
Which of the following differential equation has y = x as one

of its particular solution?
2
d y dy
2
(A) − x + xy = x
2
dx dx

2
d y dy
(B) + x + xy = x
2
dx dx

2
d y dy
2
(C) − x + xy = 0
2
dx dx

2
d y dy
(D) + x + xy = 0
2
dx dx

Answer.The given equation of curve is y = x.
Differentiating with respect to x, we get:
dy
= 1 (i)
dx

Again, differentiating with respect to x, we get:
2
d y
= 0 (ii)
2
dx
2
d y dy
Now, on substituting the values of y, dx
2
, and
dx

from equation (i) and (ii) in each of the given alternatives,
we nd that only the differential equation given in alternative C is correct.
2
d y dy
2 2
− x + xy = 0 − x ⋅ 1 + x ⋅ x
2
dx dx
2 2
= −x + x

= 0

Hence, the correct answer is C.

Page : 391 , Block Name : Exercise 9.3

dy
Q1
1−cos x
=
dx 1+cos x

The given differential equation is:

dy 1−cos x
=
dx 1+cos x

2 x
dy 2 sin
2 2 x
⇒ = x = tan
dx 2 cos2 2
2

dy x
2
⇒ = (sec − 1)
dx 2

Separating the variables, we get:

2 x
dy = (sec − 1) dx
2

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

2 x 2 x
∫ dy = ∫ (sec − 1) dx = ∫ sec dx − ∫ dx
2 2

x
⇒ y = 2 tan − x + C
2

This is the required general solution of the given differential equation.

Page : 395 , Block Name : Exercise 9.4

dy
Q2 dx
= √4 − y (−2 < y < 2)
2

The given differential equation is:

dy 1−cos x
=
dx 1+cos x

2 x
dy 2 sin
2 2 x
⇒ = x = tan
dx 2 2
2 cos
2

dy x
2
⇒ = (sec − 1)
dx 2

Separating the variables, we get:

2 x
dy = (sec − 1) dx
2

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

2 x 2 x
∫ dy = ∫ (sec − 1) dx = ∫ sec dx − ∫ dx
2 2

x
⇒ y = 2 tan − x + C
2

This is the required general solution of the given differential

equation.

Page : 395 , Block Name : Exercise 9.4

dy
Q3 dx
+ y = 1(y ≠ 1)

The given differential equation is:

dy
+ y = 1
dx

⇒ dy + ydx = dx

⇒ dy = (1 − y)dx

Separating the variables, we get:

dy
⇒ = dx
1−y

Now, integrating both sides, we get:

dy
∫ = ∫ dx
1 − y

⇒ log(1 − y) = x + log C

⇒ − log C − log(1 − y) = x

⇒ log C(1 − y) = −x

−x
⇒ C(1 − y) = e
1 −x
⇒ 1 − y = e
C

1 −x
⇒ y = 1 − e
C

−x 1
⇒ y = 1 + Ae ( where A = − )
C

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

Q4 sec x tan ydx + sec y tan xdy = 0
2 2

The given differential equation is:

2 2
sec x tan ydx + sec y tan xdy = 0

2 2
sec x tan ydx+sec y tan xdy
⇒ = 0
tan x tan y

2 2
sec x sec y
⇒ dx = − dy
tan x tan y

2 2
sec x sec y
⇒ dx = − dy
tan x tan y

Integrating both sides of this equation, we get:

2 2
sec x sec y
∫ dx = − ∫ dy . . . (i)
tan x tan y

Let tan x = t

d dt
∴ (tan x) =
dx dx

2 dt
⇒ sec x =
dx

2
⇒ sec xdx = dt
2
sec x 1
Now, ∫ dx = ∫ dt
tan x t

= log t

= log(tan x)
2
sec x
Similarly, ∫ dy = log(tan y)
tan x

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

log(tan x) = − log(tan y) + log C

C
⇒ log(tan x) = log( )
tan y

C
⇒ tan x =
tan y

⇒ tan x tan y = C

This is the required general solution of the given differential

equation.

Page : 396 , Block Name : Exercise 9.4

Q5 (e x
+ e
−x
) dy − (e
x
− e
−x
) dx = 0

The given differential equation is:
x −x x −x
(e + e ) dy − (e − e ) dx = 0

x −x x −x
⇒ (e + e ) dy = (e − e ) dx

x −x
e −e
⇒ dy = [ ] dx
ex +e−x

Integrating both sides of this equation, we get:
x −x
e −e
∫ dy = ∫ [ ] dx + C
ex +e−x

x −x
e −e
⇒ y = ∫ [ ] dx + C . . . (i)
e +e−x
x

x −x
Let (e + e ) = t

Differentiating both sides with respect to x, we get:

d x −x dt
(e + e ) =
dx dx

x −x dt
⇒ e − e =
dt

x −x
⇒ (e − e ) dx = dt

Substituting this value in equation (i), we get:
1
y = ∫ dt + C
t

⇒ y = log(t) + C

x −x
⇒ y = log(e + e ) + C

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

dy
Q6 dx
= (1 + x ) (1 + y )
2 2

The given differential equation is:

dy
2 2
= (1 + x ) (1 + y )
dx

dy
2
⇒ 2
= (1 + x ) dx
1+y

Integrating both sides of this equation, we get:
dy
2
∫ 2
= ∫ (1 + x ) dx
1+y

−1 2
⇒ tan y = ∫ dx + ∫ x dx

3
−1 x
⇒ tan y = x + + C
3

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

Q7 y log ydx − xdy = 0

The given differential equation is:

y log ydx − xdy = 0

⇒ y log ydx = xdy

dy dx
⇒ =
y log y x

Integrating both sides, we get:

dy dx
= ∫ . . . (i)
y log y x

Let log y = t

d dt
∴ (log y) =
dy dy

1 dt
⇒ =
y dy

1
⇒ dy = dt
y

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

dt dx
∫ = ∫
t x

⇒ log t = log x + log C

⇒ log(log y) = log Cx

⇒ log y = Cx

C
⇒ y = e

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

dy
Q8 x 5

dx
= −y
5

The given differential equation is:

dy
5 5
x = −y
dx

dy dx
⇒ 5
= − 5
y x

dx dy
⇒ 5
+ 5
= 0
x y

Integrating both sides, we get:

dx dy
∫ 5
+ ∫ 5
= k (where k is any constant)
x y

−5 −5
⇒ ∫ x dx + ∫ y dy = k

−4
−4 y
x
⇒ + = k
−4 −4

−4 −4
⇒ x + y = −4k
−4 −4
⇒ x + y = C (C = −4k)

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

dy
Q9 dx
= sin
−1
x

The given differential equation is:

dy
−1
= sin x
dx

−1
⇒ dy = sin xdx

Integrating both sides, we get:

−1
dy = ∫ sin x ⋅ 1)dx

−1
⇒ y = ∫ (sin x ⋅ 1) dx

−1 d −1
⇒ y = sin x ⋅ ∫ ( (sin x) ⋅ ∫ (1)dx)]dx
dx

−1 1
⇒ y = sin x ⋅ x − ∫ ( ⋅ x) dx
√1−x2

−1 1
⇒ y = x sin x + ∫ dx . . . (i)
√1−x2

2
Let 1 − x = t

d 2 dt
⇒ (1 − x ) =
dx dx

dt
⇒ −2x =
dx

1
⇒ xdx = − dt
2

Substituting this value in equation (i), we get:
−1 1
y = x sin x + ∫ dt
2√t

1
−1 1
⇒ y = x sin x + ⋅ ∫ (t) 2 dt
2

1
−1 1
⇒ y = x sin x + ⋅ ∫ (t) 2 + C
2

−1
⇒ y = x sin x + √t + C

−1 2
⇒ y = x sin x + √1 − x + C

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

Q10 e x
tan ydx + (1 − e ) sec
x 2
ydy = 0

The given differential equation is:

x x 2
e tan ydx + (1 − e ) sec ydy = 0

x 2 x
(1 − e ) sec ydy = −e tan ydx

Separating the variables, we get:
2 x
sec y −e
dy = dx
tan y 1−ex

Integrating both sides, we get:
2 x
sec y −e
∫ dy = ∫ dx . . . (i)
tan y 1−ex

Let tan y = u

d du
⇒ (tan y) =
dy dy

2 du
⇒ sec y =
dy

2
⇒ sec ydy = du

2
sec y du
∴ ∫ dy = ∫ = log u = log(tan y)
tan y u

x
Now, let 1 − e = t

d x dt
∴ (1 − e ) =
dx dx

x dt
⇒ −e =
dx

x
⇒ −e dx = dt
x
−e dt x
⇒ ∫ x
dx = ∫ = log t = log(1 − e )
1−e t

2 x
sec y −e
Substituting the values of ∫ dy and ∫ dx
tan y 1−ex

x
⇒ log(tan y) = log(1 − e ) + log C

x
⇒ log(tan y) = log[C (1 − e )]

x
⇒ tan y = C (1 − e )

This is the required general solution of the given differential equation.

Page : 396 , Block Name : Exercise 9.4

dy
Q11 (x 3
+ x
2
+ x + 1)
dx
= 2x
2
+ x; y = 1 when x = 0

The given differential equation is:

dy
3 2 2
(x + x + x + 1) = 2x + x
dx

2
dy 2x +x
⇒ = 3 2
dx (x +x +x+1)

2
2x +x
⇒ dy = 2
dx
(x+1)(x +1)

Integrating both sides, we get:
2
2x +x
∫ dy = ∫ 2
dx . . . (i)
(x+1)(x +1)

2
2x +x A Bx+C
Let 2
= + 2
. . . (ii)
(x+1)(x +1) x+1 x +1

2
2 Ax +A+(Bx+C)(x+1)
2x +x
⇒ 2
= 2
(x+1)(x +1) (x+1)(x +1)

2 2
⇒ 2x + x = Ax + A + Bx + Cx + C

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

2
Comparing the coefficients of x and x, we get:

A + B = 2

B + C = 1

A + C = 0

Solving these equations, we get:

1 3 −1
A = ,B = and C =
2 2 2

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

2
2x +x 1 1 1
2
= ⋅ + 2
(x+1)(x +1) 2 (x+1) 2(x +1)

Therefore, equation (i) becomes:

1 1 1 3x − 1
∫ dy = ∫ dx + ∫ dx
2
2 x + 1 2 x + 1

1 3 x 1 1
⇒ y = log(x + 1) + ∫ dx − ∫ dx
2 2
2 2 x + 1 2 x + 1

1 3 2x 1
−1
⇒ y = log(x + 1) + ⋅ ∫ dx − tan x + C
2
2 4 x + 1 2

1 3 1
2 −1
⇒ y = log(x + 1) + log(x + 1) − tan x + C
2 4 2
1 2 1 −1
⇒ y = [2 log(x + 1) + 3 log(x + 1)] − tan x + C
4 2

1 3 1
2 2 −1
⇒ y = [(x + 1) (x + 1) ] − tan x + C . . . (iii)
4 2

Now, y = 1 when x = 0 .

1 1 −1
⇒ 1 = log(1) − tan 0 + C
4 2

1 1
⇒ I = × 0 − × 0 + C
4 2

⇒ C = 1

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

1 3 1
2 2 −1
y = [log(x + 1) (x + 1) ] − tan x + 1
4 2

Page : 396 , Block Name : Exercise 9.4

dy
Q12 x (x 2
− 1)
dx
= 1; y = 0 when x = 2

dy
2
x (x − 1) = 1
dx

dx
⇒ dy =
x(x2 −1)

1
⇒ dy = dx
x(x−1)(x+1)

Integrating both sides, we get:
1
∫ dy = ∫ dx . . . (i)
x(x−1)(x+1)

1 A B C
Let = + + . . . . (ii)
x(x−1)(x+1) x x−1 x+1

1 A(x−1)(x+1)+Bx(x+1)
⇒ =
x(x−1)(x+1) x(x+1)

2
(A+B+C)x +(B−C)x−A
=
x(x−1)(x+1)

2
Comparing the coefficients of x , x, and constant, we get:

A = −1

B − C = 0

A + B + C = 0

1 1
Solving these equations, we get B = and C = .
2 2

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

1 −1 1 1
= + +
x(x−1)(x+1) x 2(x−1) 2(x+1)

Therefore, equation (i) becomes:
1 1 1 1 1
∫ dy = − ∫ dx + ∫ dx + ∫ dx
x 2 x−1 2 x+1

1 1
⇒ y = − log x + log(x − 1) + log(x + 1) + log k
2 2

2
1 k (x−1)(x+1)
⇒ y = log[ 2
] . . . (iii)
2 x

Now, y = 0 when x = 2
2
1 k (2−1)(2+1)
⇒ 0 = log[ ]
2 4

2
3k
⇒ ⇒ log( ) = 0
4

2
3k
⇒ ⇒ = 1
4

2
⇒ 3k = 4

2 4
⇒ k =
3

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

1 4(x−1)(x+1)
y = log[ 2
]
2 3x

2
4(x −1)
1
y = log[ 2
]
2 3x

Page : 396 , Block Name : Exercise 9.4

dy
Q13 cos( dx
) = a(a ∈ R); y = 1 when x = 0

dy
cos( ) = a
dx

dy
−1
⇒ = cos a
dx

−1
⇒ dy = cos a

Integrating both sides, we get:

−1
dy = cos a ∫ dx

−1
⇒ y = cos a ⋅ x + C

−1
⇒ y = x cos a + C . . . (i)

Now, y = 1 when x = 0.

−1
⇒ 1 = 0 ⋅ cos a + C

⇒ C = 1

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

−1
y = x cos a + 1

y−1
−1
⇒ = cos a
x

y−1
⇒ cos( ) = a
x

Page : 396 , Block Name : Exercise 9.4

dy
Q14 dx
= y tan x; y = 1 when x = 0

dy
= y tan x
dx

dy
⇒ = tan xdx
y

Integrating both sides, we get:
dy
∫ = − ∫ tan xdx
y

⇒ log y = log(sec x) + log C

⇒ log y = log(Csecx)

⇒ y = C sec x . . . (i)

Now , y = 1 when x = 0

⇒ 1 = C × sec 0

⇒ 1 = C × 1

⇒ C = 1

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

y = sec x

Page : 396 , Block Name : Exercise 9.4

Q15
Find the equation of a curve passing through the point (0, 0) and whose differential

′ x
equation is y = e sin x

The differential equation of the curve is:

′ x
y = e sin x

dy
x
⇒ = e sin x
dx

x
⇒ dy = e sin x

Integrating both sides, we get:
x
∫ dy = ∫ e sin xdx . . . (i)

x
Let I = ∫ e sin xdx

x d x
⇒ I = sin x ∫ e dx − ∫ (sin x) ⋅ ∫ e dx)dx
dx

x x
⇒ I = sin x ⋅ e − ∫ cos x ⋅ e dx

d
x x x
⇒ I = sin x ⋅ e − [cos x ⋅ ∫ e dx − ∫ (cos x) ⋅ ∫ e dx] dx]
dx

x x x
⇒ I = sin x ⋅ e − [cos x ⋅ e − ∫ (− sin x) ⋅ e dx]

x x
⇒ I = e sin x − e cos x − I
x
⇒ 2I = e (sin x − cos x)
x
e (sin x − cos x)
⇒ I =
2
Substituting this value in equation (i), we get:
x
e (sin x−cos x)
y = + C . . . (ii)
2

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

0
e (sin 0−cos 0)
∴ 0 = + C
2

1(0−1)
⇒ 0 = + C
2

1
⇒ C =
2
1
Substituting C = in equation (ii), we get:
2
x
e (sin x−cos x) 1
y = +
2 2

x
⇒ 2y = e (sin x − cos x) + 1

x
⇒ 2y − 1 = e (sin x − cos x)

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

Page : 396 , Block Name : Exercise 9.4

Q16
dy
For the differential equation x = (x + 2)(y + 2) , find the solution curve passing
dx

through the point (1, −1) .

The differential equation of the given curve is:

dy
xy = (x + 2)(y + 2)
dx

y x+2
⇒ ( ) dy = ( ) dx
y+2 x

2 2
⇒ (1 − ) dy = (1 + ) dx
y+2 x

Integrating both sides, we get:

2 2
∫ (1 − ) dy = ∫ (1 + ) dx
y+2 x

1 1
⇒ ∫ dy − 2 ∫ dy = ∫ dx + 2 ∫ dx
y+2 x

⇒ y − 2 log(y + 2) = x + 2 log x + C

2 2
⇒ y − x − C = log x + log(y + 2)

2 2
⇒ y − x − C = log[x (y + 2) ] . . . (i)

Now, the curve passes through point (1, −1) .
2 2
⇒ −1 − 1 − C = log[(1) (−1 + 2) ]

⇒ −2 − C = log 1 = 0

⇒ C = −2

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

2 2
y − x + 2 = log[x (y + 2) ]

This is the required solution of the given curve.

Page : 396 , Block Name : Exercise 9.4

Q17 Find the equation of a curve passing through the point (0, –2) given that at any point 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.

Answer. 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:

dy
y ⋅ = x
dx

⇒ ydy = xdx

Integrating both sides, we get:

∫ ydy = ∫ xdx

2 2
y x
⇒ = + C
2 2

2 2
⇒ y − x = 2C . . . (i)

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

2 2
∴ (−2) − 0 = 2C

⇒ 2C = 4

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

2 2
y − x = 4

This is the required equation of the curve.

Page : 396 , Block Name : Exercise 9.4

Q18 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).

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

y+3
The slope (m1 ) of the line segment joining (x, y) and (−4, −3) is
x+4

We know that the slope of the tangent to the curve is given by the relation,

dy

dx
dy
∴ Slope (m2 ) of the tangent =
dx

According to the given information:

m2 = 2m1

dy 2(y+3)
⇒ =
dx x+4

dy 2dx
⇒ =
y+3 x+4

Integrating both sides, we get:
dy dx
∫ = 2∫
y+3 x+4

⇒ log(y + 3) = 2 log(x + 4) + log C

2
⇒ log(y + 3) log C(x + 4)

2
⇒ y + 3 = C(x + 4) . . . (i)

This is the general equation of the curve.

It is given that it passes through point (-2, 1).
2
⇒ 1 + 3 = C(−2 + 4)

⇒ 4 = 4C

⇒ C = 1

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

2
y + 3 = (x + 4)

This is the required equation of the curve.

Page : 396 , Block Name : Exercise 9.4

Q19 The volume of spherical balloon being in ated 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.

Answer. Let the rate of change of the volume of the balloon be k (where k is a constant).
dv
⇒ = k
dt

d 4 3 4 3
⇒ ( πr ) = k [ Volume of sphere = πr ]
dt 3 3

4 2 dr
⇒ π ⋅ 3r ⋅ = k
3 dt

2
⇒ 4πr dr = kdt

Integrating both sides, we get:

2
4π ∫ r dr = k ∫ dt

3
r
⇒ 4π ⋅ = kt + C
3
3
⇒ 4πr = 3(kt + C) . . . (i)

Now, at t = 0, r = 3 :

⇒ 4π × 33 = 3 (k × 0 + C)
⇒ 108π = 3C
⇒ C = 36π
At t = 3, r = 6:
⇒ 4π × 63 = 3 (k × 3 + C)
⇒ 864π = 3 (3k + 36π)
⇒ 3k = –288π – 36π = 252π
⇒ k = 84π
Substituting the values of k and C in equation (i), we get:
3
4πr = 3[84πt + 36π]

3
⇒ 4πr = 4π(63t + 27)

3
⇒ r = 63t + 27
1

⇒ r = (63t + 27) 3

1

(63t+27) 3
Thus, the radius of the balloon after t seconds is

Page : 396 , Block Name : Exercise 9.4

Q20 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).

Answer. Let p, t, and r represent the principal, time, and rate of interest respectively.
It is given that the principal increases continuously at the rate of r% per year.
dp r
⇒ = ( )p
dt 100

dp r
⇒ = ( ) dt
p 100

Integrating both sides, we get:
dp r
∫ = ∫ dt
p 100

rt
⇒ log p = + k
100
n

+k
⇒ p = e 100 + k . . . (i)

It is given that when t = 0, p = 100 .

k
⇒ 100 = e … (ii)

Now, if t = 10, then p = 2 × 100 = 200.
Therefore, equation (i) becomes:

r
+k
200 = e 10
r
t
⇒ 200 = e 10 ⋅ e
′

⇒ 200 = e 10 ⋅ 100 . . . [f rom(ii)]
r

⇒ e 10 = 2
r
⇒ = log 2
10 e

r
⇒ = 0.6931
10

⇒ r = 6.931

Hence, the value of r is 6.93%.

Page : 397 , Block Name : Exercise 9.4

Q21
In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs

0.5
1000 is deposited with this bank, how much will it worth after 10 years (e = 1.648)

Answer. 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 5
⇒ = ( )p
dt 100

dp p
⇒ =
dt 20

dp dt
⇒ =
p 20

Integrating both sides, we get:

dp 1
= ∫ dt
p 20
t
⇒ log p = + C
20

t
+C
⇒ p = e 20 . . . (i)

Now, when t = 0, p = 1000.
⇒ 1000 = eC … (ii)
At t = 10, equation (i) becomes:
1
C
+
p = e2

0.5 c
⇒ p = e × e

⇒ p = 1.648 × 1000

⇒ p = 1648

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

Page : 397 , Block Name : Exercise 9.4

Q22 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?

Answer. 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
∴ ∝ y
dt

dy
⇒ = ky (where k is a constant)
dt

dy
⇒ = kdt
y

Integrating both sides, we get:

dy
∫ = k ∫ dt
y

⇒ log y = kt + C . . . (i)

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

⇒ log y0 = c

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

log y = kt + log y0

⇒ log y − log y0 = kt

y
⇒ log( ) = kt
y0

y
⇒ kt = log( )
y0

y
⇒ kt = log( ) . . . (ii)
y0

Also, it is given that the number of bacteria increases by 10% in 2 hours.
110
⇒ y = y0
100

y 11
⇒ = . . . (iii)
y0 10

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

11
k ⋅ 2 = log( )
10

1 11
⇒ k = log( )
2 10

1 11
⇒ k = log( )
2 10

Therefore, equation (ii) becomes:

1 11 y
log( ) ⋅ t = log( )
2 10 y0

y
2 log( )
y
0
⇒ t = . . . (iv)
11
log( )
10

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

⇒ y = 2y0 at t = t1

From equation (iv), we get:
y
2 log( )
y
0 2 log 2
t1 = =
11 11
log( ) log( )
10 10

2 log 2
hence, in
11
log( )
10
hours the number of bacteria increases from 100000 to 200000.

Page : 397 , Block Name : Exercise 9.4

Q23

dy
x+y
The general solution of the differential equation = e is
dx

x −y
A. e + e = C

x y
B. e + e = C

−x y
C. e + e = C

−x y
D. e + e = C

dy
x+y x y
= e = e ⋅ e
dx

dy
x
⇒ y = e dx
e

−y x
⇒ e dy = e dx

Integrating both sides, we get:

−y x
∫ e dy = ∫ e dx
−y x
⇒ −e = e + k

x −y
⇒ e + e = −k

x −y
⇒ e + e = c (c = −k)

Hence, the correct answer is A .

Page : 397 , Block Name : Exercise 9.4

Q1 (x 2
+ xy) dy = (x
2
+ y ) dx
2

2 2 2
The given differential equation i.e., (x + xy) dy = (x + y ) dx can be written as:

2 2
dy x +y
= . . . (i)
2
dx x +xy

2 2
x +y
Let F (x, y) =
2
x +xy

2 2 2 2
(λx) +(λy) x +y 0
Now, F (λx, λy) = 2
= 2
= λ ⋅ F (x, y)
(λx) +(λx)(λy) x +xy

This shows that equation (i) is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:
dy dv
= v + x
dx dx

dy
Substituting the values of v and in equation (i), we get:
dx
2 2
dv x +(vx)
v + x = 2
dx x +x(vx)

2
dv 1+v
⇒ v + x =
dx 1+v

2
2 (1+v )−v(1+v)
dv 1+v
⇒ x = − v =
dx 1+v 1+v

dv 1−v
⇒ x =
dx 1+v

1+v dx
⇒ ( ) = dv =
1−v x

2−1+v dx
⇒ ( ) dv =
1−v x

2 dx
⇒ ( − 1) dv =
1−v x

Integrating both sides, we get:

⇒ v = −2 log(1 − v) − log x + log k

k
⇒ v = log[ ]
x(1−v)2

y k
⇒ = log[ 2
]
x y
x(1− )
x

y kx
⇒ = log[ ]
x (x−y)2

kx x
⇒ = e
(x−y)2

y
2 −
⇒ (x − y) = kxe x

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q2 y =
x+y
′
x

x+y
′
y =
x

dy x+y
⇒ = . . . (i)
dx x

x+y
Let F (x, y) =
x

λx+λy x+y
0
Now, F (λx, λy) = = = λ F (x, y)
λx x

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy dv
= v + x
dx dx

dy
Substituting the values of y and in equation (i), we get:
dx
dv x+vx
v + x =
dx x

dv
⇒ v + x = 1 + v
dx

dv
x = 1
dx

dx
⇒ dv =
x

Integrating both sides, we get:

v = logx + C
y
⇒ = log x + C
x

⇒ y = x log x + Cx

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q3 (x − y)dy − (x + y)dx = 0

The given differential equation is:

(x − y)dy − (x + y)dx = 0

dy x+y
⇒ = . . . (i)
dx x−y

x+y
Let F (x, y) =
x−y

λx+λy x+y
0
∴ F (λx, λy) = = = λ ⋅ F (x, y)
λx−λy x−y

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx
d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

dy
Substituting the values of y and in equation (i), we get:
dx

dv x+vx 1+v
v + x = =
dx x−vx 1−v

dv 1+v 1+v−v(1−v)
x = − v =
dx 1−v 1−v

2
dv 1+v
⇒ x =
dx 1−v

1−v dx
⇒ 2
dv =
(1+v ) x

1 v dx
⇒ ( − ) dv =
1+v2 1−v2 x

Integrating both sides, we get:

−1 1 2
tan v − log(1 + v ) = log x + C
2

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

2 2
y 1 x +y
−1
⇒ tan ( ) − log( ) = log x + C
x 2
2 x

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

⇒ this is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q4 (x 2 2
− y ) dx + 2xydy = 0

The given differential equation is:

2 2
(x − y ) dx + 2xydy = 0

2 2
dy −(x −y )

⇒ = . . . (i)
dx 2xy

2 2
−(x −y )

Let F (x, y) =
2xy

2 2 2 2
(λx) −(λy) −(x −y )
0
∴ F (λx, λy) = [ ] = = λ ⋅ F (x, y)
2(λx)(λy) 2xy

To solve it, we given differential equation is a homogeneous equation.

y = vx
d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

dy
Substituting the values of y and equation (i), we get:
dx

2 2
dv x −(vx)
v + x = − ]
dx 2x⋅(vx)

2
dv v −1
v + x =
dx 2v

2 2 2
dv v − 1 v − 1 − 2v
⇒ x = − v =
dx 2v 2v
2
dv (1 + v )
⇒ x = −
dx 2v

2v dx
⇒ dv = −
2
1 + v x

Integrating both sides, we get:

C
2
log(1 + v ) = − log x + log C = log
x
2 C
⇒ 1 + v =
x

2
y C
⇒ [1 + ] =
2 x
x

2 2
⇒ x + y = Cx

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

dy
Q5 x 2

dx
− x
2
− 2y
2
+ xy

The given differential equation is:

dy
2 2 2
x = x − 2y + xy
dx
2 2
dy x −2y +xy
= . . . (i)
2
dx x

2 2
x −2y +xy
Let F (x, y) =
2
x

2 2 2 2
(λx) −2(λy) +(λx)(λy) x −2y +xy 0
∴ F (λx, λy) = 2
= 2
= λ ⋅ F (x, y)
(λx) x

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

dy dv
⇒ = v + x
dx dx

dy
Substituting the values of y and in equation (i), we get:
dx
2 2
dv x −2(vx) +x⋅(vx)
v + x =
dx x2

dv 2
⇒ v + x = 1 − 2v + v
dx

dv 2
⇒ x = 1 − 2v
dx

dv dx
⇒ =
1−2v2 x

1 dv dx
⇒ ⋅ =
2 1 2 x
−v
2

⎡ ⎤
1 dv dx
⇒ ⋅ =
2 2 x
⎣ ( 1 ) −v2 ⎦
√2

Integrating both sides, we get :
1
∣ √ +v ∣
1 1 2
⋅ log∣ ∣ = log |x| + C
1 1
2 2×
√2
∣ √2 −v ∣
1 y

∣ √ + ∣
x
1 2
⇒ log∣ y ∣ = log |x| + C
1
2√2 ∣ √2 − x ∣

1 ∣ x+√2y ∣
⇒ log = log |x| + C
2√2 ∣ x−√2y ∣

This is the required solution for the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q6 xdy − ydx = √x 2
+ y dx
2

2 2
xdy − ydx = √x + y dx

2 2
⇒ xdy = [y + √x + y ] dx

2 2
dy y+√x +y
= 2 2
. . . (i)
dx x +y

2 2
y+√x +y
Let F (x, y) = 2
x
2 2 2 2
λx+√(λx) +(λy) y+√x +y
0
∴ F (λx, λy) = = = λ ⋅ F (x, y)
λx x

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: y=vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

Substituting the values of v and dx in equation (i), we get:

2 2
dv vx+√x +(vx)
v + x =
dx x

dv 2
⇒ v + x = v + √1 + v
dx

dv dx
⇒ =
x
√1+v2

Integrating both sides, we get:

log∣
∣v +
√1 + v2 ∣ = log |x| + log C
∣

2
y y
⇒ log | + √1 + | = log |Cx|
x x2

2 2
∣ y+√x +y ∣
⇒ log∣ ∣ = log |Cx|
x
∣ ∣

2 2 2
⇒ y + √x + y = Cx

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q7 {x cos(
y y y y
) + y sin( )} ydx = {y sin( ) − x cos( )} xdy
x x x x

Answer. The given differential equation is:
y y y y
{x cos( ) + y sin( )} ydx = {y sin( ) − x cos( )} xdy
x x x x

y y
{x cos( )+y sin( )}y
dy x x

= . . . (i)
dx y y
{y sin( )−x cos( )}x
x x

y y
{x cos( )+y sin( )}
x x

Let F (x, y) = .
y y
{y sin( )−x cos( )}x
x x

λy λy
{λx cos( ) + λy sin( )} λy
λx λx

∴ F (λx, λy) =
λy λy
{λy sin( ) − λx sin( )}
λx λx
λx
y y
{x cos( ) + y sin( )} y
x x
=
y y
{y sin( ) − x cos( )} x
x x

0
= λ ⋅ F (x, y)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

dy dv
⇒ = v + x =
dx dx

dy
Substituting the values of y and in equation (i), we get:
dx

dv (x cos v+vx sin v)⋅vx
v + x =
dx (vx sin v−x cos v)⋅x

2
dv v cos v+v sin v
⇒ v + x =
dx v sin v−cos v

2
dv v cos v+v sin v
⇒ x = − v
dx v sin v−cos v

2 2
dv v cos v+v sin v−v sin v+v cos v
⇒ x =
dx v sin v−cos v

dv 2v cos v
⇒ x =
dx v sin v − cos v

v sin v − cos v 2dx
⇒ [ ] dv =
v cos v x

1 2dx
⇒ (tan v − ) dv =
v x

Integrating both sides, we get:

log(sec v) − log v = 2 log x + log C

sec v
2
⇒ ⇒ log( ) = log(Cx )
v
sec v
2
⇒ ⇒ ( ) = Cx
v
2
⇒ sec v = Cx v
y y
2
⇒ sec( ) = C ⋅ x ⋅
x x
y
⇒ sec( ) = Cxy
x

y 1 1 1
⇒ cos( ) = = ⋅
x Cxy C xy

y 1
⇒ xy cos( ) = k (k = )
x c

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

dy y
Q8 x dx
− y + x sin(
x
) = 0

dy y
x − y + x sin( ) = 0
dx x

dy y
⇒ x = y − x sin( )
dx x

y
y−x sin( )
dy x

⇒ =
dx x

y
y−x sin( )
dy x

⇒ = . . . (i)
dx x

y
y−x sin( )
x

Let F (x, y) =
x
λy y
λy−λx sin( ) y−x sin( )
λx x
0
∴ F (λx, λy) = = = λ ⋅ F (x, y)
λx x

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: y=vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx
dy
Substituting the values of y and in equation (i), we get:
dx

dv wx−x sin v
v + x =
dx x

dv
⇒ v + x = v − sin v
dx

dv dx
⇒ − =
sin v x

dx
⇒ csc vdv = −
x

Integrating both sides, we get:

C
log | csc v − cot v| = − log x + log C = log
x

y y C
⇒ csc( ) − cot( ) =
x x x

y
cos( )
x
1 C
⇒ − =
y y x
sin( ) sin( )
x x

y y
⇒ x [1 − cos( )] = Csin( )
x x

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q9 ydx + x log(
y
)dy − 2xdy = 0
x

y
ydx + x log( )dy − 2xdy = 0
x

y
⇒ ydx = [2x − x log( )] dy
x

dy y
⇒ = . . . (i)
y
dx
2x−x log( )
x

y
Let F (x, y) =
y
2x−x log( )
x

λy y
∘
∴ F (λx, λy) = = = λ ⋅ F (x, y)
λy y
2(λx)−(λx) log( ) 2x−log( )
λx x

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx
dy d
⇒ = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

dy
Substituting the values of y and in equation (i), we get:
dx

dv vx
v + x =
dx 2x−x log v

dv v
⇒ v + x =
dx 2 − log v

dv v
⇒ x = − v
dx 2 − log v

dv v − 2v + v log v
⇒ x =
dx 2 − log v

dv v log v − v
⇒ x =
dx 2 − log v

2 − log v dx
⇒ dv =
v(log v − 1) x

1 + (1 − log v) dx
⇒ [ ] dv =
v(log v − 1) x

1 1 dx
⇒ [ − ] dv =
v(log v − 1) v x

Integrating both sides, we get:

1 1 1
∫ dv − ∫ dv = ∫ dx
v(log v − 1) v x

dv
⇒ ∫ − log v = log x + log C . . . (ii)
v(log v − 1)

⇒ Let log v − 1 = t

d dt
⇒ (log v − 1) =
dv dv

1 dt
⇒ =
v dv

dv
⇒ = dt
v

Therefore, equation ( i ) becomes:
dt
⇒ ∫ − log v = log x + log C
t

y
⇒ log t − log( ) = log(Cx)
x

y y
⇒ log[log( ) − 1] − log( ) = log(Cx)
x x

y
log( )−1
x

⇒ log[ y
] = log(Cx)
x

x y
⇒ [log( ) − 1] = Cx
y x

y
⇒ log( ) − 1 = Cy
x

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

x x

Q10 (1 + e ) dx + e y J (1 −
x

y
) dy = 0

x x
x
(1 + e y ) dx + e y (1 − ) dy = 0
y

y y x
⇒ (1 + e ) dx = −e (1 − ) dy
y

r
x
−e r (1− )
y
dx
⇒ = x
. . . (i)
dy y
1+e
x
x
−e r (1− y )

F (x, y) = z
y
1+e
2x x
2y λx y x
−e (1− ) −e (1− )
λy y
0
∴ F (λx, λy) = = r
= λ ⋅ F (x, y)
dx
y
dy 1+e
1+e

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

x = vy

d d
⇒ (x) = (yy)
dy dy

dx dv
⇒ = v + y
dy dy

Substituting the values of x and dy in equation (i), we get:
y
dv −e (1 − v)
v + y =
x
dy 1 + e
r y
dv −e + ve
⇒ y = − v
y
dy 1 + e
v v v
dv −e +ve −v−ve
⇒ y = γ
dy 1+e

r
dv v+e
⇒ y = −[ r
]
dy 1+e

v dy
1+e
⇒ [ y ] dv = −
v+e y

Integrating both sides, we get:

C
v
⇒ log(v + e ) = − log y + log C = log( )
y

x x
C
⇒ [ + ey ] =
y y
x

y
⇒ x + ye = C

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q11 (x + y)dy + (x − y)dy = 0; y = 1 when x = 1

(x + y)dy + (x − y)dx = 0

⇒ (x + y)dy = −(x − y)dx

dy −(x−y)
⇒ = . . . (i)
dx x+y

−(x−y)
Let F (x, y) =
x+y

−(λx−λy) −(x−y)
0
∴ F (λx, λy) = = = λ ⋅ F (x, y)
λx−λy x+y

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

Substituting the values of y and dx in equation (i), we get:

dv −(x−vx)
v + x =
dx x+vx

dv v−1
⇒ v + x =
dx v+1

dv v−1 v−1−v(v+1)
⇒ x = − v =
dx v+1 v+1

2
2 −(1+v )
dv v−1−v −v
⇒ x = =
dx v+1 v+1

(v+1) dx
⇒ 2
dv = −
1+v x

Integrating both sides, we get:

1
2 −1
log(1 + v ) + tan v = − log x + k
2
2 −1
⇒ ⇒ log(1 + v ) + 2 tan v = −2 log x + 2k

2 2 −1
⇒ log[(1 + v ) ⋅ x ] + 2 tan v = 2k

2
y y
2 −1
⇒ log[(1 + ) ⋅ x ] + 2 tan = 2k
2
x x

y
2 2 −1
⇒ log(x + y ) + 2 tan = 2k . . . (ii)
x
Now, y = 1 at x = 1

−1
⇒ log 2 + 2 tan 1 = 2k
π
⇒ log 2 + 2 × = 2k
4

π
⇒ + log 2 = 2k
2

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

y π
2 2 −1
log(x + y ) + 2 tan ( ) = + log 2
x 2

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q12 x dy + (xy + y ) dx = 0; y = 1 when x = 1
2 2

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

2 2
⇒ x dy = − (xy + y ) dx

2
dy −(xy+y )

⇒ = 2
. . . (i)
dx x

2
−(xy+y )

Let F (x, y) = 2
x

2 2
[λx⋅λy+(λy) ] −(xy+y )
0
∴ F (λx, λy) = 2
= 2
= λ ⋅ F (x, y)
(λx) x

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

Substituting the values of y and dx in equation (i), we get:

2
−[x⋅vx+(vx) ]
dv 2
v + x = 2
= −v − v
dx x

dv 2
⇒ x = −v − 2v = −v(v + 2)
dx

dv dx
⇒ = −
v(v+2) x

1 (v+2)−v dx
⇒ [ ] dv = −
2 v(v+2) x

1 1 1 dx
⇒ [ − ] dv = −
2 v v+2 x

Integrating both sides, we get:

1
[log v − log(v + 2)] = − log x + log C
2

1 v C
⇒ log( ) = log
2 v+2 x

2
v C
⇒ = ( )
v+2 x

y
2

x C
⇒ = ( )
y
+ 2 x
x

2
y C
⇒ =
2
y + 2x x
2
x y 2
⇒ = C . . . (ii)
y + 2x

Now, y = 1 at x = 1
1 2
⇒ = C
1+2

2 1
⇒ C =
3
2 1
C = in equation (ii), we get:
3

2
x y 1
=
Substituting y+2x 3

2
⇒ y + 2x = 3x y

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q13 [x sin ( 2 x

y
− y)] dx + xdy = 0; y
π

4
when x = 1

2 y
[x sin ( ) − y] dx + xdy = 0
x

y
2
−[x sin ( )−y]
dy x

⇒ =
dx x

y
2
−[x sin ( )−y]
x

Let F (x, y) =
x

λx y
2 2
−[λx⋅sin ( )−λy] −[x sin ( )−y]
λy x
0
∴ F (λx, λy) = = = λ ⋅ F (x, y)
λx x

To solve this diven differential equation is a homogeneous equation.

y = vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x =
dx dx

dy
Substituting the values of y and in equation (i), we get:
dx

2
dv −[x sin v−vx]
v + x =
dx x

dv 2 2
⇒ v + x = − [sin v − v] = v − sin v
dx

dv 2
⇒ x = − sin v
dx

dv dx
⇒ = −
2
sin v dx

2 dx
⇒ csc vdv = −
x

Integrating both sides, we get:
− cot v = − log |x| − C

⇒ cot v = log |x| + C

y
⇒ cot( ) = log |x| + log C
x

y
⇒ cot( ) = log |Cx| . . . (ii)
x

π
now, y = at x = 1
4

π
⇒ cot( ) = log |C|
4

⇒ 1 = log C

1
⇒ C = e = e

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

y
cot( ) = log |ex|
x

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

dy y y
Q14 dx
−
x
+ csc(
x
) = 0; y = 0 when x = 1

dy y y
− + csc( ) = 0
dx x x

dy y y
⇒ = − csc( ) . . . (i)
dx x x

y y
Let F (x, y) = − csc( )
x x

y y
∴ F (λx, λy) = − csc( )
λx x

y y
0
⇒ F (λx, λy) = − csc( ) = F (x, y) = λ ⋅ F (x, y)
x x

Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:

y = vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

¯
¯¯¯
¯¯
Substituting the values of y and dx in equation (i), we get:

dv
v + x = v − csc v
dx

dv dx
⇒ − = −
csc v x

dx
⇒ − sin vdv =
x

Integrating both sides, we get:

cos v = log x + log C = log |Cx|

y
⇒ cos( ) = log |Cx| . . . (ii)
x

This is the required solution of the given differential equation.

Now, y = 0 at x = 1

⇒ cos(0) = log C

⇒ 1 = log C

1
⇒ C = e = e

Substituting C = e in equation (ii), we get:
y
cos( ) = log(ex)
x

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

dy
Q15 2xy + y 2
− 2x
2

dx
= 0; y = 2 when x = 1

dy
2 2
2xy + y − 2x = 0
dx

dy
2 2
⇒ 2x = 2xy + y
dx

2
dy 2xy+y
⇒ = . . . (i)
2
dx 2x

2
2xy+y
Let F (x, y) =
2
2x

2 2
2(λx)(λy)+(λy) 2xy+y 0
∴ F (λx, λy) = 2
= 2
= λ ⋅ F (x, y)
2(λx) 2x

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx

dy
Substituting the value of y and in equation (i), we get:
dx
2
dv 2x(vx)+(vx)
v + x = 2
dx 2x

2
dv 2v+v
⇒ v + x =
dx 2

2
dv v
⇒ v + x = v +
dx 2

2 dx
⇒ dv =
v2 x

Integrating both sides, we get:

−2+1
v
2 ⋅ = log |x| + C
−2 + 1

2
⇒ − = log |x| + C
v
2
⇒ − = log |x| + C
y

2x
⇒ − = log |x| + C . . . (ii)
y

Now, y = 2 at x = 1

⇒ −1 = log(1) + C

⇒ C = −1

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

2x
− = log |x| − 1
y

2x
⇒ = 1 − log |x|
y

2x
⇒ y = , (x ≠ 0, x ≠ e)
1−log |x|

This is the required solution of the given differential equation.

Page : 406 , Block Name : Exercise 9.5

Q16
dx x
A homogeneous differential equation of the form = h( ) can be solved by making the
dy y

substitution

A. y = vx

B. v = yx

C. x = vy

D. x = v

Answer. 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.

Page : 406 , Block Name : Exercise 9.5

Q17
Which of the following is a homogeneous differential equation?

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

3 3
B. (xy)dx − (x + y ) dy = 0

3 2
C. (x + 2y ) dx + 2xydy = 0

2 2 2 2
D. y dx + (x − xy − y ) dy = 0

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

n
F(λx, λy) = λ F(x, y) for any non-zero constant (λ) .

Consider the equation given in alternativeD:

2 2 2
y dx + (x − xy − y ) dy = 0

2 2
dy −y y
⇒ = =
dx x2 −xy−y 2 y 2 +xy−x2

2
y
Let F (x, y) =
y 2 +xy−x2

2
(λy)
⇒ F (λx, λy) = 2 2
(λy) +(λx)(λy)−(λx)

2 2
λ y
=
2 2 2
λ (y +xy−x )

2
y
0
= λ ( )
y +xy−x2
2

2
y
0
= λ ( )
y +xy−x2
2

0
= λ ⋅ F (x, y)

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

Page : 407 , Block Name : Exercise 9.5

dy
Q1 dx
+ 2y = sin x

dy
Answer. The given differential equation is dx
+ 2y = sin x
dy
This is in the form of dx
+ py = Q( where p = 2 and Q = sin x)

Now, I.F = e = e
∫ p⋅dt
= e
∫ 2dx 2x

The solution of the given differential equation is given by the relation,
y(IF) = ∫ (Q × IF)dx + C

2x 2x
⇒ ye = ∫ sin x ⋅ e dx + C . . . (i)
2x
Let I = ∫ sin x ⋅ e

2x d 2x
⇒ I = sin x ⋅ ∫ e dx − ∫ ( (sin x) ⋅ ∫ e dx) dx
dx

2x 2x
e e
⇒ I = sin x ⋅ − ∫ (cosx ⋅ )dx
2 2
2x
e sin x 1 2x d 2x
⇒ I = − [cos x ⋅ ∫ e − ∫ (cos x) ⋅ ∫ e dx) dx]
2 2 dx

2x 2x 2x
e sin x 1 e e
⇒ I = − [cos x ⋅ − ∫ (− sin x) ⋅ ] dx]
2 2 2 2

2x 2x
e sin x e cos x 1 2x
⇒ I = − − ∫ (sin xe ) dx
2 4 4

2x
e 1
⇒ I = (2 sin x − cos x) − I
4 4

2x
5 e
⇒ I = (2 sin x − cos x)
4 4

2x
e
⇒ I = (2 sin x − cos x)
5

Therefore, equation (i) becomes:
2x
2x e
ye = (2 sin x − cos x) + C
5

1 −2x
⇒ y = (2 sin x − cos x) + Ce
5

This is the required general solution of the given differential equation.

Page : 413 , Block Name : Exercise 9.6

dy
Q2 dx
+ 3y = e
−2x

dy
−2x
The given differential equation is + py = Q ( where p = 3 and Q = e )
dx

Now, I.F = e ∫ pΔx
= e
∫ 3at
= e
3x
.
The solution of the given differential equation is given by the relation,

y( I.F. ) = ∫ (Q × I.F. )dx + C

3x −2x 3x
⇒ ye = ∫ (e × e ) + C

3x x
⇒ ye = ∫ e dx + C

3x x
⇒ ye = e + C

−2x −3x
⇒ y = e + Ce

This is the required general solution of the given differential equation

Page : 413 , Block Name : Exercise 9.6

dy
Q3
y
2
+ = x
dx x

The given differential equation is:

dy
1 2
+ py = Q ( where p = and Q = x )
dx x

Now, I.F = e
1
∫ pdx ∫ dx log x
= e x = e = x

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

y( I.F. ) = ∫ (Q × I.F. )dx + C

2
⇒ y(x) = ∫ (x ⋅ x) dx + C

3
⇒ xy = ∫ x dx + C

4
x
⇒ xy = + C
4

This is the required general solution of the given differential equation.

Page : 413 , Block Name : Exercise 9.6

dy
Q4
π
+ sec xy = tan x (0 ≤ x < )
dx 2

The given differential equation is:

dy
+ py = Q (where p = sec x and Q = tan x)
dx

Now I.F = e ∫ pΔx
= e
∫ sec xdx
= e
log(sec x+tan x)
= sec x + tan x

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

y( I.F. ) = ∫ (Q × 1. F . )dx + C

⇒ y(sec x + tan x) = ∫ tan x(sec x + tan x)dx + C

2
⇒ y(sec x + tan x) = ∫ sec x tan xdx + ∫ tan xdx + C

2
⇒ y(sec x + tan x) = sec x + ∫ (sec x − 1) dx + C

⇒ y(sec x + tan x) = sec x + tan x − x + C

Page : 413 , Block Name : Exercise 9.6
π

Q5 ∫ 0
2
cos 2xdx

π

2
Let I = ∫ cos 2xdx
j

sin 2x
∫ cos 2xdx = ( ) = F(x)
2

By second fundamental theorem of calculus, we obtain
π
I = F( ) − F(0)
2
1 π
= [sin 2 ( ) − sin 0]
2 2
1
= [sin π − sin 0]
2
1
= [0 − 0] = 0
2

Page : 413 , Block Name : Exercise 9.6

dy
Q6 x + 2y = x
2
log x
dx

The given differential equation is:

dy
2
x + 2y = x log x
dx

dy 2
⇒ + y = x log x
dx x

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

dy
2
+ py = Q ( where p = and Q = x log x)
dx x

2

Now I.F = e
2
∫ pdx ∫ dx 2 log x log x 2
= e = e = e = x x

The general solution of the given differential equation is given by the relation,
y( I.F. ) = ∫ (Q × I. F.)dx + C

2 2
⇒ y ⋅ x = ∫ (x log x ⋅ x ) dx + C

2 3
⇒ x y = ∫ (x log x) dx + C

2 3 d 3
⇒ x y = log x ⋅ ∫ x dx − ∫ [ (log x) ⋅ ∫ x dx] dx + C
dx

4 4
2 x 1 x
⇒ x y = log x ⋅ − ∫ ⋅ )dx + C
4 x 4

4
2 x log x 1 3
⇒ x y = − ∫ x dx + C
4 4
4 4
2 x log x 1 x
⇒ x y = − ⋅ + C
4 4 4

2 1 4
⇒ x y = x (4 log x − 1) + C
16

1 2 −2
⇒ y = x (4 log x − 1) + Cx
16

Page : 413 , Block Name : Exercise 9.6

dy
Q7 x log x
2
+ y = log x
dx x

The given differential equation is:

dy 2
x log x + y = log x
dx x

dy y 2
⇒ + = 2
dx x log x x

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

dy 1 2
+ py = Q ( where p = and Q = )
dx x log x x2

1

Now, I.F = e ∫ ρdx ∫ dx log(log x)
= e = e = log x x log

The general solution of the given differential equation is given by the relation,
y(I. F.) = ∫ (Q × I. F)dx + C

2
⇒ y log x = ∫ ( log x) dx + C . . . (i)
x2

2 1
Now. ∫ ( log x) dx = 2 ∫ (log x ⋅ ) dx
x2 x2

1 d 1
= 2 [log x ⋅ ∫ dx − ∫ { (log x) ⋅ ∫ dx} dx]
2 2
x dx x

1 1 1
= 2 [log x (− ) − ∫ ( ⋅ (− )) dx]
x x x

log x 1
= 2 [−2 [− + ∫ dx]
2
x x

log x 1
= 2 [− − ]
x x

log x 1
= 2 [− − ]
x x

2
= − (1 + log x)
x
2
∫ Substituting the value of ∫ ( log x) dx in Equation (i) we get :
x2

2
y log x = − (1 + log x) + C
x

This is the required general solution of the given differential equation.

Page : 413 , Block Name : Exercise 9.6

Q8 (1 + x ) dy + 2xydx = cot xdx(x ≠ 0)
2

2
(1 + x ) dy + 2xydx = cot xdx

dy 2xy cot x
⇒ + 2
= 2
dx 1+x 1+x

This equation is a linear differential equation of the form:

dy 2x cot x
+ py = Q (where p = 2
and Q = 2
)
dx 1+x 1+x
2x
∫ dx 2
∫ pdx 2 log(1+x ) 2
= e = e 1+x
= e = 1 + x

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

y( L.F. ) = ∫ (Q × 1. F. )dx + C

2 cot x 2
⇒ y (1 + x ) = ∫ × (1 + x )]dx + C
1+x2

2
⇒ y (1 + x ) = ∫ cot xdx + C

2
⇒ y (1 + x ) = log | sin x| + C

Page : 413 , Block Name : Exercise 9.6

dy
Q9 x dx
+ y − x + xy cot x = 0(x ≠ 0)

dy
x + y − x + xy cot x = 0
dx

dy
⇒ x + y(1 + x cot x) = x
dx

dy 1
⇒ + ( + cot x) y = 1
dx x

This equation is a linear differential equation of the form:

dy 1
+ py = Q ( where p = + cot x and Q = 1)
dx x

1
∫( cot x)dx
Now, I.F = e = e
∫ pdx
= e
x
= e
log x+log(sin x)
= x sin x
log(x sin x)

The general solution of the given differential equation is given by the relation,
y(IF. ) = ∫ (Q × I. F.)dx + C

⇒ y(x sin x) = ∫ (1 × x sin x)dx + C

⇒ y(x sin x) = ∫ (x sin x)dx + C

d
⇒ y(x sin x) = x ∫ sin xdx − ∫ [ (x) ⋅ ∫ sin xdx] + C
dx

⇒ y(x sin x) = x(− cos x) − ∫ 1 ⋅ (− cos x)dx + C

⇒ y(x sin x) = −x cos x + sin x + C

−x cos x sin x C
⇒ y = + +
x sin x x sin x x sin x

1 C
⇒ y = − cot x + +
x x sin x

Page : 414 , Block Name : Exercise 9.6

dy
Q10 (x + y) = 1
dx

dy
(x + y) = 1
dx

dy 1
⇒ =
dx x+y

dx
⇒ = x + y
dy

dx
⇒ − x = y
dy

This is a linear differential equation of the form:

dy
+ px = Q( where p = −1 and Q = y)
dx

Now, I.F = e = e
∫ pdy
= e
∫ −dy −y

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

x(I. F.) = ∫ (Q × L. F.)dy + C
−y −y
⇒ xe = ∫ (y ⋅ e ) dy + C

−y −y d −y
⇒ xe = y ⋅ ∫ e dy − ∫ [ (y) ∫ e dy] dy + C
dy

−y −y −y
⇒ xe = y (−e ) − ∫ (−e ) dy + C

−y −y −y
⇒ xe = −ye + ∫ e dy + C

y
⇒ x = −y − 1 + Ce

y
⇒ x + y + 1 = Ce

Page : 414 , Block Name : Exercise 9.6

Q11 ydx + (x − y ) dy = 0 2

2
ydx + (x − y ) dy = 0

2
⇒ ydx = (y − x) dy

2
dx y −x x
⇒ = = y −
dy y y

dx x
⇒ + = y
dy y

This is a linear differential equation of the form:
dy 1
+ px = Q ( where p = and Q = y)
dx y

1

Now, I.F = e ∫ ρdy ∫ dy log y
y
= e = e = y

The general solution of the given differential equation is given by the relation,
x( I.F. ) = ∫ (Q × I. F. )dy + C

⇒ xy = ∫ (y ⋅ y)dy + C

2
⇒ xy = ∫ y dy + C

3
y
⇒ xy = + C
3

2
y C
⇒ x = +
3 y

Page : 414 , Block Name : Exercise 9.6

dy
Q12 (x + 3y ) 2

dx
= y(y > 0)

dy
2
(x + 3y ) = y
dx

dy y
⇒ =
dx x+3y 2

2
dx x+3y x
⇒ = = + 3y
dy y y

dx x
⇒ − = 3y
dy y

This is a linear differential equation of the form:

dx 1
+ px = Q (where p = − and Q = 3y)
dy y

dy 1

Now, I.F = e
−∫ log( ) 1
∫ pdy − log y y
= e y
= e = e =
y

The general solution of the given differential equation is given by the relation,
x( I.F. ) = ∫ (Q × I. F.)dy + C

1 1
⇒ x × = ∫ (3y × )dy + C
y y

x
⇒ = 3y + C
y

2
⇒ x = 3y + Cy

Page : 414 , Block Name : Exercise 9.6

dy
Q13
π
+ 2y tan x = sin x; y = 0 when x =
dx 3

dy
The given differential equation is dx + 2y tan x = sin x
dx

This is a linear equation of the form:

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

Now, I.F = e
2
∫ pdx ∫ 2 tan xdx 2 log | sec x| log(sec x) 2
= e = e = e = sec x

The general solution of the given differential equation is given by the relation,
y(I. F.) = ∫ (Q × I. F.)dx + C
2 2
⇒ y (sec x) = ∫ (sin x ⋅ sec x) dx + C

2
⇒ y sec x = ∫ (sec x ⋅ tan x)dx + C

2
⇒ y sec x = sec x + C . . . (i)
π
y = 0 at x =
3

Therefore,

2 π π
0 × sec = sec + C
3 3

⇒ 0 = 2 + C

⇒ C = −2

Substituting C = −2 in equation (i), we get:
2
y sec x = sec x − 2

2
⇒ y = cos x − 2 cos x

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

Page : 414 , Block Name : Exercise 9.6

dy
Q14 (1 + x ) 2

dx
+ 2xy =
1
2
; y = 0 when x = 1
1+x

dy 1
2
(1 + x ) + 2xy = 2
dx 1+x

dy 2xy 1
⇒ + 2
=
2
dx 1+x (1+x )
2

This is a linear differential equation of the form:

dy 2x 1
+ py = Q (where p = 2
and Q = )
2
dx 1+x (1+x )
2

2xdx
∫
Now, I.F = e
2
∫ pdx 2 log(1+x ) 2
= e 1+x
= e = 1 + x

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

y( I.F. ) = ∫ (Q × I.F. )dx + C

2 1 2
⇒ y (1 + x ) = ∫ 2
⋅ (1 + x )]dx + C
2
(1+x )

2 1
⇒ y (1 + x ) = ∫ 2
dx + C
1+x

2 −1
⇒ y (1 + x ) = tan x + C . . (i)

Now, y = 0 at x = 1

Therefore,

−1
0 = tan 1 + C
π
⇒ C = −
4
π
C = − in equation (i), we get:
4

2 −1 π
y (1 + x ) = tan x −
4

This is the required general solution of the given differential equation.

Page : 414 , Block Name : Exercise 9.6

dy
Q15 dx
− 3y cot x = sin 2x; y = 2 when x =
π

2

Answer. The given differential equation is
dy
− 3y cot x = sin 2x
dx

This is a linear differential equation of the form:

dy
+ py = Q (where p = −3 cot x and Q = sin 2x)
dx

∣ 1 ∣

Now, I.F e
log 1
∫ pdx −3 ∫ cot xdx −3 log | sin x| ∣ sin3 x ∣
= e = e = e =
3
sin x

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

y( I.F. ) = ∫ (Q × I.F. )dx + C

1 1
⇒ y ⋅ = ∫ [sin 2x ⋅ ] dx + C
3 3
sin x sin x

3
⇒ y csc x = 2 ∫ (cot x csc x)dx + C

3
⇒ y csc x = 2 csc x + C

2 3
⇒ y = − 2
+ 3
csc x csc x

2 3
⇒ y = −2 sin x + Csin x . . . (i)
π
y = 2 at x =
2

Therefore, we get:

2 = −2 + C

⇒ C = 4

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

2 3
y = −2 sin x − 4 sin x

3 2
⇒ y = 4 sin x − 2 sin x

This is the required particular solution of the given differential equation.

Page : 414 , Block Name : Exercise 9.6

Q16 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.

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

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

dy
According to the given inforve will be
dx

dy
= x + y
dx
dy
⇒ − y = x
dx

This is a linear differential equation of the form:

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

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

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

y( I.F. ) = ∫ (Q × I.F. )dx + C

−x −x
⇒ ye = ∫ xe dx + C … (i)

d
−x −x −x
Now, ∫ xe dx = x ∫ e dx − ∫ [ (x) ⋅ ∫ e dx] dx
dx

−x −x
= −xe − ∫ −e dx

−x −x
= −xe + (−e )

−x
= −e (x + 1)

Substituting in equation (i), we get:

−x −x
ye = −e (x + 1) + C

x
⇒ y = −(x + 1) + Ce

x
⇒ x + y + 1 = Ce . . . (ii)

The curve passes through the origin.

Therefore, equation (ii) becomes:

1 = c

⇒ C = 1

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

x
x + y + 1 = e

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

Page : 414 , Block Name : Exercise 9.6

Q17 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.

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

dy
to the curve at (x, y) is
dx

According to the given information:

dy
+ 5 = x + y
dx

dy
⇒ − y = x − 5
dx

This is a linear differential equation of the form:

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

Now, I.F = e ∫ ρdx
= e
[(−1)dx
= e
−x

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

y( I.F. ) = ∫ (Q × 1. F . )dx + C

−x −x
⇒y ⋅ e = ∫ (x − 5)e dx + C . . . (i)

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

−x −x
= (x − 5) (−e ) − ∫ (−e ) dx

−x −x
= (5 − x)e + (−e )

−x
= (4 − x)e

Therefore, equation (i) becomes:

−x −x
ye = (4 − x)e + C

x
⇒ y = 4 − x + Ce

x
⇒ x + y − 4 = Ce

The curve passes through point (0, 2).

Therefore, equation (ii) becomes:

0
0 + 2 − 4 = Ce

⇒ −2 = c

⇒ C = −2

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

x
x + y − 4 = −2e

x
⇒ y = 4 − x − 2e

This is the required equation of the curve.

Page : 414 , Block Name : Exercise 9.6

Q18
dy
2
The integrating factor of the differential equation x − y = 2x
dx

−x
A. e

−y
B. e

1
C.
x

D. x

The given differential equation is:

dy
2
x − y = 2x
dx

dy y
⇒ − = 2x
dx x

This is a linear differential equation of the form:

dy
1
+ py = Q (where p = − and Q = 2x)
dx x

The integrating factor (I.F) is given by the relation,
∫ pdx
e
1 −1
∫ dx −logx log(x ) −1 1
∴ I. F = e x = e = e = x =
x

Hence, the correct answer is C.

Page : 414 , Block Name : Exercise 9.6

Q19
The integrating factor of the differential equation.

2 dx
(1 − y ) + yx = ay(−1 < y < 1)
dy

1
A. 2
y −1

1
B.
√y 2 −1

2
C. 1 − y

1
D.
√1−y 2

The given differential equation is:

2 dx
(1 − y ) + yx = ay
dy

dy yx ay
⇒ + 2
= 2
dx 1−y 1−y

This is a linear differential equation of the form:

dx y ay
+ py = Q (where p = and Q = )
dy 1−y 2 1−y 2

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

1
y log[ ]
∫ dy 1 2
∫ pdy 2 − log(1−y ) √1−y 2 1
∴ I. F = e = e 1−y
= e 2 = e =
√1−y 2

Hence, the correct answer is D.

Page : 414 , Block Name : Exercise 9.6

Q1
For each of the differential equations given below, indicate its order and degree (if

defined).

2 2
d y dy
(i) 2
+ 5x( ) − 6y = log x
dx dx

3 2
dy dy
(ii) ( ) − 4( ) + 7y = sin x
dx dx

4 3
d y d y
(iii) − sin( 3
) = 0
4
dx dx

(i) The differential equation is given as:

2 2
d y dy
+ 5x( ) − 6y = log x
2
dx dx

2 2
d y dy
⇒ + 5x( ) − 6y − log x = 0
2
dx dx

The highest order derivative present in the differential equation is one.

2
d y
The highest power raised to is one. Hence, its degree is one.
2
dx

(ii) The differential equation is given as:

3 2
dy dy
( ) − 4( ) + 7y = sin x
dx dx

3 2
dy dy
⇒ ( ) − 4( ) + 7y − sin x = 0
dx dx

dy
The highest order derivative present in the differential equation is . Thus, its order is one.
dx

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

(iii) The differential equation is given as:

4 3
d y d y

4
− sin( 3
) = 0
dx dx

4
d y
The highest order derivative present in the differential equation is 4
, Thus, its order is four.
dx

However, the given differential equation is not a polynomial equation. Hence, its order is not

defined.

Page : 419 , Block Name : Miscellaneous Exercise

Q2 For each of the exercises given below,
verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
2
d y dy
x −x 2 2
(i) y = ae + be + x : x + 2 − xy + x − 2 = 0
2
dx dx

2
d y dy
x
(ii) y = e (a cos x + b sin x) : − 2 + 2y = 0
2
dx dx
2
d y
(iii) y = x sin 3x : + 9y − 6 cos 3x = 0
2
dx

dy
2 2 2 2
(iv) x = 2y log y : (x + y ) − xy = 0
dx

x −x 2
(i) y = ae + be + x

Differentiating both sides with respect to x, we get:

dy d d d
x −x 2
= a (e ) + b (e ) + (x )
dx dx dx dx

dy
x −x
⇒ = ae − be + 2x
dx

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

2
d y
x −x
= ae + be + 2
2
dx
2
dy d y
Now, on substituting the values of and in the differential equation, we get:
2
dx dx

L.H.S.
2
d y dy
2
x + 2 − xy + x − 2
2
dx dx

x −x x −x x −x 2 2
= x (ae + be + 2) + 2 (ae − be + 2x) − x (ae + be + x ) + x − 2

x −x x −x x −x 3 2
= (axe + bxe + 2x) + (2ae − 2be + 4x) − (axe + bxe + x ) + x − 2

x −x 2
= 2ae − 2be + x + 6x − 2

≠ 0

⇒ L.H.S. ≠ R.H.S.
Hence, the given function is not a solution of the corresponding differential equation.
x x x
(ii) y = e (a cos x + b sin x) = ae cos x + be sin x

Differentiating both sides with respect to x, we get:

dy d d
x x
= a ⋅ (e cos x) + b ⋅ (e sin x)
dx dx dx

dy
x x x x
⇒ = a (e cos x − e sin x) + b ⋅ (e sin x + e cos x)
dx

dy
x x
⇒ = (a + b)e cos x + (b − a)e sin x
dx

Again, dif f erentiating both sides with respect to x, we get :

2
d y d d
x x
⇒ = (a + b) ⋅ (e cos x) + (b − a) (e sin x)
2
dx dx dx
2
d y
x x x x
⇒ = (a + b) ⋅ [e cos x − e sin x] + (b − a) [e sin x + e cos x]
2
dx

2
d y
x
⇒ = e [(a + b)(cos x − sin x) + (b − a)(sin x + cos x)]
2
dx

2
d y
x
⇒ = e [a cos x − a sin x + b cos x − b sin x + b sin x + b cos x − a sin x − a cos x]
2
dx

2
d y
x
⇒ = [2e (b cos x − a sin x)]
2
dx
2
d y dy
Now, on substituting the values of and in the L.H.S. of the given differential
2
dx dx

equation, we get:

2
d y dy
+ 2 + 2y
2
dx dx
x x x
= 2e (b cos x − a sin x) − 2e [(a + b) cos x + (b − a) sin x] + 2e (a cos x + b sin x)
x
= e [(2b sin x − 2a sin x) − (2a cos x + 2b cos x)
x
= e [(2b sin x − 2a sin x) + (2a cos x + 2b sin x)]
x x
= e [(2b − 2a − 2b + 2a) cos x] + e [(−2a − 2b + 2a + 2b) sin x]

= 0

Hence, the given function is a solution of the corresponding differential equation.
(iii) y = x sin 3x

Differentiating both sides with respect to x, we get:

dy d
= (x sin 3x) = sin 3x + x ⋅ cos 3x ⋅ 3
dx dx

dy
⇒ = sin 3x + 3x cos 3x
dx

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

2
d y d d
= (sin 3x) + 3 (x cos 3x)
2
dx dx dx
2
d y
⇒ = 3 cos 3x + 3[cos 3x + x(− sin 3x) ⋅ 3]
2
dx

2
d y
⇒ = 6 cos 3x − 9x sin 3x
2
dx

2
d y
Substituting the value of in the L.H.S. of the given differential equation, we get:
2
dx

2
d y
+ 9y − 6 cos 3x
2
dx

= (6 ⋅ cos 3x − 9x sin 3x) + 9x sin 3x − 6 cos 3x

= 0

Hence, the given function is a solution of the corresponding differential equation.
2 2
(iv) x = 2y log y

Differentiating both sides with respect to x, we get:

d 2
2x = 2 ⋅ = [y log y]
dx

dy 1 dy
2
⇒ x = [2y ⋅ log y ⋅ + y ⋅ ⋅ ]
dx y dx

dy
⇒ x = (2y log y + y)
dx

dy x
⇒ =
dx y(1+2 log y)

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

dy
2 2
(x + y ) − xy
dx
2 2 x
= (2y log y + y ) ⋅ − xy
y(1+2 log y)

2 x
= y (1 + 2 log y) ⋅ − xy
y(1+2 log y)

= xy − xy

= 0

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

Page : 420 , Block Name : Miscellaneous Exercise

Q3
Form the differential equation representing the family of curves given by

2 2 2
(x − a) + 2y = a where a is an arbitrary constant.

2 2 2
(x − a) + 2y = a

2 2 2 2
⇒ x + a − 2ax + 2y = a

2 2
⇒ 2y = 2ax − x . . . (i)

Differentiating with respect to x, we get:

dy 2a−2x
2y =
dx 2

dy 2
2ax−2x
⇒ =
dx 2xy

dy 2
2ax−2x
⇒ = . . . (ii)
dx 4xy

From equation (i), we get:

2 2
2ax = 2y + x

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

2 2 2
dy 2y +x −2x
=
dx 4xy

2 2
dy 2y −x
⇒ =
dx 4xy

2 2
dy
Hence, the differential equation of the family of curves is given as
2y −x
=
dx 4xy

Page : 420 , Block Name : Miscellaneous Exercise

Q4
2
2 2 2 2
Prove that x − y = c(x + y ) is the general solution of differential equation

3 2 3 2
(x − 3xy ) dx = (y − 3x y) dy

where c is a parameter.

3 2 3 2
(x − 3xy ) dx = (y − 3x y) dy

3 2
dy x −3xy
⇒ = . . . (i)
dx y 3 −3x2 y

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

d d
⇒ (y) = (vx)
dx dx

dy dv
⇒ = v + x
dx dx
dv
Substituting the values of y and in equation (i), we get:
dx

3 2
dv x −3x(vx)
v + x = 3 2
dx (vx) −3x (vx)

2 2
dv 1−3v 1−3v
⇒ v + x = = 3
dx dx v −3v

2
dv 1−3v
⇒ x = 3
− v
dx v −3v
2 3
1−3v −v(v −3v)
dv
⇒ x = 3
dx v −3v

4
dv 1−v
⇒ x = 3
dx v −3v

3
v −3v dx
⇒ ( ) dv =
1−v4 x

Integrating both sides, we get:

3
v −3v ′
∫ ( ) dv = log x + log C (ii)
1−v4

3 3
v −3v v dv vdv
Now, ∫ ( ) dv = ∫ − 3∫
1−v4 1−v4 1−v4

3 3
v −3v v dv vdv
⇒ ∫ ( ) dv = I1 − 3I2 , where I1 = ∫ and I2 = ∫ . . . (iii)
1−v4 1−v4 1−v4

4
Let 1 − v = t

d 4 dt
∴ (1 − v ) =
dv dv

3 dt
⇒ −4v =
dv

3 dt
⇒ v dv = −
4

−dt 1 1 4
Now, I1 = ∫ = − log t = − log(1 − v )
4t 4 4

vdv vdv
And, I2 = ∫ = ∫
2
dv 1−(v )
2

d dp
2
∴ (v ) =
dv dv

dp
⇒ 2v =
dv

dp
⇒ vdv =
2

dp 2
1+p
1 1 ∣ ∣ 1 ∣ 1+v ∣
⇒ I2 = ∫ = log = log
2 1−p2 2×2 ∣ 1−p ∣ 4 ∣ 1−v2 ∣

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

2
3 ∣
∣−v
v −3v 1 4 3
∫( 4
)dv = − log(1 − v ) − log 2
|
1−v 4 4 1+v

Therefore, equation (ii) becomes:
2
1 3 |1+v | ′
4
log(1 − v ) − log 2
= log x + log C
4 4 1−v

2
3
1 4 1+v ′
⇒ − log[(1 − v ) ( ) ] = log C x
4 1−v2

4
2
(1+v ) −4
′
⇒ = (C x)
2
(1−v2 )

4
2
y
(1+ )
2
x
1
⇒ 2
=
C ′4 x4
y2
(1− )
2
x

4
2 2
(x +y )
1
⇒ =
2 ′4 4
x4 (x2 −y 2 ) C x

2
2 2 ′4 2 2
⇒ (x − y ) = C (x + y )

2
2 2 ′2 2 2
⇒ x − y = C (x + y )

2
2 2 2 2 ′2
⇒ x − y = C(x + y ) , where C = C

Hence, the given result is proved.

Page : 420 , Block Name : Miscellaneous Exercise

Q5 Form the differential equation of the family of circles in the rst quadrant which touch the coordinate axes.

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

touches the coordinate axes is:

2 2 2
(x − a) + (y − a) = a . . . (i)

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

dy
2(x − a) + 2(y − a) = 0
dx

′
⇒ (x − a) + (y − a)y = 0

′ ′
⇒ x − a + yy − ay = 0

′ ′
⇒ x + yy − a (1 + y ) = 0

′
x+yy
⇒ a =
1+y ′

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

′ 2 ′ 2 ′ 2
x+yy x+yy x+yy
[x − ( )] + [y − ( )] = ( )
1+y ′ 1+y ′ 1+y ′

′ 2 2 ′ 2
(x−y)y y−x x+yy
⇒ [ ] + [ ′
] = [ ′
]
′ 1+y 1+y
(1+y )

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

2 ′ 2 ′ 2
⇒ (x − y) [1 + (y ) ] = (x + yy )

H ence, the required dif f erential equation of the f amily of circles are

2 ′ 2 ′ 2
(x − y) [1 + (y ) ] = (x + yy )

Page : 420, Block Name : Miscellaneous Exercise

dy
1−y 2
Q6 Find the general solution of the differential equation dx dx
+ √ 2
= 0
1−x

2
dy 1−y
+ √ = 0
dx 1−x2

dy √1−y 2

⇒ = −
dx √1−x2

dy −dx
⇒ =
√1−y 2 √1−x2

Integrating both sides, we get:

−1 −1
sin y = − sin x + C

−1 −1
⇒ sin x + sin y = C

Page : 420 , Block Name : Miscellaneous Exercise

Q7
2
dy y +y+1
Show that the general solution of the differential equation + = 0
2
dx x +x+1

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

2
dy y +y+1
+ = 0
2
dx x +x+1

2
dy (y +y+1)

⇒ = − 2
dx x +x+1

dy −dx
⇒ 2
= 2
y +y+1 x +x+1

dy dx
⇒ 2
+ 2
= 0
y +y+1 x +x+1

Integrating both sides, we get:
dy dx
∫ 2
+ ∫ 2
= C
y +y+1 x +x+1

dy dx
⇒ ∫ 2
+ ∫ 2
= C
2 2
√3 √3
1 1
(y+ ) +( ) (x+ ) +( )
2 2 2 2

1 1
y+ x+
2 −1 2 2 −1 2
⇒ tan [ ] + tan [ ] = C
√3 2 √3 2

2y+1 2x+1 √3C
−1 −1
⇒ tan [ ] + tan [ ] =
√3 √3 2

2y+1 2x+1
+
√3 √3 √3C
−1
⇒ tan [ ] =
(2y+1) (2x+1) 2
1− ⋅
√3 √3

2x+2y+2

√3 √3C
−1
⇒ tan [ ] =
4xy+2x+2y+1 2
1−( )
3

2√3(x+y+1) √3C
−1
⇒ tan [ ] =
3−4xy−2x−2y−1 2

√3(x+y+1) √3C
−1
⇒ tan [ ] =
2(1−x−y−2xy) 2

√3(x+y+1) √3C √3C
⇒ = tan( ) = B, where B = tan( )
2(1−x−y−2xy) 2 2

2B
⇒ x + y + 1 = (1 − xy − 2xy)
√3

2B
⇒ x + y + 1 = A(1 − x − y − 2xy), where A =
√3

Hence, the given result is proved.

Page : 420 , Block Name : Miscellaneous Exercise

Q8
π
Find the equation of the curve passing through the point (0, )whose differential equation is,
4

sin x cos ydx + cos x sin ydy = 0

The differential equation of the given curve is:

sin x cos ydx + cos x sin ydy = 0

sin x cos ydx+cos x sin ydy
⇒ = 0
cos x cos y

⇒ tan xdx + tan ydy = 0

log(sec x) + log(sec y) = log C

log(sec x ⋅ sec y) = log C

⇒ sec x ⋅ sec y = C . . . (i)

π
The curve passes through point (0, )
4

∴ 1 × √2 = C

⇒ C = √2

On substituting C = √2 in equation (i), we get:

⇒ sec x ⋅ cos y = √2

1
⇒ sec x ⋅
√2

1
⇒ cos y =
√2
sec x
Hence, the required equation of the curve is cos y =
√2

Page : 420, Block Name : Miscellaneous Exercise

Q9
Find the particular solution of the differential equation

2x 2 x
(1 + e ) dy + (1 + y ) e dx = 0, given that y = 1 when x = 0

2x 2 x
(1 + e ) dy + (1 + y ) e dx = 0

dy x
e dx
⇒ 2
+ 2x
= 0
1+y 1+e

Integrating both sides, we get:
x
−1 e dx
tan y + ∫ = C . . . (i)
1+e2x

x dt
Let e = t ⇒
dx

x dt
⇒ e dx =
dx

x
⇒ e dx = dt

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

−1 dt
tan y + ∫ = C
1+t2

−1 −1
⇒ tan y + tan t = C

−1 −1 x
⇒ tan y + tan (e ) = C . . . (ii)

Therefore, equation (ii) becomes:

−1 −1
tan 1 + tan 1 = C
π π
⇒ + = C
4 4

π
⇒ C =
2
π
Substituting C = in equation (ii), we get:
2

−1 −1 x π
tan y + tan (e ) =
2

This is the required particular solution of the given differential equation.

Page : 420, Block Name : Miscellaneous Exercise

x x

Q10 Solve the differential equation ye dx = (xe y y 2
+ y ) dy(y ≠ 0)

x
y 2
ye dx = (xe y + y ) dy

x x
dx 2
y y
⇒ ye = xe + y
dy

x
dx 2
⇒ e y [y ⋅ − x] = y
dy

dx
x [y⋅ −x]
dy

⇒ ey ⋅ = 1 . . . (i)
2
y
x

y
Let e = z

Differentiating it with respect to y, we get:
x
d dz
(e y ) =
dy dy

x
d x dz
⇒ ey ⋅ ( ) =
dy y dy

dx
x y⋅ −x
dy dz
y
⇒ e ⋅ [ 2
] = . . . (ii)
y dy

From equation (i) and equation (ii), we get:

dz
= 1
dy

⇒ dz = dy

Integrating both sides, we get:

z = y + C
x

y
⇒ e = y + C

Page : 420 , Block Name : Miscellaneous Exercise

Q11
Find a particular solution of the differential equation (x − y)(dx + dy) = dx − dy , given that y = −1,

when x = 0 (Hint: put x − y = t)

(x − y)(dx + dy) = dx − dy

⇒ (x − y + 1)dy = (1 − x + y)dx

dy 1−x+y
⇒ =
dx x−y+1

dy 1−(x−y)
⇒ = . . . (i)
dx 1+(x−y)

Let x − y = t

d dt
⇒ (x − y) =
dx dx

dy dt
⇒ 1 − =
dx dx

dt dy
⇒ 1 − =
dx dx

dy
Substituting the values of x − y and in equation (i), we get:
dx

dt 1−t
1 − =
dx 1+t

dt 1−t
⇒ = 1 − ( )
dx 1+t

dt (1+t)−(1−t)
⇒ =
dx 1+t

dt 2t
⇒ =
dx 1+t

1+t
⇒ ( ) dt = 2dx
t

1
⇒ (1 + ) dt = 2dx . . . (ii)
t

Integrating both sides, we get:

t + log |t| = 2x + C

⇒ (x − y) + log |x − y| = 2x + C

⇒ log |x − y| = x + y + C . . . (iii)

Now, y = −1 at x = 0

Therefore, equation ( iii ) becomes:

Tog 1 = 0 − 1 + C

⇒ C = 1

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

log |x − y| = x + y + 1

This is the required particular solution of the given differential equation.

Page : 420 , Block Name : Miscellaneous Exercise

−2√x y
Q12
e dx
Solve the differential equation [ − ] = 1(x ≠ 0)
√x √x dy

e
−2√x y dx
[ − ] = 1
√x √x dy

dy e
−2√x y
⇒ = −
dx √x √x

dy y e
−2√x

⇒ + =
dx √x √x

dy 1 e
−2√x

+ P y = Q, where P = and Q =
dx √x √x
1
∫ dx
Now, I.F = e ∫ pdx
= e √x
= e
2√x

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

y( I.F. ) = ∫ (Q × I.F. )dx + C
−2√x
2√x e 2√x
⇒ ye = ∫ ( × e ) dx + C
√x

2√x 1
⇒ ye = ∫ dx + C
√x

2√x
⇒ ye = 2 √x + C

Page : 421 , Block Name : Miscellaneous Exercise

Q13

dy
Find a particular solution of the differential equation + y cot x = 4x csc x(x ≠ 0)
dx

π
given that y = 0 when x =
2

The given differential equation is:

dy
+ y cot x = 4x csc x
dx

This equation is a linear differential equation of the form

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

Now, I.F = e ∫ pdx
= e
∫ cot xdx
= e
log | sin x|
= sin x

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

y( I.F. ) = ∫ (Q × I. F.)dx + C

⇒ y sin x = ∫ (4x csc x ⋅ sin x)dx + C

⇒ y sin x = 4 ∫ xdx + C

2
x
⇒ y sin x = 4 ⋅ + C
2

2
⇒ y sin x = 2x + C . . . (i)
π
y = 0 at x =
2

Therefore, equation (i) becomes:

2
π
0 = 2 × + C
4

2
π
⇒ C = −
2
2
π
C = − in equation (i), we get:
2

2
2 π
y sin x = 2x −
2

This is the required particular solution of the given differential equation.

Page : 421 , Block Name : Miscellaneous Exercise

Q14
dy
−y
(x+1) =2e −1
Find a particular solution of the differential equation dx
, given that y = 0 when

x = 0

dy
−y
(x + 1) = 2e − 1
dx

dy dx
⇒ −y
=
2e −1 x+1

y
e dy dx
⇒ =
2−ey x+1

Integrating both sides, we get:
y
e dy
∫ = log |x + 1| + log C . . . (i)
2−ey

y
Let 2 − e = t

d y dt
∴ (2 − e ) =
dy dy

y dt
⇒ −e =
dy

y
⇒ e dt = −dt

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

−dt
∫ = log |x + 1| + log C
t

⇒ − log |t| = log |C(x + 1)|

y
⇒ − log|⋅ − e | = log |C(x + 1)|

1
⇒ y
= C(x + 1)
2−e

y 1
⇒ 2 − e = . . . (ii)
C(x+1)

Now, at x = 0 and y = 0, equation (ii) becomes:
1
⇒ 2 − 1 =
C

⇒ C = 1

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

y 1
2 − e =
x+1

y 1
⇒ e = 2 −
x+1

y 2x+2−1
⇒ e =
x+1

y 2x+1
⇒ e =
x+1

2x+1
⇒ y = log∣
∣
∣, (x ≠ −1)
∣
x+1

This is the required particular solution of the given differential equation.

Page : 421 , Block Name : Miscellaneous Exercise

Q15 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?

Answer. 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
∴ ∝ y
dt

dy
⇒ = ky (k is constent)
dt

dy
⇒ = kdt
y

Integrating both sides, we get:
log y = kt + C … (i)
In the year 1999, t = 0 and y = 20000.
Therefore, we get:
log 20000 = C … (ii)
In the year 2004, t = 5 and y = 25000.
Therefore, we get:

log 25000 = k ⋅ 5 + C

⇒ log 25000 = 5k + log 20000

25000 5
⇒ 5k = log( ) = log( )
20000 4

1 5
⇒ k = log( ) . . . (iii)
5 4

In the year 2009, t = 10 years.

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

1 5
log y = 10 × log( ) + log(20000)
5 4

2
5
⇒ log y = log[20000 × ( ) ]
4

5 5
⇒ y = 20000 × ×
4 4

⇒ y = 31250

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

Page : 421 , Block Name : Miscellaneous Exercise

Q16
ydx−xdy
The general solution of the differential equation y = 0
y

A. xy = c

2
B. x = cy

C. y = cx

2
D. y = cx

The given differential equation is:

ydx−xdy
= 0
y

ydx−xdy
⇒ = 0
xy

1 1
⇒ dx − dy = 0
x y

Integrating both sides, we get:

log |x| − log |y| = log k

x
⇒ log∣
∣
∣ = log k
∣ y

x
⇒ = k
y

1
⇒ y = x
k

1
⇒ y = Cx where C =
k

Hence, the correct answer is C.

Page : 421 , Block Name : Miscellaneous Exercise

Q17 The general solution of a differential equation of the type is
dx
+ P1 x = Q
dy 1

∫ p dy ∫ p dy
A. ye 1
= ∫ (Q1 e 1
) dy + C

∫ P1 dx ∫ p dx
B ⋅ e = ∫ (Q e 1
) dx + C
l

∫ p1 dy ∫ p1 dy
C. xe = ∫ (Q e ) dy + C
1

∫ pdx ∫ p1 dx
D. xe = ∫ (Q e ) dx + C
1

Answer. The integrating factor of the given differential equation
dx ∫ P1dy
+ P1 x = Q is e .
dy 1

The general solution of the differential equation is given by,
x( I.F. ) = ∫ (Q × I. F.)dy + C

Hence, the correct answer is C.

Page : 421 , Block Name : Miscellaneous Exercise

Q18
x x
The general solution of the differential equation e dy + (ye + 2x) dx = 0 is

y 2
A. xe + x = c

y 2
B. xe + y = c

x 2
C. ye + x = c

y 2
D. ye + x = c

The given differential equation is:
x x
e dy + (ye + 2x) dx = 0

dy
x x
⇒ e + ye + 2x = 0
dx

dy
−x
⇒ + y = −2xe
dx

This is a linear differential equation of the form

dy
−x
+ P y = Q, where P = 1 and Q = −2xe .
dx

Now, I.F = e = e
∫ pdx
= e
∫ dx x

The general solution of the given differential equation is given by,
y( I.F. ) = ∫ (Q × I. F. )dx + C
x −x x
⇒ ye = ∫ (−2xe ⋅ e ) dx + C

x
⇒ ye = − ∫ 2xdx + C

x 2
⇒ ye = −x + C

x 2
⇒ ye + x = C

Hence, the correct answer is C .

Page : 421 , Block Name : Miscellaneous Exercise