Class 12
Chapter 9 - Differential Equations
Exercise 9.1
Determine order and degree (if defined) of differential equations
given in Exercise 1 to 10:
1.
d4 y
+ sin (y′′′) = 0
dx 4
d4 y
+ sin y′′′ = 0
dx 4
The highest order derivative present in the differential equation
Sol. The given D.E. is
is
d4 y
and its order is 4.
dx 4
The given differential equation is not a polynomial equation in
derivatives (... The term sin y′′′ is a T-function of derivative y′′′).
Therefore degree of this D.E. is not defined.
Ans. Order 4 and degree not defined.
2. y′′ + 5y = 0
Sol. The given D.E. is y′ + 5y = 0.
 dy 
The highest order derivative present in the D.E. is y′  =
 and
 dx 
so its order is one. The given D.E. is a polynomial equation in
derivatives (y′ here) and the highest power raised to highest order
derivative y′ is one, so its degree is one.
Ans. Order 1 and degree 1.
MathonGo
1
Class 12
Chapter 9 - Differential Equations
4
d 2s
= 0
dt 2
4
d2 s
 ds 
Sol. The given D.E. is   + 3s
= 0.
dt 2
 dt 
 ds 
3. 

 dt 
+ 3s
d2 s
and its
dt 2
order is 2. The given D.E is a polynomial equation in derivatives
d2 s
and the highest power raised to highest order derivative
is
dt 2
one. Therefore degree of D.E. is 1.
Ans. Order 2 and degree 1.
2
 d2 y 
dy
4. 
2
 + cos dx = 0
dx


2
 d2 y 
 dy 


Sol. The given D.E. is  2  + cos 
 = 0.
dx
 dx 


The highest order derivative present in the differential equation
The highest order derivative present in the D.E. is
is
d2 y
dx2
and its order is 2.
The given D.E. is not a polynomial equation in derivatives
dy
dy
is a T-function of derivative
).
(... The term cos
dx
dx
Therefore degree of this D.E. is not defined.
Ans. Order 2 and degree not defined.
5.
d2 y
dx 2
= cos 3x + sin 3x
Sol. The given D.E. is
d2 y
= cos 3x + sin 3x.
dx 2
The highest order derivative present in the D.E. is
d2 y
and its
dx 2
order is 2.
The given D.E. is a polynomial equation in derivatives and the
1
 d2 y 
d2 y


highest power raised to highest order
2 =  dx 2  is one, so
dx


its degree is 1.
Ans. Order 2 and degree 1.
Remark. It may be remarked that the terms cos 3x and sin 3x
present in the given D.E. are trigonometrical functions (but not
T-functions of derivatives).
dy 

It may be noted that  cos 3
is not a polynomial function of
dx 

derivatives.
′′′
′′
6. ( y′′′
′′′)2 + ( y′′
′′)3 + ( y′′ )4 + y5 = 0
Sol. The given D.E. is ( y′′′)2 + ( y′′)3 + ( y′)4 + y5 = 0.
...(i)
The highest order derivative present in the D.E. is y′′′ and its
order is 3.
MathonGo
2
Class 12
7.
Sol.
8.
Sol.
9.
Sol.
10.
Sol.
11.
Chapter 9 - Differential Equations
The given D.E. is a polynomial equation in derivatives y′′′, y′′ and
y′ and the highest power raised to highest order derivative y′′′ is
two, so its degree is 2.
Ans. Order 3 and degree 2.
′′′ + 2y′′
′′ + y′′ = 0
y′′′
The given D.E. is y′′′ + 2y′′ + y′ = 0.
...(i)
The highest order derivative present in the D.E. is y′′′ and its
order is 3.
The given D.E. is a polynomial equation in derivatives y′′′, y′′
and y′ and the highest power raised to highest order derivative
y′′′ is one, so its degree is 1.
Ans. Order 3 and degree 1.
y′′ + y = ex
The given D.E. is y′ + y = ex.
...(i)
The highest order derivative present in the D.E. is y′ and its
order is 1.
The given D.E. is a polynomial equation in derivative y′. (It may
be noted that ex is an exponential function and not a polynomial
function but is not an exponential function of derivatives) and the
highest power raised to highest order derivative y′ is one, so its
degree is 1.
Ans. Order 1 and degree 1.
′′ + ( y′′ )2 + 2y = 0
y′′
The given D.E. is y′′ + ( y′)2 + 2y = 0.
...(i)
The highest order derivative present in the D.E. is y′′ and its
order is 2.
The given D.E. is a polynomial equation in derivatives y′′ and y′
and the highest power raised to highest order derivative y′′ is
one, so its degree is 1.
Ans. Order 2 and degree 1.
′′ + 2y′′ + sin y = 0
y′′
The given D.E. is y′′ + 2y′ + sin y = 0.
...(i)
The highest order derivative present in the D.E. is y′′ and its
order is 2.
The given D.E. is a polynomial equation in derivatives y′′ and y′.
(It may be noted that sin y is not a polynomial function of y, it
is a T-function of y but is not a T-function of derivatives) and the
highest power raised to highest order derivative y′′ is one, so its
degree is one.
Ans. Order 2 and degree 1.
The degree of the differential equation
 d2 y 

2

 dx 
(A) 3
3
 dy 
+ 

 dx 
(B) 2
2
 dy 
+ sin 
 + 1 = 0 is
 dx 
(C) 1
MathonGo
(D) Not defined.
3
Class 12
Chapter 9 - Differential Equations
Sol. The given D.E. is
3
2
 d2 y 
 dy 
 dy 
 2  + 
+ sin 
 + 1 = 0

 dx 
 dx 
 dx 
This D.E. (i) is not a polynomial equation in derivatives.
...(i)

dy 
 dy 
∵ sin  dx  is a T-function of derivative dx 




∴ Degree of D.E. (i) is not defined.
Answer. Option (D) is the correct answer.
12. The order of the differential equation
2x2
(A) 2
dy
d2 y
2 – 3 dx + y = 0 is
dx
(B) 1
(C) 0
(D) Not defined
dy
d2 y
– 3
+ y = 0
dx
dx 2
The highest order derivative present in the differential equation
Sol. The given D.E. is 2x2
is
d2 y
dx2
and its order is 2.
Answer. Order of the given D.E. is 2.
MathonGo
4
Class 12
Chapter 9 - Differential Equations
Exercise 9.2
In each of the Exercises 1 to 6 verify that the given functions
(explicit) is a solution of the corresponding differential equation:
′′ – y′′ = 0
1. y = ex + 1 : y′′
Sol. Given: y = ex + 1
...(i)
To prove: y given, by (i) is a solution of the D.E. y′′ – y′ = 0 ...(ii)
From (i), y′ = ex + 0 = ex and y′′ = ex
∴ L.H.S. of D.E. (ii) = y′′ – y′ = ex – ex = 0 = R.H.S. of D.E. (ii)
∴ y given by (i) is a solution of D.E. (ii).
2. y = x2 + 2x + C : y′′ – 2x – 2 = 0
...(i)
Sol. Given: y = x2 + 2x + C
To prove: y given by (i) is a solution of the D.E.
y′ – 2x – 2 = 0
...(ii)
From (i), y′ = 2x + 2
∴ L.H.S. of D.E. (ii) = y′ – 2x – 2
= (2x + 2) – 2x – 2 = 2x + 2 – 2x – 2 = 0 = R.H.S. of D.E. (ii)
∴ y given by (i) is a solution of D.E. (ii).
3. y = cos x + C : y′′ + sin x = 0
Sol. Given: y = cos x + C
...(i)
To prove: y given by (i) is a solution of D.E. y′ + sin x = 0 ...(ii)
From (i), y′ = – sin x
∴ L.H.S. of D.E. (ii) = y′ + sin x = – sin x + sin x
= 0 = R.H.S. of D.E. (ii)
MathonGo
5
Class 12
Chapter 9 - Differential Equations
∴ y given by (i) is a solution of D.E. (ii).
4. y =
xy
1 + x 2 : y′′ = 1 + x 2
1 + x2
...(i)
xy
To prove: y given by (i) is a solution of D.E. y′ =
...(ii)
1 + x2
d
d
2 1/2
From (i), y′ =
1 + x2 = dx (1 + x )
dx
x
1
d
1
=
(1 + x2)– 1/2
(1 + x2) =
(1 + x2)– 1/2 . 2x =
...(iii)
1 + x2
2
dx
2
Sol. Given: y =
R.H.S. of D.E. (ii) =
xy
x
=
1 + x2
1 + x2
x
1 + x2
(By (i))

t
t
1 
=
=
∵

t
1+ x
t t
t 

= y′ [By (iii)] = L.H.S. of D.E. (ii)
∴ y given by (i) is a solution of D.E. (ii).
5. y = Ax : xy′′ = y (x ≠ 0)
Sol. Given: y = Ax
...(i)
To prove: y given by (i) is a solution of the D.E. xy′ = y (x ≠ 0)
...(ii)
From (i), y′ = A(1) = A
L.H.S. of D.E. (ii) = xy′ = xA
= Ax = y [By (i)] = R.H.S. of D.E. (ii)
∴ y given by (i) is a solution of D.E. (ii).
=
2
x 2 – y 2 (x ≠ 0 and x > y or x < – y)
Sol. Given: y = x sin x
...(i)
To prove: y given by (i) is a solution of D.E.
6. y = x sin x : xy′′ = y + x
xy′ = y + x
x 2 − y2
...(ii) (x ≠ 0 and x > y or x < – y)
dy
d
d
x = x cos x + sin x
(= y′) = x
(sin x) + sin x
dx
dx
dx
L.H.S. of D.E. (ii) = xy′ = x (x cos x + sin x)
= x2 cos x + x sin x
...(iii)
From (i),
R.H.S. of D.E. (ii) = y + x
x 2 − y2
Putting y = x sin x from (i),
= x sin x + x
= x sin x + x
= x sin x + x2
From (iii) and
∴ y given by
x 2 − x 2 sin 2 x
= x sin x + x
x 2 (1 − sin 2 x)
x2 cos2 x = x sin x + x . x cos x
cos x = x2 cos x + x sin x
(iv), L.H.S. of D.E. (ii) = R.H.S. of D.E. (ii)
(i) is a solution of D.E. (ii).
MathonGo
...(iv)
6
Class 12
Chapter 9 - Differential Equations
In each of the Exercises 7 to 10, verify that the given functions
(Explicit or Implicit) is a solution of the corresponding
differential equation:
7. xy = log y + C : y′′ =
y2
(xy ≠ 1)
1 – xy
Sol. Given: xy = log y + C
...(i)
To prove that Implicit function given by (i) is a solution of the
y2
1 − xy
Differentiating both sides of (i) w.r.t. x, we have
D.E.
y′ =
xy′ + y(1) =
⇒
xy′ –
...(ii)
1
y y′ + 0
y′
y = – y
 xy − 1 
 = – y
⇒ y′ 
 y 

1
⇒ y′  x −  = – y
y

⇒
y′(xy – 1) = – y2
− y2
− y2
y2
=
− (1 − xy)
xy − 1
1 − xy
which is same as differential equation (ii), i.e., Eqn. (ii) is proved.
∴ Function (Implicit) given by (i) is a solution of D.E. (ii).
8. y – cos y = x : ( y sin y + cos y + x) y′ = y
Sol. Given: y – cos y = x
...(i)
To prove that function given by (i) is a solution of D.E.
( y sin y + cos y + x) y′ = y
...(ii)
Differentiating both sides of (i) w.r.t. x, we have
y′ + (sin y) y′ = 1
⇒ y′ (1 + sin y) = 1
⇒
y′ =
⇒
y′ =
1
1 + sin y
...(iii)
Putting the value of x from (i) and value of y′ from (iii) in L.H.S.
of (ii), we have
L.H.S. = ( y sin y + cos y + x) y′
= ( y sin y + cos y + y – cos y)
= y (sin y + 1)
1
1
= ( y sin y + y)
1 + sin y
1 + sin y
1
= y = R.H.S. of (ii).
(1 + sin y)
∴ The function given by (i) is a solution of D.E. (ii).
9. x + y = tan– 1 y : y2 y′′ + y2 + 1 = 0
Sol. Given: x + y = tan– 1 y
To prove that function given by (i) is a solution of D.E.
y2 y′ + y2 + 1 = 0
Differentiating both sides of (i), w.r.t. x, 1 + y′ =
MathonGo
...(i)
...(ii)
1
y′
1 + y2
7
Class 12
Chapter 9 - Differential Equations
Cross-multiplying
(1 + y′)(1 + y2) = y′
⇒ 1 + y2 + y′ + y′y2 = y′
2
2
⇒ y y′ + y + 1 = 0 which is same as D.E. (ii).
∴ Function given by (i) is a solution of D.E. (ii).
10. y =
dy
a 2 – x 2 , x ∈ (– a, a) : x + y dx = 0 ( y ≠ 0)
Sol. Given: y = a2 − x2 , x ∈ (– a, a)
To prove that function given by (i) is a solution of D.E.
x + y
From (i),
dy
= 0
dx
dy
1
d
=
(a2 – x2)– 1/2
(a2 – x2)
dx
2
dx
−x
1
=
2
2 (– 2x) =
2
2 a −x
a − x2
Putting these values of y and
Sol.
12.
Sol.
...(ii)
...(iii)
dy
from (i) and (iii) in L.H.S. of (ii),
dx

− x 

2
 a − x2 


= x – x = 0 = R.H.S. of D.E. (ii).
∴ Function given by (i) is a solution of D.E. (ii).
Choose the correct answer:
The number of arbitrary constants in the general solution
of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4.
Option (D) 4 is the correct answer.
Result. The number of arbitrary constants (c1, c2, c3 etc.) in the
general solution of a differential equation of nth order is n.
The number of arbitrary constants in the particular
solution of a differential equation of third order are
(A) 3
(B) 2
(C) 1
(D) 0.
The number of arbitrary constants in a particular solution of a
differential equation of any order is zero (0).
[ . . . By definition, a particular solution is a solution which
contains no arbitrary constant.]
∴ Option (D) is the correct answer.
L.H.S. = x + y
11.
...(i)
dy
= x +
dx
a2 − x 2
MathonGo
8
Class 12
Chapter 9 - Differential Equations
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.
x
y
1.
+
= 1
a
b
MathonGo
9
Class 12
Chapter 9 - Differential Equations
x
y
+
= 1
...(i)
a
b
Here there are two arbitrary constants a and b. So we shall
differentiate both sides of (i) two times w.r.t. x.
Sol. Equation of the given family of curves is
1
1 dy
1
1 dy
. 1 +
= 0 or
= –
...(ii)
a
b dx
a
b dx
2
1 d y
Again diff. (ii) w.r.t. x, 0 = –
b dx 2
d2 y
Multiplying both sides by – b,
= 0.
dx 2
Which is the required D.E.
Remark. We need not eliminate a and b because they have
already got eliminated in the process of differentiation.
2. y2 = a(b2 – x2)
Sol. Equation of the given family of curves is
y2 = a(b2 – x2)
...(i)
Here there are two arbitrary constants a and b. So, we are to
differentiate (i) twice w.r.t. x.
From (i),
dy
= a(0 – 2x) = – 2ax.
dx
dy
Dividing by 2, y
= – ax
dx
Again differentiating both sides of (ii) w.r.t. x,
From (i), 2y
...(ii)
2
dy
dy
d2 y
d2 y
 dy 
or
y
+ 
 = – a ...(iii)
2 + dx . dx = – a
dx
dx 2
 dx 
Putting this value of – a from (iii) in (ii), (To eliminate a, as b is
already absent in both (ii) and (iii)), we have
2
 d 2 y  dy 2 
dy
dy
d2 y
 dy 
or
xy
+
x
= y
=  y dx 2 +  dx   x
y


2
dx
dx
dx
dx
 


y
2
dy
d2 y
 dy 
+ x 
or xy
 – y dx = 0.
dx 2
 dx 
3. y = ae3x + be–2x
Sol. Equation of the family of curves is
y = a e3x + b e– 2x
Here are two arbitrary constants a and b.
dy
From (i),
= 3 ae3x – 2 be– 2x
dx
Again differentiating both sides of (ii), w.r.t. x,
d2 y
= 9 ae3x + 4 be– 2x
dx 2
Let us eliminate a and b from (i), (ii) and (iii).
Equation (ii) – 3 × eqn. (i) gives (To eliminate a),
MathonGo
...(i)
...(ii)
...(iii)
10
Class 12
Chapter 9 - Differential Equations
dy
– 3y = – 5 be– 2x
dx
Again Eqn. (iii) – 3 × eqn (ii) gives (again to eliminate a)
...(iv)
dy
d2 y
= 10 be– 2x
– 3
dx
dx 2
Now Eqn. (v) + 2 × eqn. (iv) gives (To eliminate b)
...(v)
dy
 dy

d2 y
− 3 y  = 10 be– 2x – 10 be– 2x
2 – 3 dx + 2  dx


dx
dy
dy
d2 y
or
– 3
+ 2
– 6y = 0
dx
dx
dx 2
2
dy
d y
or
2 – dx – 6y = 0
dx
which is the required D.E.
4. y = e2x (a + bx)
Sol. Equation of the given family of curves is
y = e2x (a + bx)
Here are two arbitrary constants a and b.
...(i)
dy
d
 d 2x 
e  (a + bx) + e2x
(a + bx)
= 
dx
dx
 dx

dy
or
= 2 e2x (a + bx) + e2x . b
dx
dy
or
= 2y + be2x
dx
(By (i))
Again differentiating both sides of (ii), w.r.t. x
From (i),
d2 y
= 2
dx 2
Let us eliminate
absent in both (ii)
...(ii)
dy
+ 2 be2x
...(iii)
dx
b from eqns. (ii) and (iii), (as a is already
and (iii))
dy
– 2y = be2x
dx
Putting this value of be2x in (iii), we have
From eqn. (ii)
dy
 dy

d2 y
− 2y ⇒
= 2
+ 2 
dx
 dx

dx 2
2
dy
d y
or
– 4
+ 4y = 0
dx
dx 2
which is the required D.E.
5. y = ex (a cos x + b sin x)
Sol. Equation of family of curves is
y = ex (a cos x + b sin x)
∴
dy
dy
d2 y
= 2
+ 2
– 4y
dx
dx
dx 2
...(i)
dy
 d x
e  (a cos x + b sin x) + ex (– a sin x + b cos x)
= 
dx
 dx 
MathonGo
11
Class 12
Chapter 9 - Differential Equations
dy
= ex (a cos x + b sin x) + ex (– a sin x + b cos x)
dx
dy
or
= y + ex (– a sin x + b cos x)
...(ii)
dx
(By (i))
Again differentiating both sides of eqn. (ii), w.r.t. x, we have
or
dy
d2 y
=
+ ex (– a sin x + b cos x) + ex (– a cos x – b sin x)
dx
dx 2
dy
 dy

d2 y
− y  – ex (a cos x + b sin x)
or
2 = dx +  dx


dx
(By (ii))
dy
d2 y
= 2
– y – y
or
dx
dx 2
(By (i))
dy
d2 y
– 2
+ 2y = 0 which is the required D.E.
dx
dx 2
6. Form the differential equation of the family of circles
touching the y-axis at the origin.
Sol. Clearly, a circle which touches y-axis at the origin must have its
centre on x-axis.
[... x-axis being at right angles to tangent y-axis is the normal or
line of radius of the circle.]
∴ The centre of circle is (r, 0) where r is the radius of the
circle.
∴ Equation of required circles is
(x – r)2 + ( y – 0)2 = r2
[(x – α)2 + ( y – β)2 = r2]
or
x2 + r2 – 2rx + y2 = r2
or
x2 + y2 = 2rx
...(i)
where r is the only arbitrary constant.
∴ Differentiating both sides of (i) only once w.r.t. x, we have
or
2x + 2y
dy
= 2r
dx
...(ii)
To eliminate r, putting the value of 2r from (ii) in (i),
dy 

x2 + y2 =  2 x + 2 y
x
dx 

or
x2 + y2 = 2x2 + 2xy
Y
dy
dx
( r, 0 )
dy
r
– x2 + y2 = 0
dx
dy
Multiplying by – 1, 2xy
+ x2 – y2 = 0
dx
dy
or 2xy
+ x2 = y2 which is the required D.E.
dx
O
or – 2xy
MathonGo
X
C
12
Class 12
Chapter 9 - Differential Equations
Remark. The above question can also be stated as : Form the
D.E. of the family of circles passing through the origin and
having centres on x-axis.
7. Find the differential equation of the family of parabolas
having vertex at origin and axis along positive y-axis.
Sol. We know that equation of parabolas having vertex at origin and
axis along positive y-axis is x2 = 4ay
...(i)
Here a is the only arbitrary constant. So differentiating both
sides of Eqn. (i) only once w.r.t. x, we have
dy
2x = 4a
...(ii)
dx
Y
To eliminate a, putting
4a =
x2
from (i) in (ii), we have
y
2x =
x2
y
⇒ 2xy = x2
dy
dx
X
O
(VERTEX)
dy
dx
Dividing both sides by x, 2y = x
dy
dx
dy
+ 2y = 0
dx
dy
⇒
x
– 2y = 0 which is the required D.E.
dx
8. Form the differential equation of family of ellipses having
foci on y-axis and centre at the origin.
Y
Sol. We know that equation of
ellipses having foci on y-axis
Major Axis
i.e., vertical ellipses with
(0, a)
major axis as y-axis is
⇒
– x
y2
x2
+
= 1 ...(i)
a2
b2
Here a and b are two arbitrary X′
constants.
So we shall differentiate eqn.
(i) twice w.r.t. x.
Differentiating both sides of
(i) w.r.t. x, we have
F
b
b
(– b, 0) O
F′
Focus
X
(b , 0)
Minor Axis
Focus
(0, – a)
Y′
1
dy
1
2y
+ 2 2x = 0
dx
a2
b
2
dy
2
or
y
= – 2 x
dx
a2
b
Dividing both sides by 2,
MathonGo
13
Class 12
Chapter 9 - Differential Equations
−1
1
dy
y
= 2 x
dx
a2
b
Again differentiating both sides of (ii) w.r.t. x, we have
1
a2
...(ii)
 d2 y dy dy 
−1
+
.
y

2
dx dx  = b2
 dx
...(iii)
To eliminate a and b, putting this value of
the required differential equation is
1
dy
1
y
= 2
dx
a2
a
Multiplying both sides by a2, y
−1
from (iii) in (ii),
b2
 d 2 y  dy 2 
+
y
  x
2
 dx  
 dx
dy
d2 y
 dy 
= xy
2 + x  dx 
dx
dx
 
2
2
dy
d2 y
 dy 
+ x 
 – y dx = 0
dx
dx 2
 
which is the required differential equation.
9. Form the differential
Y
equation of the family of
Conjugate
hyperbolas having foci on
Axis
x-axis and centre at the
(– a, 0)
origin.
F1(Focus)
X
Sol. We know that equation of
O
F2
hyperbolas having foci on xTransverse
(Focus)
(a, 0)
axis and centre at origin is
Axis
or xy
x2
y2
= 1
...(i)
2 –
a
b2
Here a and b are two arbitrary constants. So we shall differentiate eqn. (i) twice w.r.t. x.
From (i),
1
1
dy
= 0 or
2 . 2x –
2 . 2y
dx
a
b
2
2
dy
2 x =
2 y
dx
a
b
1
1
dy
x = 2 y
dx
a2
b
Again differentiating both sides of (ii), w.r.t. x,
Dividing both sides by 2,
...(ii)
 d2 y dy dy 
1
1
. 1 = 2 .  y 2 + dx . dx 
 dx
a2
b

or
1
1
= 2
a2
b
 d 2 y  dy 2 
+
y
 
2
 dx  
 dx
...(iii)
Dividing eqn. (iii) by eqn. (ii), we have (To eliminate a and b)
MathonGo
14
Class 12
Chapter 9 - Differential Equations
2
1
=
x
Cross-multiplying,
d2 y  dy 
+

dx2  dx 
dy
y
dx
 d 2 y  dy 2 
dy
x  y dx 2 +  dx   = y
dx


y
2
dy
d2 y
 dy 
+ x 
 – y dx = 0
dx 2
 dx 
which is the required differential equation.
10. Form the differential equation of the family of circles
having centres on y-axis and radius 3 units.
Sol. We know that on y-axis, x = 0.
∴ Centre of the circle on y-axis is (0, β).
∴ Equation of the circle having centre on y-axis and radius 3
units is
[(x – α)2 + (y – β)2 = r2]
(x – 0)2 + ( y – β)2 = 32
2
2
or
x + ( y – β) = 9
...(i)
Here β is the only arbitrary constant. So we shall differentiate
both sides of eqn. (i) only once w.r.t. x,
d
From (i), 2x + 2 (y – β)
( y – β) = 0
dx
dy
or 2x + 2 ( y – β)
= 0
dx
−x
dy
− 2x
=
or
2 ( y – β)
= – 2x
∴
y – β =
dx
dy
dy
2
dx
dx
Putting this value of ( y – β) in (i) (To eliminate β), we have
x2
x2 +
2 = 9
 dy 
 dx 


or
xy
2
 dy 
L.C.M. = 
 . Multiplying both sides by this L.C.M.,
 dx 
 dy 
x2 

 dx 
2
2
 dy 
+ x2 = 9 

 dx 
2
2
2
 dy 
 dy 
 dy 
2
2
⇒ x2 
(x2 – 9) 
 – 9  dx  + x = 0 or
 +x =0
dx
 
 
 dx 
which is the required differential equation.
11. Which of the following differential equation has y = c1 ex
+ c2 e– x as the general solution?
d2 y
d2 y
(A)
(B)
– y = 0
2 + y = 0
dx
dx 2
MathonGo
15
Class 12
Chapter 9 - Differential Equations
d2 y
d2 y
– 1 = 0
(D)
2 + 1 = 0
dx
dx 2
x
–x
Sol. Given: y = c1 e + c2 e
dy
= c1 ex + c2 e– x (– 1) = c1 ex – c2 e– x
∴
dx
(C)
∴
d2 y
= c1 ex – c2 e– x (– 1) = c1 ex + c2 e– x
dx 2
or
d2 y
= y
dx 2
...(i)
[By (i)]
d2 y
– y = 0 which is differential equation given in option (B)
dx 2
∴ Option (B) is the correct answer.
12. Which of the following differential equations has y = x as
one of its particular solutions?
dy
dy
d2 y
d2 y
2
(A)
+ xy = x
(B)
+ x
+ xy = x
2 – x
dx
dx
dx
dx 2
2
2
dy
dy
d y
d y
+ xy = 0
(D)
(C)
– x2
+ x
+ xy = 0
dx
dx
dx 2
dx 2
Sol. Given: y = x
or
dy
d2 y
= 1 and
= 0
dx
dx 2
dy
d2 y
and
These values of y,
clearly satisfy the D.E. of option (C).
dx
dx 2
∴
dy
d2 y
2
+ xy
[... L.H.S. of D.E. of option (C) =
2 – x
dx
dx
2
2
2
= 0 – x (1) + x (x) = – x + x = 0 = R.H.S. of option (C)]
∴ Option (C) is the correct answer.
Exercise 9.4 (Page No. 395-397)
For each of the differential equations in Exercises 1 to 4, find
the general solution:
1 – cos x
dy
=
1 + cos x
dx
Sol. The given differential equation is
1 − cos x
1 − cos x
dy
=
or dy =
dx.
1 + cos x
1 + cos x
dx
x
2 sin 2
2
dy = ∫
Integrating both sides,
x dx
2 cos2
2
x
tan


2 x
2 x
2
− 1  dx =
or y = ∫ tan
dx = ∫  sec
– x+ c
2
1
2


2
1.
∫
MathonGo
16
Class 12
Chapter 9 - Differential Equations
Exercise 9.4
For each of the differential equations in Exercises 1 to 4, find
the general solution:
1 – cos x
dy
=
1 + cos x
dx
Sol. The given differential equation is
1 − cos x
1 − cos x
dy
=
or dy =
dx.
1 + cos x
1 + cos x
dx
x
2 sin 2
2
dy = ∫
Integrating both sides,
x dx
2 cos2
2
x
tan


2 x
2 x
2
− 1  dx =
or y = ∫ tan
dx = ∫  sec
– x+ c
2
1
2


2
1.
∫
MathonGo
17
Class 12
Chapter 9 - Differential Equations
x
– x + c
2
which is the required general solution.
dy
= 4 – y2 (– 2 < y < 2)
dx
dy
The given D.E. is
= 4 − y2
⇒ dy = 4 − y2 dx
dx
dy
= dx
Separating variables,
4 − y2
dy
Integrating both sides, ∫
dy =
1 dx
2
2 − y2

1
x
y
∵ ∫
dx = sin − 1 
∴ sin– 1
=x+c
a

2
a2 − x 2


y
⇒
= sin (x + c)
2
⇒
y = 2 sin (x + c) which is the required general solution.
dy
+ y = 1 ( y ≠ 1)
dx
dy
The given differential equation is
+ y = 1
dx
dy
⇒
= 1 – y
⇒ dy = (1 – y) dx ⇒ dy = – ( y – 1) dx
dx
or y = 2 tan
2.
Sol.
∫
3.
Sol.
Separating variables,
Integrating both sides,
dy
y − 1 = – dx
∫
dy
y−1 = –
∫1
dx
⇒ log | y – 1 | = – x + c
⇒
| y – 1 | = e– x + c
[... If log x = t, then x = et]
– x + c
⇒ y = 1 ± e– x ec
⇒
y – 1 = ± e
c – x
⇒
y= 1 ±e e
⇒
y = 1 + Ae– x where A = ± ec
which is the required general solution.
4. sec2 x tan y dx + sec2 y tan x dy = 0
Sol. The given differential equation is
sec2 x tan y dx + sec2 y tan x dy = 0
Dividing by tan x tan y, we have
sec 2 x
sec 2 y
dx +
dy = 0
tan x
tan y
Integrating both sides,
∫
sec 2 x
dx +
tan x
MathonGo
(Variables separated)
∫
sec 2 y
dy = log c
tan y
18
Class 12
Chapter 9 - Differential Equations
or log | tan x | + log | tan y | = log c

∵


f ′(x)
dx = log| f (x)|
f (x)

∫
or log | (tan x tan y) | = log c
or
| tan x tan y | = c
∴ tan x tan y = ± c = C where C = ± c.
[... | t | = a(a ≥ 0) ⇒ t = ± a]
which is the required general solution.
For each of the differential equations in Exercises 5 to 7,
find the general solution:
5. (ex + e– x) dy – (ex – e– x) dx = 0
Sol. The given D.E. is (ex + e– x) dy = (ex – e– x) dx
 e x − e− x 
or dy =  x
−x 
 dx
e +e 
∫
Integrating both sides,
dy =
 e x − e− x 
∫  e x + e− x 


∵

or y = log | ex + e– x | + c
dx

∫

f ′(x)
dx = log| f (x)|
f (x)

which is the required general solution.
dy
= (1 + x2)(1 + y2)
6.
dx
dy
Sol. The given differential equation is
= (1 + x2)(1 + y2)
dx
⇒ dy = (1 + x2)(1 + y2) dx
dy
Separating variables,
= (1 + x2) dx
1 + y2
Integrating both sides,
1
x3
2
⇒
tan– 1 y =
+ x + c
∫ y2 + 1 dy = ∫ ( x + 1) dx
3
which is the required general solution.
7. y log y dx – x dy = 0
Sol. The given differential equation is y log y dx – x dy = 0
⇒ – x dy = – y log y dx
Separating variables,
∫
Integrating both sides
dy
dx
y log y = x
dy
dy
y log y = ∫ x
...(i)
For integral on left hand side, put log y = t.
∴
dt
1
y = dy
⇒
dy
= dt
y
dt
dx
= ∫
t
x
⇒ log | t | = log | x | + log | c |*
= log | xc |
∴ Eqn. (i) becomes
∫
MathonGo
...(ii)
19
Class 12
Chapter 9 - Differential Equations
⇒
| t | = | xc |
⇒
t = ± xc
[... | x | = | y | ⇒ x = ± y]
⇒
log y = ± xc = ax where a = ± c
∴ y = eax which is the required general solution.
For each of the differential equations in Exercises 8 to 10,
find the general solution:
dy
8. x5
= – y5
dx
dy
Sol. The given differential equation is x5
= – y5
dx
⇒
x5 dy = – y5 dx
dy
dx
= –
⇒ y– 5 dy = – x– 5 dx
Separating variables,
( y5 )
( x5 )
∫
Integrating both sides,
y− 5 dy = –
∫
x − 5 dx
y− 4
x− 4
= –
+ c
−4
−4
– 4
– 4
Multiplying by – 4,
y = – x – 4c
⇒ x– 4 + y– 4 = – 4c ⇒ x– 4 + y– 4 = C where C = – 4c
which is the required general solution.
9.
dy
= sin– 1 x
dx
dy
= sin– 1 x
dx
Sol. The given differential equation is
or
dy = sin– 1 x dx
∫1
Integrating both sides,
or
y =
∫ sin
−1
y = (sin– 1 x)
−1
x dx
II
∫1
= x sin– 1 x –
∫
∫ sin
x . 1 dx
I
Applying product rule,
To evaluate
dy =
x
2
dx –
∫
∫
d
(sin– 1 x)
dx
1
1 − x2
dx = –
1
2
∫1
dx dx
x dx
∫
− 2x
1−x
1 − x2
2
Put 1 – x = t. Differentiate – 2x dx = dt
...(i)
dx
*Remark. To explain * in eqn. (ii)
If all the terms in the solution of a D.E. involve logs, it is better to use
log c or log | c | instead of c in the solution.
MathonGo
20
Class 12
∴
Chapter 9 - Differential Equations
x
∫
1−x
2
1
2
dx = –
∫
dt
1
= –
t
2
∫t
− 1/ 2
dt
1 t1 / 2
= – t = – 1 − x2
2 1/2
x
Putting this value of ∫
dx in (i), the required general
1 − x2
solution is
= –
1 − x2 + c.
10. ex tan y dx + (1 – ex) sec2 y dy = 0
Sol. The given equation is ex tan y dx + (1 – ex) sec2 y dy = 0
Dividing every term by (1 – ex) tan y, we have
y = x sin– 1 x +
ex
sec 2 y
dy = 0
x dx +
tan y
1−e
Integrating both sides,
or
–
∫
ex
dx +
1 − ex
∫
(Variables separated)
∫
sec 2 y
dy = c
tan y
− ex
dx + log | tan y | = c
1 − ex


f ′(x)
dx = log| f (x)|
– log | 1 – ex | + log | tan y | = c ∵ ∫
f (x)


|tan y|
or log
= log c′
(See Remark at the end of page 612)
|1 − e x |
or
|tan y|
= c′
|1 − e x |
or
tan y = C (1 – ex). [... | t | = c′
⇒ t = ± c′ = C (say)]
For each of the differential equations in Exercises 11 to 12,
find a particular solution satisfying the given condition:
or
11. (x3 + x2 + x + 1)
dy
= 2x2 + x, y = 1, when x = 0
dx
Sol. The given differential equation is (x3 + x2 + x + 1)
dy
= 2x2 + x
dx
∴ (x3 + x2 + x + 1) dy = (2x2 + x) dx
Separating variables dy =
(2 x2 + x)
dx
x3 + x2 + x + 1
or
2 x2 + x
dx
( x + 1)( x2 + 1)
dy =
[... x3 + x2 + x + 1 = x2(x + 1) + (x + 1) = (x + 1)(x2 + 1)]
Integrating both sides, we have
MathonGo
21
Class 12
∫
Chapter 9 - Differential Equations
1 dy =
Let
∫
2 x2 + x
dx
( x + 1)( x2 + 1)
or
y=
∫
2 x2 + x
dx ...(i)
( x + 1)( x2 + 1)
Bx + C
2 x2 + x
A
=
+ 2
(Partial fractions)
x +1
x+1
( x + 1)( x2 + 1)
...(ii)
2
Multiplying both sides by L.C.M. = (x + 1)(x + 1), we have
2x2 + x = A(x2 + 1) + (Bx + C)(x + 1)
or
2x2 + x = Ax2 + A + Bx2 + Bx + Cx + C
Comparing coeff. of x2 on both sides, we have
A + B = 2
Comparing coeff. of x on both sides, we have
B + C = 1
Comparing constants A + C = 0
Let us solve eqns. (iii), (iv) and (v) for A, B, C
eqn. (iii) – eqn. (iv) gives to eliminate B,
A – C = 1
1
Adding (v) and (vi), 2A = 1 or A =
2
1
From (v),
C = – A = –
2
1
1
1
3
Putting C = –
in (iv), B –
= 1 or B = 1 +
=
2
2
2
2
Putting these values of A, B, C in (ii), we have
1
3
...(iii)
...(iv)
...(v)
...(vi)
1
x−
2 x2 + x
2
+ 22 2
=
2
x +1
( x + 1)( x + 1)
x +1
x
1
1 1
3
1
=
+
–
. 2
2
x +1
2 x+1
2
2 x +1
2x
1
1 1
3
1
=
+
. 2
–
2
x +1
2 x+1
4
2 x +1
Putting this value in (i)
2x
1
1
1
3
1
dx +
dx –
dx
∫
∫
2
2
∫
+
+1
+
x
1
x
x
1
2
4
2
1
3
1
y =
log (x + 1) +
log (x2 + 1) –
tan– 1 x + c
...(vii)
2
4
2


2x
f ′(x)
dx = ∫
dx = log f (x) 
∵ ∫ 2
f (x)
x +1


To find c
When x = 0, y = 1 (given)
Putting x = 0 and y = 1 in (vii),
y =
1 =
1
3
1
log 1 +
log 1 –
tan– 1 0 + c
2
4
2
MathonGo
22
Class 12
Chapter 9 - Differential Equations
or
1 = c
[... log 1 = 0 and tan– 1 0 = 0]
Putting c = 1 in eqn. (vii), the required solution is
1
3
1
log (x2 + 1) –
y = 2 log (x + 1) +
tan– 1 x + 1.
4
2
1
1
y =
[2 log (x + 1) + 3 log (x2 + 1)] –
tan– 1 x + 1
4
2
1
1
=
[log (x + 1)2 + log (x2 + 1)3] –
tan– 1 x + 1
4
2
1
1
[log (x + 1)2 (x2 + 1)3] –
=
tan– 1 x + 1
4
2
which is the required particular solution.
dy
= 1; y = 0 when x = 2.
12. x (x2 – 1)
dx
dy
= 1
Sol. The given differential equation is x(x2 – 1)
dx
dx
⇒ x(x2 – 1) dy = dx
⇒ dy =
x( x2 − 1)
1
Integrating both sides, ∫ 1 dy = ∫
dx
2
x( x − 1)
1
dx + c
x( x + 1)( x − 1)
A
1
B
C
Let the integrand
=
+
+
x
x( x + 1)( x − 1)
x+1
x −1
⇒ y=
∫
...(i)
...(ii)
(By Partial Fractions)
Multiplying by L.C.M. = x(x + 1)(x – 1),
1 = A(x + 1)(x – 1) + Bx(x – 1) + Cx(x + 1)
or 1 = A(x2 – 1) + B(x2 – x) + C(x2 + x)
or 1 = Ax2 – A + Bx2 – Bx + Cx2 + Cx
Comparing coefficients of x2, x and constant terms on both sides,
we have
x2:
A + B + C = 0
...(iii)
x:
– B + C = 0
⇒ C = B
...(iv)
Constants – A = 1
or A = – 1
Putting A = – 1 and C = B from (iv) in (iii),
– 1 + B + B = 0
∴ From (iv), C = B =
or
2B = 1
⇒
B =
1
2
1
2
Putting these values of A, B, C in (ii),
1
1
1
−1
2
2
x( x + 1)( x − 1) = x + x + 1 + x − 1
MathonGo
23
Class 12
∴
Chapter 9 - Differential Equations
∫
1
x( x + 1)( x − 1) dx = –
= – log | x | +
∫
1
1
dx +
2
x
∫
1
1
x + 1 dx + 2
∫
1
x − 1 dx
1
1
log | x + 1 | +
log | x – 1 |
2
2
1
[– 2 log | x | + log | x + 1 | + log | x – 1 |]
2
1
=
[– log | x |2 + log | (x + 1)(x – 1) |]
2
=
⇒
∫
| x 2 − 1| 1
| x 2 − 1|
1
1 
= log
 log
dx =
2 
x( x + 1)( x − 1)
x2
| x|  2
2 
Putting this value in (i),
y =
x2 − 1
x2
1
log
2
+ c
...(v)
To find c for the particular solution
Putting y = 0, when x = 2 (given) in (v),
−1
1
3
3
log
+ c
⇒ c =
log
2
4
2
4
Putting this value of c in (v), the required particular solution is
0 =
y =
x2 − 1
x2
1
log
2
To evaluate
∫
1
1
3
log
2
4
OR
–
dx =
2
∫
1
x
dx =
2
2
x ( x − 1)
2
∫
2x
dx
x2 ( x2 − 1)
x( x − 1)
Put
x2 = t.
For each of the differential equations in Exercises 13 to 14,
find a particular solution satisfying the given condition:
 dy 
13. cos 
 = a (a ∈ R); y = 1 when x = 0
 dx 
Sol. The given differential equation is
dy
cos
= a (a ∈ R); y = 1 when x = 0
dx
dy
∴
= cos– 1 a
⇒ dy = (cos– 1 a) dx
dx
Integrating both sides
∫1
dy = ∫ (cos a) dx
⇒ y = (cos– 1 a) ∫ 1 dx
– 1
⇒
y = (cos a) x + c
To find c for particular solution
y = 1 when x = 0 (given)
∴ From (i), 1 = c.
Putting c = 1 in (i), y = x cos– 1 a + 1
y −1
⇒ y – 1 = x cos– 1 a
⇒
= cos– 1 a
x
−1
MathonGo
...(i)
24
Class 12
Chapter 9 - Differential Equations
 y −1
⇒ cos 
 = a which is the required particular solution.
 x 
dy
= y tan x; y = 1 when x = 0
14.
dx
dy
Sol. The given differential equation is
= y tan x
dx
⇒ dy = y tan x dx
dy
y = tan x dx
Separating variables,
Integrating both sides
∫
1
y dy =
∫
tan x dx
⇒
log | y | = log | sec x | + log | c |
⇒
log | y | = log | c sec x |
⇒
| y | = | c sec x |
∴
y = ± c sec x
or
y = C sec x
...(i)
where C = ± c
To find C for particular solution
Putting y = 1 and x = 0 in (i), 1 = C sec 0 = C
Putting C = 1 in (i), the required particular solution is y = sec x.
15. Find the equation of a curve passing through the point (0, 0)
and whose differential equation is y′′ = ex sin x.
Sol. The given differential equation is y′ = ex sin x
⇒
dy
= ex sin x
dx
⇒ dy = ex sin x dx
∫1
Integrating both sides,
or
y = I + C
where I =
∫
dy =
∫
e x sin x dx
...(i)
x
e sin x dx
I
II

 Applying Product Rule

...(ii)
 d


∫ I . II dx = I ∫ II dx − ∫  dx (I) ∫ II dx  dx 
∫
= ex (– cos x) –
e x (− cos x) dx
∫
e x cos x dx
I
II
Again applying product rule,
⇒
I = – ex cos x +
x
I = – ex cos x + ex sin x – ∫ e sin x dx
x
⇒
I = e (– cos x + sin x) – I
Transposing 2I = ex (sin x – cos x)
[By (ii)]
ex
(sin x – cos x)
2
Putting this value of I in (i), the required solution is
∴
I =
MathonGo
25
Class 12
Chapter 9 - Differential Equations
1 x
e (sin x – cos x) + c
...(iii)
2
To find c. Given that required curve (i) passes through the point
(0, 0).
Putting x = 0 and y = 0 in (iii),
−1
1
1
0 =
(– 1) + c or 0 =
+ c ∴ c =
2
2
2
1
in (iii), the required equation of the curve is
Putting c =
2
1
1 x
y =
e (sin x – cos x) +
2
2
L.C.M. = 2 ∴ 2y = ex (sin x – cos x) + 1 or 2y – 1 = ex(sin x – cos x)
which is the required equation of the curve.
dy
16. For the differential equation xy
= (x + 2)( y + 2), find
dx
the solution curve passing through the point (1, – 1).
dy
= (x + 2)( y + 2)
Sol. The given differential equation is xy
dx
⇒
xy dy = (x + 2)( y + 2) dx
y =
Separating variables
∫
Integrating both sides,
∫
⇒
y+2−2
y + 2 dy =
 y+2
2 
∫  y + 2 − y + 2 
⇒
⇒
y
x+2
y + 2 dy = x dx
y
x+2
y + 2 dy = ∫ x dx

2 
∫  1 − y + 2 
dy =

dy =
2
∫  1 + x 
x
2

2
∫  x + x 
∫  1 + x 
dx
dx
dx
⇒ y – 2 log | y + 2 | = x + 2 log | x | + c
⇒ y – x = log ( y + 2)2 + log x2 + c
| ... | x |2 = x2
⇒ y – x = log (( y + 2)2 x2) + c
...(i)
To find c. Curve (i) passes through the point (1, – 1).
Putting x = 1 and y = – 1 in (i), – 1 – 1 = log (1) + c
or – 2 = c
(... log 1 = 0)
Putting c = – 2 in (i), the particular solution curve is
y – x = log (( y + 2)2 x2) – 2
or y – x + 2 = log (( y + 2)2 x2).
17. Find the equation of the 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.
MathonGo
26
Class 12
Chapter 9 - Differential Equations
Sol. Let P(x, y) be any point on the required curve.
According to the question,
(Slope of the tangent to the curve at P(x, y)) × y = x
⇒
dy
. y = x ⇒ y dy = x dx
dx
Now variables are separated.
Integrating both sides
∫
y dy =
∫
x dx
∴
y2
x2
=
+ c
2
2
Multiplying by L.C.M. = 2, y2 = x2 + 2c
or
y2 = x2 + A
...(i)
where A = 2c.
Given: Curve (i) passes through the point (0, – 2).
Putting x = 0 and y = – 2 in (i), 4 = A.
Putting A = 4 in (i), equation of required curve is
y2 = x2 + 4 or y2 – x2 = 4.
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).
Sol. According to question, slope of
the tangent at any point P(x, y)
of the required curve.
= 2 . (Slope of the line
joining the point of
contact P(x, y) to the
given point A(– 4, – 3)).
 y − (− 3) 
y2 − y1
dy

= 2 
x2 − x1
dx
 x − (− 4) 
2( y + 3)
dy
=
⇒
( x + 4)
dx
Cross-multiplying, (x + 4) dy = 2( y + 3) dx
⇒
Separating variables,
Integrating both sides,
dy
2
y + 3 = x + 4 dx
∫
1
y + 3 dy = 2
∫
1
x + 4 dx
⇒ log | y + 3 | = 2 log | x + 4 | + log | c |
(For log | c |, see Foot Note page 612)
⇒ log | y + 3 | = log | x + 4 |2 + log | c | = log | c | (x + 4)2
⇒
| y + 3 | = | c | (x + 4)2
⇒
y + 3 = ± | c | (x + 4)2
⇒
y + 3 = C(x + 4)2
...(i) where C = ± | c |
MathonGo
27
Class 12
Chapter 9 - Differential Equations
To find C. Given that curve (i) passes through the point (– 2, 1).
Putting x = – 2 and y = 1 in (i),
1 + 3 = C(– 2 + 4)2 or 4 = 4C
⇒
C =
4
= 1.
4
Putting C = 1 in (i), equation of required curve is
y + 3 = (x + 4)2 or (x + 4)2 = y + 3.
19. The volume of a 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.
Sol. Let x be the radius of the spherical balloon at time t.
Given: Rate of change of volume of spherical balloon is constant
= k (say)
d  4π 3 
4π
dx
dx
x  = k ⇒
⇒
3x2
= k ⇒ 4πx2
= k

dt  3
3
dt
dt

2
Separating variables,
4πx dx = k dt
Integrating both sides,
4π
∫
x 2 dx = k
∫1
dt
x3
= kt + c
3
To find c: Given: Initially radius is 3 units.
⇒ When t = 0, x = 3
Putting t = 0 and x = 3 in (i), we have
⇒ 4π
...(i)
4π
(27) = c or c = 36π
3
...(ii)
To find k: Given: When t = 3 sec, x = 6 units
Putting t = 3 and x = 6 in (i),
4π
(6)3 = 3k + c.
3
4π
(216) = 3k + 36π
3
or 4π (72) – 36π = 3k
⇒ 288π – 36π = 3k
or
3k = 252π
⇒ k = 84π
...(iii)
Putting values of c and k from (ii) and (iii) in (i), we have
4π 3
x = 84πt + 36π
3
Putting c = 36π from (ii),
x3
= 21t + 9
3
3
⇒ x = 63t + 27
⇒ x = (63t + 27)1/3.
20. In a bank principal increases at the rate of r % per year.
Find the value of r if ` 100 double itself in 10 years.
(loge 2 = 0.6931)
Dividing both sides by 4π,
MathonGo
28
Class 12
Chapter 9 - Differential Equations
Sol. Let P be the principal (amount) at the end of t years.
According to given, rate of increase of principal per year
= r% (of the principal)
dP
r
=
× P
⇒
dt
100
dP
r
Separating variables,
=
dt
P
100
r
Integrating both sides, log P =
t + c
...(i)
100
(Clearly P being principal is > 0, and hence log | P | = log P)
To find c. Initial principal = ` 100 (given)
i.e., When t = 0, P = 100
Putting t = 0 and P = 100 in (i), log 100 = c.
r
t + log 100
Putting c = log 100 in (i), log P =
100
r
r
P
⇒ log P – log 100 =
=
t
...(ii)
t ⇒ log
100
100
100
Putting P = double of itself = 2 × 100 = ` 200
When t = 10 years (given) in (ii),
200
r
r
=
× 10
⇒ log 2 =
100
100
10
⇒ r = 10 log 2 = 10 (0.6931) = 6.931% (given).
21. In a bank, principal increases at the rate of 5% per year. An
amount of ` 1000 is deposited with this bank, how much
will it worth after 10 years (e0.5 = 1.648).
Sol. Let P be the principal (amount) at the end of t years.
According to given rate of increase of principal per year
= 5% (of the principal)
dP
5
dP
P
⇒
=
× P
⇒
=
dt
100
dt
20
⇒ 20 dP = P dt
dP
dt
Separating variables,
=
20
P
Integrating both sides, we have
log
1
t+ c
...(i)
20
To find c. Given: Initial principal deposited with the bank is
` 1000.
⇒ When t = 0, P = 1000
Putting t = 0 and P = 1000 in (i), we have log 1000 = c
log P =
t
+ log 1000
20
t
P
t
⇒ log P – log 1000 =
⇒ log
=
20
1000
20
Putting t = 10 years (given), we have
Putting c = log 1000 in (i), log P =
MathonGo
29
Class 12
Chapter 9 - Differential Equations
P
=
1000
P
⇒
=
1000
⇒
P =
log
10
1
= 0.5
=
20
2
e0.5
1000 e0.5
[... If log x = t, then x = et]
= 1000 (1.648)
[... e0.8 = 1.648 (given)]
 1648 
= 1000 
 = ` 1648.
 1000 
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.
Sol. Let x be the bacteria present in the culture at time t hours.
According to given,
Rate of growth of bacteria is proportional to the number present.
i.e.,
∴
dx
is proportional to x.
dt
dx
= kx where k is the constant of proportionality (k > 0
dt
because rate of growth (i.e., increase) of bacteria is given.)
dx
⇒ dx = kx dt
⇒
= k dt
x
1
Integrating both sides, ∫
dx = k ∫ 1 dt
x
⇒
log x = kt + c
...(i)
To find c. Given: Initially the bacteria count is x 0 (say) =
1,00,000.
⇒ When t = 0, x = x0.
Putting these value in (i), log x0 = c.
Putting c = log x0 in (i), log x = kt + log x0
x
⇒ log x – log x0 = kt
⇒ log x = kt
...(ii)
0
To find k: According to given, the number of bacteria is increased
by 10% in 2 hours.
10
∴ Increase in bacteria in 2 hours =
× 1,00,000 = 10,000
100
∴ x, the amount of bacteria at t = 2
= 1,00,000 + 10,000 = 1,10,000 = x1 (say)
Putting x = x1 and t = 2 in (ii),
x1
x1
1
log x = 2k
⇒ k =
log x
2
0
0
⇒ k =
1,10,000
1
1
11
log
1,00,000 = 2 log 10
2
MathonGo
30
Class 12
Chapter 9 - Differential Equations
x
11 
1 
Putting this value of k in (ii), we have log x =
 log 10  t
2 

0
When x = 2,00,000 (given);
2,00,000
11 
1
then log
=  log
t
1,00,000
10 
2
⇒ log 2 =
1
 11 
log   t
2
 10 
2 log 2
11 

Cross-multiplying 2 log 2 =  log
t ⇒ t=
hours.
10 

11 

 log

10 
23. The general solution of the differential
dy
equation
= ex + y is
dx
(A) ex + e– y = c (B) ex + ey = c (C) e– x + ey = c (D) e– x + e–y = c
Sol. The given D.E. is
⇒
dy
= ex . ey
dx
dy
= ex + y
dx
⇒ dy = ex . ey dx
Separating variables,
dy
= ex dx
(e y )
Integrating both sides
∫
e− y dy =
or
∫
e– y dy = ex dx
e x dx
e− y
= ex + c ⇒ – e– y – ex = c
−1
Dividing by – 1, e– y + ex = – c
or ex + e– y = C where C = – c which is the required solution.
∴ Option (A) is the correct answer.
⇒
MathonGo
31
Class 12
Chapter 9 - Differential Equations
Exercise 9.5
In each of the Exercises 1 to 5, show that the given differential
equation is homogeneous and solve each of them:
1. (x2 + xy) dy = (x2 + y2) dx
Sol. The given D.E. is
(x2 + xy) dy = (x2 + y2) dx
...(i)
This D.E. looks to be homogeneous as degree of each coefficient of
dx and dy is same throughout (here 2).

y2 
x2  1 + 2 
x 

x2 + y2
dy
From (i),
= 2
=
dx
y
x + xy

x2  1 + 
x

2
 y
1+ 
dy
 y
 x
or
=
= F 
...(ii)
dx
 x
 y
1+ 
 x
MathonGo
32
Class 12
Chapter 9 - Differential Equations
∴ The given D.E. is homogeneous.
Put
y
= v. Therefore y = vx.
x
dy
dv
= v . 1 + x
= v + x
dx
dx
y
and
Putting these values of
x
∴
v + x
dv
dx
dy
in (ii), we have
dx
1 + v2
dv
=
dx
1+v
Transposing v to R.H.S., x
dv
1 + v2
=
– v
dx
1+v
1−v
dv
1 + v2 − v − v2
=
=
1+v
dx
1+v
Cross-multiplying x(1 + v) dv = (1 – v) dx
1+v
dx
Separating variables
dv =
1−v
x
1+v
1
Integrating both sides ∫
dv = ∫ dx
1−v
x
1+1−1+ v
2 − (1 − v)
⇒ ∫
dv = log x + c ⇒ ∫
dv = log x + c
1−v
1−v
⇒
⇒
⇒
x
 2

∫  1 − v − 1 
y
Put v =
,
x
dv = log x + c ⇒
– 2 log (1 – v) – v = log x + c
y
y

– 2 log  1 −  –
= log x + c
x
x


Dividing by – 1,
 x − y
⇒ log 

 x 
2 log (1 − v)
– v = log x + c
−1
y
 x − y
2 log 
= – log x – c
 +
x
x


2
+ log x = –
y
 ( x − y)2 x 
y
y
 = –
– c ⇒ log 
–c
x2
x
x


y
y
− −c
−
−
( x − y)2
= e x
= e x e– c ⇒ (x – y)2 = Cx e x where C = e–
x
which is the required solution.
⇒
2. y′′ =
c
x+y
x
Sol. The given differential equation is y′ =
x+ y
x
dy
x
y
dy
y
 y
=
+
⇒
= 1 +
= f  
dx
x
x
dx
x
 x
∴ Differential equation (i) is homogeneous.
⇒
MathonGo
...(i)
33
Class 12
Chapter 9 - Differential Equations
y
= v
x
Put
∴
y = vx
dy
dv
dv
= v . 1 + x
= v + x
dx
dx
dx
dy
and y in (i),
Putting these values of
dx
dv
dv
v + x
= 1 + v
⇒
x
= 1
⇒ x dv = dx
dx
dx
dx
Separating variables, dv =
x
dx
Integrating both sides, ∫ 1 dv = ∫
v = log | x | + c
x
y y
Putting v =
,
= log | x | + c
∴
y = x log | x | + cx
x x
which is the required solution.
3. (x – y) dy – (x + y) dx = 0
Sol. The given differential equation is
...(i)
(x – y) dy – (x + y) dx = 0
Differential equation (i) looks to be homogeneous because each
coefficient of dx and dy is of degree 1.
From (i), (x – y) dy = (x + y) dx
y
y

1+
x 1 + 
x
x+ y
dy
dy
 y
x

 or
∴
=
=
=
y = f   ...(ii)
x− y
dx
dx
−
1
 x
y

x
x 1 − 
∴

x
∴ Differential equation (i) is homogeneous.
Put
∴
y
= v
∴ y = vx
x
dy
dv
dv
= v . 1 + x
= v + x
dx
dx
dx
Putting these values in (ii), v + x
Shifting v to R.H.S., x
1+v
dv
=
1−v
dx
1+v
dv
1 + v − v + v2
=
– v =
1−v
dx
1−v
1 + v2
dv
=
dx
1−v
Cross-multiplying,
⇒ x
x (1 – v) dv = (1 + v2) dx
(1 − v)
dx
Separating variables,
dv =
1 + v2
x
1−v
1
Integrating both sides, ∫
dv = ∫
dx + c
1 + v2
x
MathonGo
34
Class 12
⇒
∫
Chapter 9 - Differential Equations
1
dv –
1 + v2
⇒ tan– 1 v –
1
2
∫
∫
v
1
dv = ∫
dx + c
1 + v2
x
2v
dv = log x + c
1 + v2


f ′(v)
∵ ∫ f (v) dv = log f (v)


2

y
y
y
1
Putting v =
, tan– 1
–
log  1 + 2  = log x + c
x
x
x
2


 x 2 + y2 
y
1
 = log x + c
⇒
tan– 1
–
log 
2
x
2
 x

y
1
⇒
tan– 1
–
[log (x2 + y2) – log x2] = log x + c
x
2
y
1
1
–
log (x2 + y2) +
⇒ tan– 1
2 log x = log x + c
x
2
2
y
1
y
1
⇒ tan– 1
–
log (x2 + y2) = c ⇒ tan– 1
=
log (x2 + y2) + c
x
2
x
2
which is the required solution.
4. (x2 – y2) dx + 2 xy dy = 0
Sol. The given differential equation is
(x2 – y2) dx + 2xy dy = 0
...(i)
This differential equation looks to be homogeneous because degree
of each coefficient of dx and dy is same (here 2).
From (i), 2xy dy = – (x2 – y2) dx
⇒ tan– 1 v –
1
log (1 + v2) = log x + c
2
y2 − x 2
− ( x 2 − y2 )
dy
=
=
2 xy
dx
2 xy
Dividing every term in the numerator and denominator of
R.H.S. by x2,
⇒
2
dy
=
dx
 y
 x −1
 
y
2
x
 y
= f  
 x
...(ii)
∴ The given differential equation is homogeneous.
y
dy
dv
dv
Put
= v. Therefore y = vx ∴
=v . 1 + x
= v+ x
x
dx
dx
dx
y
dy
Putting these values of
in differential equation (ii),
and
x
dx
we have
v + x
⇒
dv
dv
v2 − 1
v2 − 1
v2 − 1 − 2v2
=
⇒
x
=
– v =
dx
dx
2v
2v
2v
2
2
dv
− v −1
(v + 1)
x
=
∴ x 2v dv = – (v2 + 1) dx
=–
dx
2v
2v
MathonGo
35
Class 12
Chapter 9 - Differential Equations
2v dv
dx
= –
v2 + 1
x
⇒
2v
dv = –
v2 + 1
⇒
log (v2 + 1) = – log x + log c
⇒ log (v2 + 1) + log x = log c
⇒
log (v2 + 1) x = log c
⇒
(v2 + 1) x = c
Integrating both sides,
Put v =
∫
2

y y
,  2 + 1  x = c
x x

or
y2 + x 2
=c
x
or
or
∫
1
dx
x
 y2 + x 2 

 x = c
2
 x

x2 + y2 = cx
which is the required solution.
 dy 
5. x2 
 = x2 – 2y2 + xy
 dx 
dy
= x2 – 2y2 + xy
dx
The given differential equation looks to be Homogeneous as all
terms in x and y are of same degree (here 2).
Sol. The given differential equation is x2
Dividing by x2,
dy
xy
x2
2 y2
+ 2
= 2 –
dx
x
x
x2
or
dy
 y
= 1 – 2  
dx
 x
2
 y
+  
 x
...(i)
 y
= F 
 x
∴ Differential equation (i) is homogeneous.
So put
∴
y
= v
x
∴ y = vx
dy
dv
dv
= v . 1 + x
= v+ x
dx
dx
dx
Putting these values of
v + x
y
dy
and
in (i),
x
dx
dv
dv
= 1 – 2v2 + v or x
= 1 – 2v2 ⇒ x dv = (1 – 2v2) dx
dx
dx
Separating variables,
dv
dx
=
1 − 2v2
x
MathonGo
36
Class 12
Chapter 9 - Differential Equations
Integrating both sides,
⇒
∫
1
dv =
1 − ( 2v)2
2
∫
1
dx
x
1 + 2v
1 − 2v
= log | x | + c
2 → Coefficient of v
log
1
2 .1

∵

1
1
a+x 
dx =
log

2
2a
a–x 
a –x
y
1+ 2
x
1
y
y = log | x | + c
,
Putting v =
log
x
1− 2
2 2
x
Multiplying within logs by x in L.H.S.,
∫
2
1
log
2 2
x + 2y
= log | x | + c.
x − 2y
In each of the Exercises 6 to 10, show that the given D.E. is
homogeneous and solve each of them:
6. x dy – y dx = x 2 + y 2 dx
Sol. The given differential equation is
x 2 + y2 dx or x dy = y dx +
Dividing by dx
x dy – y dx =
x
dy
= y +
dx
x 2 + y2
or x
x 2 + y2 . dx
dy
= y + x
dx
 y
1+ 
 x
2
2
 y
= F 
 x
∴ Given differential equation is homogeneous.
Dividing by x,
dy
y
=
+
dx
x
 y
1+ 
 x
y
= v i.e., y = vx.
x
dy
Differentiating w.r.t. x,
= v
dx
y
Putting these values of
and
x
dv
v + x
= v + 1 + v2
dx
...(i)
Put
∴
x dv =
1 + v2 dx
Integrating both sides,
∫
+ x
dv
dx
dy
in (i), it becomes
dx
dv
or
x
= 1 + v2
dx
dx
dv
or
=
x
2
1+v
dv
2
1+v
MathonGo
=
∫
dx
x
37
Class 12
∴
Chapter 9 - Differential Equations
log (v +
Replacing v by
1 + v2 ) = log x + log c
y
, we have
x
y
y2 
log  + 1 + 2  = log cx or
x
x 

or
y+
= cx2
x 2 + y2
which is the required solution.
y + x2 + y2
= cx
x


 y
 y 
 y
 y 
7.  x cos   + y sin    y dx =  y sin   – x cos    x dy
 x
 x 
 x
 x 


Sol. The given D.E. is


 y
 y 
 y
 y 
 x cos   + y sin    y dx =  y sin   − x cos    x dy
 x
 x 
 x
 x 


y
y

y
y
xy cos + y2 sin
 x cos x + y sin x  y
dy


x
x
=
∴
=
y
y
y
y

dx
xy sin − x 2 cos
 y sin x − x cos x  x
x
x


Dividing every term in R.H.S. by x2,
2
y
y  y
y
cos +   sin
dy
x
x  x
x
 y
=
= F 
y
y
y
dx
 x
sin − cos
x
x
x
∴ The given differential equation is homogeneous.
y
So let us put
= v. Therefore y = vx.
x
dy
dv
dv
= v . 1 + x
= v + x
∴
dx
dx
dx
Putting these values in differential equation (i), we have
v + x
v cos v + v2 sin v
dv
=
dx
v sin v − cos v
⇒
x
...(i)
dv
v cos v + v2 sin v
=
– v
dx
v sin v − cos v
2v cos v
v cos v + v2 sin v − v2 sin v + v cos v
dv
⇒ x
=
v
v − cos v
sin
v sin v − cos v
dx
Cross-multiplying, x(v sin v – cos v) dv = 2v cos v dx
v sin v − cos v
dx
Separating variables,
dv = 2
v cos v
x
v sin v − cos v
1
Integrating both sides, ∫
dv = 2 ∫
dx
v cos v
x
=
Using
a−b
a
b
=
– , ⇒
c
c
c
 v sin v
cos v 
∫  v cos v − v cos v 
MathonGo
dv = 2
∫
1
dx
x
38
Class 12
Chapter 9 - Differential Equations

1
1
dx
x
⇒ log | sec v | – log | v | = 2 log | x | + log | c |
⇒
∫  tan v − v 
⇒
log
dv = 2
sec v
v
∫
= log | x |2 + log | c |
= log (| c | x2)
sec v
sec v
= | c | x2
⇒
= ± | c | x2
v
v
⇒
sec v = ± | c | x2 v
y
y
y
Putting v =
, sec
= Cx2
where C = ± | c |
x
x
x
y
1
= Cxy
⇒
= Cxy
or
sec
x
y
cos
x
y
1
y
=
= C1 (say)
⇒ C xy cos
= 1
⇒ xy cos
C
x
x
which is the required solution.
⇒
 y
dy
– y + x sin   = 0
dx
x
dy
 y
Sol. The given D.E. is x
– y + x sin   = 0
dx
 x
8. x
or
x
 y
dy
= y – x sin  
 x
dx
Dividing every term by x,
y
dy
 y
 y
=
– sin   = F   ...(i)
x
x
dx
 
 x
 y
dy
= F   , the given differential equation is
dx
 x
homogeneous.
y
dy
dv
= v + x
Putting
= v i.e., y = vx so that
x
dx
dx
y
dy
Putting these values of
and
in (i), we have
x
dx
Since
dv
= v – sin v
dx
dv
x
= – sin v
dx
v + x
or
or
− dx
dv
=
x
sin v
∴
x dv = – sin v dx
or cosec v dv = –
dx
x
Integrating, log | cosec v – cot v | = – log | x | + log | c |
or
log | cosec v – cot v | = log
c
x
MathonGo
39
Class 12
Chapter 9 - Differential Equations
cosec v – cot v = ±
or
c
x
y
y
y
C
, cosec
– cot
=
where C = ± c
x
x
x
x
y
y
cos
1 − cos
1
C
C
x
x
=
⇒
–
=
y
y
y
x
x
sin
sin
sin
x
x
x
Replacing v by
⇒
y
y

Cross-multiplying, x  1 − cos  = C sin
which is the required
x
x

solution.
 y
9. y dx + x log   dy – 2x dy = 0
x
y

Sol. The given differential equation is y dx + x  log  dy
x

– 2x dy = 0
y
y


∴
y dx = 2x dy – x  log  dy or y dx = x  2 − log  dy
x
x


y
dy
 y
x
=
= F 
...(i)
∴
y
dx
 x
2 − log
x
dy
 y
Since
= F   , the given differential equation is
dx
 x
homogeneous.
y
dy
dv
Putting
= v + x
= v i.e., y = vx so that
x
dx
dx
y
dy
and
Putting these values of
in (i), we have
x
dx
dv
v
v + x
=
dx
2 − log v
or x
or
∴
or
or
− v + v log v
v
v − 2v + v log v
dv
– v =
=
=
2 − log v
2 − log v
2 − log v
dx
dv
v (log v − 1)
x
=
dx
2 − log v
x(2 – log v) dv = v (log v – 1) dx
2 − log v
1 − (log v − 1)
dx
dx
or
dv =
v (log v − 1) dv = x
v (log v − 1)
x

1
1
dx
−  dv =
v
x
 (log v − 1) v 
MathonGo
40
Class 12
Chapter 9 - Differential Equations
 1/ v
1
−  dv = log | x | + log | c |

 log v − 1 v 
or log | log v – 1 | – log | v | = log | x | + log | c |
Integrating
∫

∵

or
log
log v − 1
v
= log | cx |
or
∫

f ′(v)
dv = log| f (v)|
f (v)

log v − 1
v
= | cx |
log v − 1
= ± cx = Cx where C = ± c
v
or
log v – 1 = Cx v
y
Replacing v by
, we have
x
or
y
y
 y
– 1 = Cx  
or
log
– 1 = Cy
x
x
x
 
which is a primitive (solution) of the given differential equation.
Second solution
log
 y
The given D.E. is y dx + x log   dy – 2x dy = 0
 x
Dividing every term by dy,
y

dx
x
y
– x log – 2x = 0 ∵ log = log y – log x = – ( log x – log y ) = – log
dy
y
x

Dividing every term by y,
dx
x
x
x
–
log
– 2
=0
dy
y
y
y

 x 
dx
x
x
x
=
log
+ 2
... (i)  = F   
dy
y
y
y
 y 

∴ The given differential is homogeneous.
x
Put
= v i.e. x = vy
y
⇒
dx
dv
= v + y
dy
dy
Putting these values in D. E. (i), we have
so that
v + y
dv
= v log v + 2 v
dy
dv
= v log v + v = v (log v + 1)
dy
Cross-multiplying y dv = v (log v + 1) dy
⇒ y
MathonGo
41
x

y
Class 12
Chapter 9 - Differential Equations
dv
dy
Separating variables v (log v +1) = y
1
1
Integrating both sides ∫ v
d v = ∫ dy
y
log v +1

∴ log log v + 1 = log y + log c = log cy ∵

∴ log v + 1 = ± cy = Cy where C = ± c
Replacing v by
log
∫

f ′(v)
d v = log | f (v) |
f (v)

x
, we have
y
x
+ 1 = Cy
y
or – log
y
+ 1 = Cy
x


x
y
∵ log = – log see page 632
y
x


y
– 1 = –Cy or = C1y which is a primitive
x
(solution) of the given D.E.
Dividing by – 1, log

x
10. (1 + ex/y) dx + ex/y  1 –  dy = 0
y


x
Sol. The given differential equation is (1 + ex/y) dx + ex/y  1 −  dy = 0
y

Dividing by dy, (1 + ex/y)

x
dx
x/y 1 −


y = 0
dy + e 
dx
or (1 + ex/y)
= – ex/y
dy

x
 1 −  or
y

x

ex / y  − 1 
dx
y

=
dy
1 + ex / y
which is a differential equation of the form
...(i)
 x
dx
= f .
dy
 y
∴ The given differential equation is homogeneous.
x
= v i.e., x = νy
y
dx
dv
Differentiating w.r.t. y,
= v+ y
dy
dy
x
dx
Putting these values of
and
in (i), we have
y
dy
Hence put
MathonGo
42
Class 12
Chapter 9 - Differential Equations
v + y
ev (v − 1)
dv
=
.
dy
1 + ev
Now transposing v to R.H.S.
y
vev − ev
vev − ev − v − νev
− ev − v
dv
=
– v =
=
v
v
dy
1+ e
1+ e
1 + ev
∴ y (1 + ev) dv = – (ev + v) dy
dy
1 + ev
v dv = –
y
v+ e
or
Integrating, log | (v + ev) | = – log | y | + log | c |
Replacing v by
log
∴
x
y , we have
x
x/ y 
 +e 
y


= log
c
y
x
+ ex / y
y
or
=
c
y
C
x
+ ex/y = ±
y
y
Multiplying every term by y,
x + y ex/y = C where C = ± c
which is the required general solution.
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) dx = 0; y = 1 when x = 1
Sol. The given differential equation is
(x + y) dy + (x – y) dx = 0, y = 1 when x = 1
...(i)
It looks to be a homogeneous differential equation because each
coefficient of dx and dy is of same degree (here 1).
From (i), (x + y) dy = – (x – y) dx
y

x  − 1
dy
− ( x − y)
y− x
x

∴
=
=
=
dx
x+ y
y+ x
y

x  + 1
x

y
−1
dy
 y
or
= x
= f 
...(ii)
dx
y
 x
+1
x
∴ Given differential equation is homogeneous.
Put
∴
y
= v. Therefore y = vx.
x
dy
dv
dv
= v . 1 + x
= v + x
dx
dx
dx
Putting these values in eqn. (ii),
MathonGo
v + x
dv
v−1
=
dx
v+1
43
Class 12
Chapter 9 - Differential Equations
⇒
x
v − 1 − v(v + 1) v − 1 − v2 − v
− v2 − 1
dv
v−1
=
–v =
=
=
v+1
dx
v+1
v+1
v+1
⇒
x
dv
(v2 + 1)
= –
dx
v+1
Separating variables,
∫
∴
v
dv +
v2 + 1
∫
∴ x(v + 1) dv = – (v2 + 1) dx
v+1
dx
dv = –
v2 + 1
x
1
1
dv = – ∫
dx
v2 + 1
x
2v
dv + tan– 1 v = – log x + c
v2 + 1
⇒
1
2
⇒

1
log (v2 + 1) + tan–1 v = – log x + c ∵
2

∫
Putting v =
∫

f ′(v)
dv = log f (v)
f (v)

 y2

y 1
y
= – log x + c
,
log  2 + 1  + tan– 1
x 2
x
x

⇒
 y2 + x 2 
1
y
 + tan– 1
log 
= – log x + c
2
2
x
 x

⇒
1
y
[log (x2 + y2) – log x2] + tan– 1
= – log x + c
2
x
1
1
log (x2 + y2) –
2 log x + tan– 1
2
2
1
⇒
log (x2 + y2) + tan– 1
2
To find c: Given: y = 1 when x = 1.
⇒
Putting x = 1 and y = 1 in (iii),
y
= – log x + c
x
y
= c
x
...(iii)
1
log 2 + tan– 1 1 = c
2
π
π
π
1

−1
log 2 +
∵ tan 4 = 1 ⇒ tan 1 = 4 
2
4


Putting this value of c in (iii),
π
1
y
1
log (x2 + y2) + tan– 1
=
log 2 +
2
x
2
4
Multiplying by 2,
π
y
log (x2 + y2) + 2 tan– 1
= log 2 +
x
2
which is the required particular solution.
12. x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
Sol. The given differential equation is
x2 dy + (xy + y2) dx = 0 or x2 dy = – y (x + y) dx
y

yx  1 + 
dy
y( x + y)
x

∴
= –
= –
dx
x2
x2
or
c =
MathonGo
44
Class 12
Chapter 9 - Differential Equations
dy
 y
y 1 + y 
= –
...(i)

 = F x
x
dx

 
x 
∴ The given differential equation is homogeneous.
y
Put
= v, i.e., y = vx
x
dy
dv
Differentiating w.r.t. x,
= v + x
dx
dx
y
dy
Putting these values of
and
in differential equation (i),
x
dx
dv
we have v + x
= – v(1 + v) = – v – v2
dx
dv
Transposing v to R.H.S., x
= – v2 – 2v
dx
dv
or
x
= – v (v + 2)
x dv = – v (v + 2) dx
dx
dν
dx
or
= –
v(v + 2)
x
or
1
1
v (v + 2) dv = – ∫ x dx
2
(v + 2) − v
1
v (v + 2) dv = – log | x | or 2 ∫ v (v + 2) dv = – log | x |
∫
Integrating both sides,
1
2 ∫
Separating terms
or
1
1 
∫  v − v + 2 
or
dv = – 2 log | x |
or log | v | – log | v + 2 | = log x– 2 + log | c |
or
∴
log
v
v+ 2
=
c
2
v
v+ 2
= log | cx– 2 |
∴
v
c
v + 2 = ± x2
x
y
, we have
Replacing v to
x
y
x
y
c
c
= ± 2 or
y
y + 2 x = ± x2
+2
x
x
or
x2y = C( y + 2x)
where C = ± c
To find C
Put x = 1 and y = 1 (given) in eqn. (ii), 1 = 3 C ∴ C =
Putting C =
...(ii)
1
3
1
in eqn. (ii), required particular solution is
3
MathonGo
45
Class 12
Chapter 9 - Differential Equations
x2y =
1
( y + 2x) or 3x2y = y + 2x.
3


π
2  y
13.  x sin   – y  dx + x dy = 0; y =
when x = 1
4
 x


Sol. The given differential equation is
π


2 y
 x sin x − y  dx + x dy = 0; y = , x = 1
4



2
x dy = –  x sin

dy
Dividing by dx, x
dx
dy
Dividing by x,
dx
⇒
y

− y  dx
x

y
+ y
x
y
y
+
x
x
= – x sin2
= – sin2
...(i)
 y
= F 
 x
∴ The given differential equation is homogeneous.
y
dy
dv
dv
Put
= v.1 + x
= v + x
= v
∴ y = vx
∴
x
dx
dx
dx
Putting these values in differential equation (i), we have
dv
dv
= – sin2 v + v ⇒
x
= – sin2 v
v + x
dx
dx
⇒
x dv = – sin2 v dx
dv
dx
= –
Separating variables,
x
sin 2 v
1
2
Integrating,
∫ cosec v dv = – ∫ x dx
⇒
– cot v = – log | x | + c
Dividing by – 1,
cot v = log | x | – c
y
y
Putting v =
,
cot
= log | x | – c
...(ii)
x
x
π
To find c : y =
when x = 1 (given)
4
π
π
Putting x = 1 and y =
in (ii), cot
= log 1 – c
4
4
or 1 = 0 – c or c = – 1
Putting c = – 1 in (ii), required particular solution is
y
cot
= log | x | + 1 = log | x | + log e = log | ex |.
x
dy
y
 y
14.
–
+ cosec   = 0; y = 0 when x = 1
dx
x
x
Sol. The given differential equation is
dy
y
y
–
+ cosec
= 0; y = 0 when x = 1
dx
x
x
MathonGo
46
Class 12
Chapter 9 - Differential Equations
dy
y
y
 y
...(i)
=
– cosec
= f  
dx
x
x
 x
∴ Given differential equation (i) is homogeneous.
y
dy
dv
Put
= v
∴ y = vx
∴
= v . 1 + x
x
dx
dx
Putting these values in differential equation (i),
−1
dv
dv
v + x
= v – cosec v ⇒ x
=
sin v
dx
dx
∴
x sin v dv = – dx
dx
Separating variables,
sin v dv = –
x
1
Integrating both sides,
∫ sin v dv = – ∫ x dx
– cos v = – log | x | + c
Dividing by – 1,
cos v = log | x | – c
y
y
Putting v =
,
cos
= log | x | – c
...(ii)
x
x
To find c: Given: y = 0 when x = 1
∴ From (ii), cos 0 = log 1 – c
or 1 = 0 – c = – c
∴
c = – 1
or
Putting
c = – 1 in (ii), cos
y
= log | x | + 1 = log | x | + log e
x
y
= log | ex | which is the required particular solution.
x
dy
15. 2xy + y2 – 2x2
= 0; y = 2 when x = 1
dx
Sol. The given differential equation is
dy
2xy + y2 – 2x2
= 0; y = 2 when x = 1.
...(i)
dx
The given differential equation looks to be homogeneous because
each coefficient of dx and dy is of same degree (2 here).
⇒ cos
2
y
dy
− 2 xy
=
2 – − 2 x2
dx
− 2x
2
dy
y
1  y
 y
or
=
+
= F 
...(ii)
dx
x
2  x 
 x
∴ The given differential equation is homogeneous.
y
dy
dv
dv
Put
= v ∴ y = vx
∴
= v . 1 + x
= v + x
x
dx
dx
dx
Putting these values in differential equation (ii), we have
dv
1 2
dv
1 2
v + x
= v +
v ⇒ x
=
v
⇒ 2x dv = v2 dx
dx
2
dx
2
dv
dx
Separating variables,
2 2 =
x
v
From (i), – 2x2
dy
= – 2xy – y2
dx
MathonGo
∴
47
Class 12
Chapter 9 - Differential Equations
∫
Integrating both sides, 2
⇒
2
v− 1
= log | x | + c
−1
Putting v =
or
y
,
x
v− 2 dv =
⇒
∫
1
dx
x
−2
= log | x | + c
v
−2
= log | x | + c
 y
 x
 
− 2x
= log | x | + c
y
...(iii)
To find c: Given: y = 2, when x = 1.
∴ From (iii),
−2
= log 1 + c or – 1 = c
2
Putting c = – 1 in (iii), the required particular solution is
–
2x
= log | x | – 1
y
⇒ y (log | x | – 1) = – 2x
⇒
y =
⇒ y =
− 2x
− (1 − log| x|)
⇒ y =
− 2x
log| x|− 1
2x
1 − log| x| .
16. Choose the correct answer:
A homogeneous differential equation of the form
 x
dx
= h   can be solved by making the substitution:
dy
 y
(A) y = vx
(B) v = yx
(C) x = vy
(D) x = v
Sol. We know that a homogeneous differential equation of the form
 x
dx
x
= h   can be solved by the substitution
= v i.e., x = vy.
y
dy
y
 
∴ Option (C) is the correct answer.
17. Which of the following is a homogeneous differential
equation?
(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(B) (xy) dx – (x3 + y3) dy = 0
(C) (x3 + 2y2) dx + 2xy dy = 0
(D) y2 dx + (x2 – xy – y2) dy = 0
Sol. Out of the four given options; option (D) is the only option in
which all coefficients of dx and dy are of same degree (here 2). It
may be noted that xy is a term of second degree.
Hence differential equation in option (D) is Homogeneous
differential equation.
MathonGo
48
Class 12
Chapter 9 - Differential Equations
Exercise 9.6
In each of the following differential equations given in
Exercises 1 to 4, find the general solution:
1.
dy
+ 2y = sin x
dx
dy
+ 2y = sin x
dx
| Standard form of linear differential equation
Sol. The given differential equation is
dy
+ Py = Q, we have P = 2 and Q = sin x
dx
Comparing with
∫
∫
P dx =
2 dx = 2 ∫ 1 dx = 2x
Solution is
y(I.F.) =
I.F. = e∫
P dx
= e2x
∫ Q (I.F.) dx + c
2x
∫ e sin x dx + c
y e2x =
y e2x = I + c
or
or
∫
sin x dx
I
II
Applying Product Rule of Integration
where
I=
e
...(i)
2x
....(ii)

 d
 
 ∫ I . II dx = I ∫ II dx − ∫  dx (I) ∫ II dx  dx ,

 

∫ 2e
2x
I = – e2x cos x + 2 ∫ e
= e2x (– cos x) –
or
2x
(− cos x) dx
I
Again applying Product Rule,
cos x dx
II
2x
2x
I = – e2x cos x + 2  e sin x − ∫ 2e sin x dx
⇒ I = – e2x cos x + 2e2x sin x – 4 ∫ e
or I = e2x (– cos x + 2 sin x) – 4I
Transposing 5I = e2x (2 sin x – cos x)
2x
sin x dx
e2 x
(2 sin x – cos x)
5
Putting this value of I in (i), the required solution is
∴ I =
y e2x =
e2 x
(2 sin x – cos x) + c
5
Dividing every term by e2x, y =
c
1
(2 sin x – cos x) + 2 x
5
(e )
1
(2 sin x – cos x) + c e– 2x
5
which is the required general solution.
or
y =
MathonGo
49
Class 12
2.
Chapter 9 - Differential Equations
dy
+ 3y = e– 2x
dx
dy
+ 3y = e– 2x
dx
| Standard form of linear differential equation
dy
Comparing with
+ Py = Q, we have P = 3 and Q = e– 2x
dx
Sol. The given differential equation is
∫
P dx =
∫
3 dx = 3 ∫ 1 dx = 3x
Solution is y(I.F.) =
I.F. = e∫
∫ Q (I.F.) dx + c
− 2 x + 3x
dx + c or = ∫ e
x
or y e = ∫ e
e
or
y e3x = e + c
Dividing every term by e3x,
− 2x
3x
3x
P dx
dx + c =
= e3x
∫
e x dx + c
c
ex
or
y = e– 2x + ce– 3x
3x
3x +
e
e
which is the required general solution.
dy
y
3.
+
= x2
dx
x
dy
y
Sol. The given differential equation is
+
= x2
dx
x
dy
1
It is of the form
+ Py = Q Comparing
P = , Q = x2
dx
x
1
∴ I.F. = e∫ P dx = elog x = x
∫ P dx = ∫ x dx = log x
y=
The general solution is y(I.F.) =
or
yx =
∫
x 2 . x dx + c =
∫
∫ Q (I.F.) dx + c
x3 dx + c
or
xy =
x4
+ c.
4
π

dy
+ (sec x) y = tan x  0 ≤ x < 
2
dx
dy
Sol. The given differential equation is
+ (sec x) y = tan x
dx
dy
It is of the form
+ Py = Q.
dx
Comparing
P = sec x, Q = tan x
4.
∫
P dx =
∫ sec x
I.F. = e∫
P dx
dx = log (sec x + tan x)
= elog (sec x + tan x) = sec x + tan x
The general solution is y(I.F.) =
or y (sec x + tan x) =
∫
tan x (sec x + tan x) dx + c
= ∫ (sec x tan x + tan x) dx + c =
= sec x + tan x – x + c
2
∫ Q (I.F.) dx + c
∫
MathonGo
(sec x tan x + sec 2 x − 1) dx + c
50
Class 12
Chapter 9 - Differential Equations
or y (sec x + tan x) = sec x + tan x – x + c.
For each of the following differential equations given in
Exercises 5 to 8, find the general solution:
5. cos2 x
π

dy
+ y = tan x  0 ≤ x < 
2
dx
dy
+ y = tan x
dx
dy
Dividing throughout by cos2 x to make the coefficient of
unity,
dx
dy
dy
y
tan x
+ (sec2 x) y = sec2 x tan x
+
=
⇒
dx
dx
cos2 x
cos2 x
dy
It is of the form
+ Py = Q.
dx
2
Comparing P = sec x, Q = sec2 x tan x
Sol. The given differential equation is
∫
P dx =
∫ sec
2
I.F. = e∫
x dx = tan x
The general solution is y(I.F.) =
or
Put
cos2 x
P dx
= etan x
∫ Q (I.F.) dx + c
2
yetan x = ∫ sec x tan x . etan x dx + c
tan x = t. Differentiating sec2 x dx = dt
∫ sec
2
∫
t et dt
I II
Applying integration by Product Rule,
∴
x tan x etan x dx =
...(i)
= t . et – ∫ 1 . et dt = t . et – et = (t – 1) et = (tan x – 1) etan x
Putting this value in eqn. (i), yetan x = (tan x – 1) etan x + c
Dividing every term by etan x,
y = (tan x – 1) + ce– tan x which is the required general solution.
dy
6. x
+ 2y = x2 log x
dx
dy
+ 2y = x2 log x
Sol. The given differential equation is x
dx
dy


unity 
Dividing every term by x  To make coeff. of
dx


dy
2
+
y = x log x
dx
x
dy
It is of the form
+ Py = Q.
dx
2
1
Comparing P = , Q = x log x
∫ P dx = 2 ∫ x dx = 2 log x
x
2
I.F. = e∫ P dx = e2 log x = elog x = x2 | ... elog f (x) = f (x)
∫ Q (I.F.) dx + c
. x2 dx + c = ∫ (log x) . x3 dx + c
The general solution is y(I.F.) =
or
yx2 =
∫
( x log x)
MathonGo
51
Class 12
Chapter 9 - Differential Equations
= log x .
x4
–
4
1
x4
.
dx + c
x
4
=
1
x4
log x –
4
4
∫
x3 dx + c
x4
x4
log x –
+ c.
4
16
yx2 =
or
∫
Dividing by x2, y =
c
x2
x2
log x –
+ 2
x
4
16
c
x2
(4 log x – 1) + 2 .
x
16
2
dy
7. x log x
+ y =
log x
dx
x
dy
2
Sol. The given differential equation is x log x
+ y =
log x
dx
x
y =
Dividing every term by x log x to make the coefficient of
dy
2
1
+
y= 2
dx
x log x
x
dy
Comparing with
+ Py = Q, we have
dx
1
2
P =
x log x and Q = x2
dy
dx
unity,
∫
P dx =
∫
1
x log x dx =
∫
1/x
log x dx = log (log x)

∵

I.F. = e∫
P dx
∫

f ′(x)
dx = log f (x)
f (x)

= elog (log x) = log x
The general solution is y(I.F.) =
or y log x =
∫
∫ Q(I.F.) dx + c
2
log x dx = 2
x2
∫
(log x) x − 2 dx + c
I
II
Applying Product Rule of integration,


x− 1
1 x− 1
 log x

+ ∫ x − 2 dx  + c
= 2  (log x) − 1 − ∫ x . − 1 dx  + c = 2  −
x




 log x x − 1 
−2
= 2  − x + − 1  + c or y log x =
(1 + log x) + c.
x


2
8. (1 + x ) dy + 2xy dx = cot x dx (x ≠ 0)
Sol. The given differential equation is (1 + x2) dy + 2xy dx = cot x dx
Dividing every term by dx, (1 + x2)
dy
+ 2xy = cot x
dx
Dividing every term by (1 + x2) to make coefficient of
MathonGo
dy
unity,
dx
52
Class 12
Chapter 9 - Differential Equations
cot x
2x
dy
+
y =
1 + x2
dx
1 + x2
dy
+ Py = Q, we have
Comparing with
dx
2x
cot x
P =
and Q =
1 + x2
1 + x2
2x
dx = log | 1 + x2 |
1 + x2
= log (1 + x2)
[... 1 + x2 >
∫
P dx =
∫
I.F. = e∫
Solution is
P dx


f ′(x)
∵ ∫ f (x) dx = log| f (x)|


0 ⇒ | 1 + x2 | = 1 + x2]
2
= elog (1 + x ) = 1 + x2
y(I.F.) =
∫ Q (I.F.)
dx + c
cot x
(1 + x2) dx + c
1 + x2
⇒ y(1 + x2) =
∫
⇒ y(1 + x2) =
∫ cot x
+ c ⇒ y(1 + x2) = log | sin x | + c
log|sin x|
c
+
2
1+ x
1 + x2
2 –1
2 –1
or y = (1 + x ) log | sin x | + c (1 + x )
which is the required general solution.
For each of the differential equations in Exercises 9 to 12,
find the general solution:
Dividing by 1 + x2,
y =
dy
+ y – x + xy cot x = 0, (x ≠ 0)
dx
Sol. The given differential equation is
dy
x
+ y – x + xy cot x = 0
dx
dy
⇒
x
+ y + xy cot x = x
dx
dy
+ (1 + x cot x) y = x
⇒
x
dx
9. x
Dividing every term by x to make coefficient of
∫
⇒
dy
unity,
dx
dy
(1 + x cot x)
+
y = 1
dx
x
dy
Comparing with
+ Py = Q, we have
dx
1 + x cot x
P =
and Q = 1
x
 1 x cot x 
1
(1 + x cot x)
P dx = ∫
dx = ∫  +
dx = ∫  + cot
x 
x
x
x
∫ P dx

x  dx

= log x + log sin x = log (x sin x)
MathonGo
53
Class 12
Chapter 9 - Differential Equations
I.F. = e∫
= elog (x sin x) = x sin x
P dx
Solution is y(I.F.) =
∫ Q(I.F.) dx + c
or
∫ . x sin x
y(x sin x) =
I
dx + c
II

 d
 
(I) ∫ II dx  dx 
 Applying Product Rule, ∫ I . II dx = I ∫ II dx − ∫ 
dx

 

⇒ y(x sin x) = x(– cos x) –
∫ 1 (− cos x) dx + c
∫ cos x dx + c
= – x cos x +
or y(x sin x) = – x cos x + sin x + c
Dividing by x sin x, y =
− x cos x
c
sin x
+
+
x sin x
x sin x
x sin x
c
1
+
x
x
sin
x
which is the required general solution.
dy
10. (x + y)
= 1
dx
Sol. The given differential equation is
or
y = – cot x +
(x + y)
⇒
dy
= 1
dx
⇒ dx = (x + y) dy
dx
= x + y
dy
⇒
dx
– x = y
dy
| Standard form of linear differential equation
Comparing with
∫ P dy
=
∫
∫ 1 dy = – y
x(I.F.) = ∫ Q(I.F.) dy
I.F. = e∫
− 1 dy = –
∴ Solution is
or
dx
+ Px = Q, we have, P = – 1 and Q = y
dy
xe– y =
∫
P dy
= e– y
+ c
ye− y dy + c
I II

 d
 
(I) ∫ II dy  dy 
 Applying Product Rule, ∫ I . II dy = I ∫ II dy − ∫ 
dy

 

⇒ xe– y = y
e− y
–
−1
= – ye– y +
∫
1.
e− y
dy + c
−1
= – ye– y +
∫
e− y dy + c
e− y
+ c
−1
⇒ xe– y = – ye– y – e– y + c
MathonGo
54
Class 12
Chapter 9 - Differential Equations
Dividing every term by e– y, x = – y – 1 +
c
(e− y )
or
x + y + 1 = cey
which is the required general solution.
11. y dx + (x – y2) dy = 0
Sol. The given differential equation is y dx + (x – y2) dy = 0
dx
+ x = y2
dy
dx
Dividing every term by y (to make coefficient of
unity),
dy
Dividing by dy, y
dx
1
dy + y x = y
Comparing with
P =
∫ P dy
=
dx
+ x – y2 = 0
dy
y
| Standard form of linear differential equation
dx
+ Px = Q, we have
dy
1
and Q = y
y
∫
I.F. = e∫
1
dy = log y
y
P dy
= elog y = y
Solution is x(I.F.) =
⇒ x . y =
or
∫
∫ Q(I.F.) dy
yy dy + c
⇒
+ c
xy =
∫
y2 dy + c ⇒ xy =
c
y2
+
y
3
which is the required general solution.
dy
= y ( y > 0)
12. (x + 3y2)
dx
y3
+ c
3
Dividing by y, x =
Sol. The given differential equation is (x + 3y2)
dy
= y
dx
dx
dx
2
2
dy = x + 3y ⇒ y dy – x = 3y
dx
Dividing every term by y (to make coefficient of
dy unity),
dx
1
–
| Standard form of linear differential equation
y x = 3y
dy
−1
dx
Comparing with
+ Px = Q, we have P =
and Q = 3y
y
dy
1
∫ P dy = – ∫ y dy = – log y = (– 1) log y = log y– 1
1
–1
I.F. = e∫ P dy = elog y = y– 1 =
y
⇒ y dx = (x + 3y2) dy ⇒ y
MathonGo
55
Class 12
Chapter 9 - Differential Equations
Solution is x(I.F.) =
⇒ x .
1
=
y
∫ Q(I.F.) dy
+ c
dy + c ⇒
x
= 3 ∫ 1 dy + c = 3y + c
y
1
∫ 3y . y
Cross – Multiplying, x = 3y2 + cy
which is the required general solution.
For each of the differential equations given in Exercises 13
to 15, find a particular solution satisfying the given
condition:
π
dy
+ 2y tan x = sin x; y = 0 when x =
3
dx
Sol. The given differential equation is
π
dy
+ 2y tan x = sin x; y = 0 when x = .
dx
3
(It is standard form of linear differential equation)
13.
dy
+ Py = Q, we have
dx
P = 2 tan x and Q = sin x
Comparing with
∫
∫
P dx = 2
I.F. = e∫
tan x dx = 2 log sec x = log (sec x)2
(... n log m = log mn)
P dx
2
= elog (sec x) = (sec x)2 = sec2 x
∴ Solution is y(I.F.) =
∫ Q(I.F.)
⇒
∫ sin x sec
y sec2 x =
=
y sec2 x =
or
∫
∫
dx + c
2
x dx + c
sin x
dx + c =
cos2 x
∫
sin x
dx + c
cos x . cos x
tan x sec x dx + c = sec x + c
y
1
+ c
=
cos2 x
cos x
⇒
Multiplying by L.C.M. = cos2 x,
y = cos x + c cos2 x
To find c: y = 0 when x =
∴ From (i), 0 = cos
or
0 =
1
1
+ c  
2
2
...(i)
π
(given)
3
π
π
+ c cos2
3
3
2
or 0 =
1
c
+
2
4
MathonGo
56
Class 12
Chapter 9 - Differential Equations
−1
c
=
⇒ c = – 2
4
2
Putting c = – 2 in (i), the required particular solution is
y = cos x – 2 cos2 x.
⇒
14. (1 + x2)
1
dy
+ 2xy =
; y = 0 when x = 1
1 + x2
dx
Sol. The given differential equation is
dy
1
; y = 0 when x = 1
+ 2xy =
dx
1 + x2
(1 + x2)
dy
unity,
dx
Dividing every term by (1 + x2) to make coefficient of
2x
1
dy
+
y =
(1 + x 2 )2
1 + x2
dx
Comparing with
2x
1 + x2
P =
∫
and Q =
2x
dx =
1 + x2
∫
P dx =
I.F. = e∫
Solution is
dy
+ Py = Q, we have
dx
P dx
1
(1 + x 2 )2
f ′( x)
f ( x) dx
∫
= log f (x) = log (1 + x2)
2
= elog (1 + x ) = 1 + x2
y(I.F.) =
∫ Q(I.F.)
dx + c
or
y(1 + x2) =
∫
1
(1 + x2) dx + c
(1 + x 2 )2
or
y(1 + x2) =
∫
1
dx + c = tan– 1 x + c
x +1
2
or
y(1 + x2) = tan– 1 x + c
To find c: y = 0 when x = 1
Putting y = 0 and x = 1 in (i), 0 = tan– 1 1 + c
or
0 =
π
+ c
4
Putting c = –
π


∵ tan 4 = 1 


⇒
c = –
...(i)
π
4
π
in (i), required particular solution is
4
y(1 + x2) = tan– 1 x –
π
.
4
MathonGo
57
Class 12
15.
Chapter 9 - Differential Equations
π
dy
– 3y cot x = sin 2x ; y = 2 when x =
2
dx
dy
– 3y cot x = sin 2x
dx
Sol. The given differential equation is
dy
+ Py = Q, we have
dx
P = – 3 cot x and Q = sin 2x
Comparing with
∫
P dx = – 3
I.F. = e∫
∫ cot x
P dx
dx = – 3 log sin x = log (sin x)– 3
–3
= elog (sin x)
= (sin x)– 3 =
∫ Q(I.F.)
The general solution is y(I.F.) =
or
y
or
= 2∫
1
=
sin 3 x
∫ sin 2x .
y
=
sin 3 x
∫
dx + c
1
dx + c
sin 3 x
2 sin x cos x
dx + c
sin 3 x
cos x
sin x . sin x dx + c
1
sin 3 x
=2
∫
= 2
∫
cos x
dx + c
sin 2 x
cosec x cot x dx = – 2 cosec x + c
y
2
= –
+ c
sin 3 x
sin x
Multiplying every term by L.C.M. = sin3 x
y = – 2 sin2 x + c sin3 x
or
To find c: Putting y = 2 and x =
...(i)
π
(given) in (i),
2
π
π
+ c sin3
or
2 = – 2 + c or c = 4
2
2
Putting c = 4 in (i), the required particular solution is
y = – 2 sin2 x + 4 sin3 x.
16. Find the equation of the curve passing through the origin,
given that the slope of the tangent to the curve at any point
(x, y) is equal to the sum of coordinates of that point.
Sol. Given: Slope of the tangent to the curve at any point (x, y) =
Sum of coordinates of the point (x, y).
2 = – 2 sin2
dy
= x + y
dx
⇒
Comparing with
∫
P dx =
Solution is
i.e.,
∫
⇒
dy
– y = x
dx
dy
+ Py = Q, we have
dx
− 1 dx = –
y(I.F.) =
ye– x =
∫1
dx = – x
P = – 1 and Q = x
I.F. = e∫
P dx
– x
= e
∫ Q(I.F.) dx + c
−x
∫ x e dx + c
I
II
MathonGo
58
Class 12
Chapter 9 - Differential Equations
d

 Applying Product Rule: ∫ I . II dx = I ∫ II dx − ∫ dx (I)

⇒ ye– x = x
e− x
–
−1
or ye– x = – xe– x +
e− x
∫ 1. −1
∫
(∫ II dx ) dx 
dx + c
e− x dx + c
or
ye– x = – xe– x +
e− x
+c
−1
1
y
x
=– x – x +c
e
ex
e
Multiplying by L.C.M. = ex, y = – x – 1 + cex
...(i)
To find c: Given: Curve (i) passes through the origin (0, 0).
Putting x = 0 and y = 0 in (i), 0 = 0 – 1 + c
or
– c = – 1
or
c = 1
Putting c = 1 in (i), equation of required curve is
y = – x – 1 + ex or x + y + 1 = ex.
17. Find the equation of the curve passing through the point
(0, 2) given that the sum of the coordinates of any point on
the curve exceeds the magnitude of the slope of the tangent
to the curve at that point by 5.
Sol. According to question,
Sum of the coordinates of any point say (x, y) on the curve.
= Magnitude of the slope of the tangent to the curve + 5
↓
(because of exceeds)
dy
+5
i.e., x + y =
dx
dy
dy
⇒
– y = x – 5
+ 5 = x + y
⇒
dx
dx
dy
+ Py = Q, we have
Comparing with
dx
P = – 1 and Q = x – 5
or ye– x = – xe– x – e– x + c
∫
P dx =
∫
− 1 dx = –
Solution is y(I.F.) =
or
ye– x =
∫1
or
I.F. = e∫
dx = – x
P dx
= e– x
∫ Q (I.F.) dx + c
∫
( x − 5) e− x dx + c
I II
d

 Applying Product Rule: ∫ I . II dx = I ∫ II dx − ∫ dx (I)

e− x
–
−1
or
ye– x = (x – 5)
or
ye– x = – (x – 5)e– x +
MathonGo
e− x
∫ 1. −1
∫
(∫ II dx ) dx 
dx + c
e− x dx + c
59
Class 12
Chapter 9 - Differential Equations
ye– x = – (x – 5) e– x +
or
e− x
+ c
−1
y
( x − 5)
1
–
+ c
= –
(e x )
(e x )
(e x )
Multiplying both sides by L.C.M. = ex
y = – (x – 5) – 1 + cex
or y = – x + 5 – 1 + cex or x + y = 4 + cex
To find c: Curve (i) passes through the point (0, 2).
Putting x = 0 and y = 2 in (i),
2 = 4 + ce0 or – 2 = c
Putting c = – 2 in (i), required equation of the curve is
x + y = 4 – 2ex or y = 4 – x – 2ex.
18. Choose the correct answer:
The integrating factor of the differential equation
or
x
...(i)
dy
– y = 2x2 is
dx
(A) e– x
(B) e– y
(C)
1
x
(D) x
dy
– y = 2x2
dx
dy
Dividing every term by x to make coefficient of
unity,
dx
dy
1
–
y = 2x | Standard form of linear differential equation
dx
x
−1
dy
and Q = 2x
Comparing with
+ Py = Q, we have P =
x
dx
−1
∴ ∫ P dx = ∫
dx = – log x = log x– 1 [... n log m = log mn]
x
1
–1
[ ... elog f (x) = f (x)]
I.F. = e∫ P dx = elog x = x– 1 =
x
∴ Option (C) is the correct answer.
19. Choose the correct answer:
The integrating factor of the differential equation
Sol. The given differential equation is x
(1 – y2)
dx
+ yx = ay (– 1 < y < 1)
dy
1
1
1
(B)
(C)
2
1
–
y2
–
1
y
y –1
Sol. The given differential equation is
(A)
2
(1 – y2)
(D)
1
1 – y2
dx
+ yx = ay (– 1 < y < 1)
dy
Dividing every term by (1 – y2) to make coefficient of
MathonGo
dx
unity,
dy
60
Class 12
Chapter 9 - Differential Equations
ay
dx
y
+
x =
1 − y2
dy
1 − y2
| Standard form of linear differential equation
dx
Comparing with
+ Px = Q, we have
dy
P =
∴
∫
y
1 − y2
P dy =
∫
and Q =
ay
1 − y2
y
−1
dy =
1 − y2
2
∫
− 2y
dy
1 − y2

∵

−1
log (1 – y2)
2
= log (1 – y2)– 1/2
=
2
∫

f ′( y)
= log f ( y)
f ( y)

– 1/2
I.F. = e∫ P dy = elog (1 – y )
– 1/2
= (1 – y2)
1
=
1 − y2
∴ Option (D) is the correct answer.
MathonGo
[... elog f (x) = f (x)]
61
Class 12
Chapter 9 - Differential Equations
MISCELLANEOUS EXERCISE
1. For each of the differential equations given below, indicate
its order and degree (if defined)
(i)
d2 y
 dy 
2 + 5x  dx 
dx


 dy 
(ii) 

 dx 
3
 dy 
– 4

 dx 
2
– 6y = log x
2
+ 7y = sin x
 d3 y 
d4 y


4 – sin  dx 3  = 0
dx


(i) The given differential equation is
(iii)
Sol.
d2 y
+ 5x
dx 2
The highest order
equation is
2
 dy 
 dx  – 6y = log x
 
derivative present in this differential
d2 y
and hence order of this differential
dx 2
equation is 2.
The given differential equation is a polynomial equation
in derivatives and highest power of the highest order
d2 y
is 1.
dx 2
∴ Order 2, Degree 1.
derivative
MathonGo
62
Class 12
Chapter 9 - Differential Equations
(ii) The given differential equation is
3
2
 dy 
 dy 
 dx  – 4  dx  + 7y = sin x.
 
 
The highest order derivative present in this differential
dy
equation is
and hence order of this differential
dx
equation is 1.
The given differential equation is a polynomial equation
in derivatives and highest power of the highest order
dy
is 3.
dx
∴ Order 1, Degree 3.
derivative
3

 dy  
∵ of 


 dx  

 d3 y 
d4 y


4 – sin  dx 3  = 0.
dx


The highest order derivative present in this differential
(iii) The given differential equation is
equation is
d4 y
and hence order of this differential
dx 4
equation is 4.
Degree of this differential equation is not defined because
the given differential equation is not a polynomial
equation in derivatives

 d3 y  
 because of the presence of term sin  3   .


 dx  

∴ Order 4 and Degree not defined.
2. For each of the exercises given below verify that the given
function (implicit of explicit) is a solution of the
corresponding differential equation.
(i) xy = aex + be– x + x2 : x
dy
d2 y
2
2 + 2 dx – xy + x – 2 = 0
dx
(ii) y = ex (a cos x + b sin x) :
dy
d2 y
2 – 2 dx + 2y = 0
dx
d2 y
+ 9y – 6 cos 3x = 0
dx 2
dy
(iv) x2 = 2y2 log y : (x2 + y2)
– xy = 0
dx
(i) The given function is
xy = aex + be– x + x2
...(i)
To verify: This given function (i) is a solution of differential
(iii) y = x sin 3x :
Sol.
dy
d2 y
2
2 + 2 dx – xy + x – 2 = 0
dx
Differentiating both sides of (i), w.r.t. x,
equation
MathonGo
...(ii)
63
Class 12
Chapter 9 - Differential Equations
x
dy
+ y . 1 = aex + be– x (– 1) + 2x
dx
dy
+ y = aex – be– x + 2x
dx
Again differentiating both sides, w.r.t. x
dy
dy
d2 y
x
+
. 1 +
= aex + be– x + 2
2
dx
dx
dx
or
x
dy
d2 y
+ 2
= aex + be– x + 2
dx
dx 2
∴ Putting aex + be– x = xy – x2 from (i), in R.H.S., we have
or
x
x
dy
d2 y
+2
= xy – x2 + 2
2
dx
dx
dy
d2 y
+ 2
– xy + x2 – 2 = 0
dx
dx 2
which is same as differential equation (ii).
∴ Function given by (i) is a solution of D.E. (ii).
(ii) The given function is
y = ex (a cos x + b sin x)
...(i)
To verify: Function given by (i) is a solution of differential
equation
or
x
dy
d2 y
– 2
+ 2y = 0
dx
dx 2
From (i),
...(ii)
dy d x
d
=
e . (a cos x + b sin x) + ex
(a cos x + b sin x)
dx dx
dx
or
⇒
∴
or
or
dy
= ex (a cos x + b sin x) + ex (– a sin x + b cos x)
dx
dy
= y + ex (– a sin x + b cos x)
dx
(By (i))
...(iii)
dy
d2 y
+ ex (–a sin x + b cos x) + ex (– a cos x – b sin x)
=
dx
dx 2
dy
d2 y
+ ex (– a sin x + b cos x) – ex (a cos x + b sin x)
=
dx
dx 2
dy
 dy

d2 y
− y – y
(By (iii)) and (By (i))
2 = dx +  dx


dx
d2 y
dy
dy
d2 y
= 2
– 2y
or
– 2
+ 2y = 0
2
dx
dx
dx 2
dx
which is same as given differential equation (ii).
or
MathonGo
64
Class 12
Chapter 9 - Differential Equations
∴ Function given by (i) is a solution of differential
equation (ii).
(iii) The given function is
y = x sin 3x
...(i)
To verify: Function given by (i) is a solution of differential
equation
d2 y
+ 9y – 6 cos 3x = 0
dx 2
dy
= x . cos 3x . 3 + sin 3x . 1
From (i),
dx
dy
= 3 x cos 3x + sin 3x
or
dx
∴
or
...(ii)
d2 y
= 3[x(– sin 3x) 3 + cos 3x . 1] + (cos 3x) 3
dx 2
d2 y
= – 9x sin 3x + 3 cos 3x + 3 cos 3x
dx 2
= – 9x sin 3x + 6 cos 3x
= – 9y + 6 cos 3x
[By (i)]
2
d y
+ 9y – 6 cos 3x = 0
dx 2
which is same as differential equation (ii).
∴ Function given by (i) is a solution of differential
equation (ii).
(iv) The given function is
...(i)
x2 = 2y2 log y
To verify: Function given by (i) is a solution of differential
equation
or
dy
– xy = 0
dx
Differentiating both sides of (i) w.r.t. x, we have
(x2 + y2)
...(ii)
 2 1 dy
dy 
+ (log y) 2 y
2x = 2  y .
y dx
dx 

dy
Dividing by 2, x =
( y + 2y log y)
dx
dy
x
x
∴
=
=
dx
y + 2 y log y
y (1 + 2 log y)
Putting 2 log y =
dy
=
dx
x2
from (i),
y2
x

x2 
y  1 + 2 
y 

MathonGo
=
x
 y2 + x 2 
y 

2
 y

=
xy2
y ( x2 + y2 )
65
Class 12
Chapter 9 - Differential Equations
xy
dy
= 2
dx
x + y2
⇒
Cross-multiplying, (x2 + y2)
dy
= xy
dx
dy
– xy = 0
dx
which is same as differential equation (ii).
∴ Function given by (i) is a solution of differential
equation (ii).
3. Form the differential equation representing the family of
curves (x – a)2 + 2y2 = a2, where a is an arbitrary constant.
Sol. Equation of the given family of curves is
(x – a)2 + 2y2 = a2
2
or x + a2– 2ax + 2y2 = a2
or
x2 – 2ax + 2y2 = 0
or
x2 + 2y2 = 2ax
...(i)
Number of arbitrary constants is one only (a here).
So, we shall differentiate both sides of equation (i) only once w.r.t.x.
dy
∴ From (i),
2x + 2 . 2y
= 2a
dx
dy
or
2x + 4y
= 2a
...(ii)
dx
Dividing eqn. (i) by eqn. (ii) (To eliminate a), we have
(x2 + y2)
or
2 ax
x2 + 2 y2
dy = 2 a = x
2x + 4 y
dx
dy 

Cross-multiplying, x  2 x + 4 y
 = x2 + 2y2
dx


dy
dy
2
2
2
or 2x + 4xy
or
4xy
= 2y2 – x2
= x + 2y
dx
dx
dy
2 y2 − x 2
which is the required differential equation.
=
dx
4 xy
2
2
4. Prove that x – y = c(x2 + y2)2 is the general solution of the
differential equation (x3 – 3xy2) dx = ( y3 – 3x2y) dy, where c
is a parameter.
Sol. The given differential equation is
(x3 – 3xy2) dx = ( y3 – 3x2y) dy
...(i)
Here each coefficient of dx and dy is of same degree (Here 3),
therefore differential equation (i) looks to be homogeneous
differential equation.
⇒
3
2
( x − 3xy )
dy
=
dx
y3 − 3x2 y
Dividing every term in the numerator and denominator of R.H.S.
by x3,
From (i),
MathonGo
66
Class 12
Chapter 9 - Differential Equations
2
 y
1−3 
 x
dy
 y
3
=  y
 y  = f  x 
dx
−
3
 x
 x
 
 
...(ii)
Therefore the given differential equation is homogeneous.
y
dy
dv
dv
= v. Therefore y = vx. ∴
= v . 1 + x
= v + x
x
dx
dx
dx
Putting these values in eqn. (ii),
Put
2
v + x
1 − 3v
dv
= 3
dx
v − 3v
1 − 3v2
1 − 3v2 − v4 + 3v2
1 − v4
dv
dv
– v =
⇒ x
= 3
= 3
3
dx
dx
v − 3v
v − 3v
v − 3v
Cross-multiplying,
x(v3 – 3v) dv = (1 – v4) dx
∴ x
Separating variables,
(v3 − 3v)
dx
dv =
x
1 − v4
Integrating both sides,
v3 − 3v
dv =
1 − v4
1
dx = log x + log c
...(iii)
x
Let us form partial fractions of
v3 − 3v
v3 − 3v
v3 − 3v
v3 − 3v
=
or
=
(1 − v2 )(1 + v2 )
(1 − v)(1 + v)(1 + v2 )
1 − v4
1 − v4
A
Cv + D
B
=
+
+
...(iv)
1−v
1 + v2
1+v
Multiplying both sides of (iv) by L.C.M. = (1 – v)(1 + v) (1 + v2),
v3 – 3v = A(1 + v)(1 + v2) + B(1 – v)(1 + v2) + (Cv + D)(1 – v2)
= A(1 + v2 + v + v3) + B(1 + v2 – v – v3) + Cv – Cv3 + D – Dv2
Comparing coefficients of like powers of v,
A – B – C = 1
...(v)
v3
v2
A + B – D = 0
...(vi)
v
A – B + C = – 3
...(vii)
Constants A + B + D = 0
...(viii)
Let us solve eqns. (v), (vi), (vii), (viii) for A, B, C, D.
−4
Eqn. (v) – eqn. (vii) gives, – 2C = 4 ⇒ C =
= – 2
2
Eqn. (vi) – eqn. (viii) gives, – 2D = 0 or D = 0
Putting C = – 2 in (v),
A – B + 2 = 1
⇒ A – B = – 1
...(ix)]
Putting D = 0 in (vi),
A+ B = 0
...(x)
Adding (ix) and (x),
∫
2A = – 1
∫
⇒ A =
MathonGo
−1
2
67
Class 12
Chapter 9 - Differential Equations
1
2
Putting values of A, B, C and D in (iv), we have
From (x), B = – A =
−1
1
v3 − 3v
2v
2
=
+ 2 –
1−v
1+v
1 − v4
1 + v2
∫
∴
− 1 log (1 − v)
1
v3 − 3v
dv =
+
log (1 + v)
2
2
1 − v4
−1

– log (1 + v2) ∵

=
1
1
log (1 – v) +
log (1 + v) – log (1 + v2)
2
2
=
1
[log (1 – v) + log (1 + v)] – log (1 + v2)
2
=
1
log (1 – v)(1 + v) – log (1 + v2)
2
⇒
∫
∫

f ′ (v)
dv = log f (v)
f (v)

 1 − v2 
v3 − 3v

dv = log (1 – v2)1/2 – log (1 + v2) = log 
4
1−v
 1 + v2 


Putting this value in eqn. (iii),
 1 − v2 
1 − v2
 = log xc
log 
∴
= xc
2
 1+v 
1 + v2


Squaring both sides and cross-multiplying, 1 – v2 = c2x2 (1 + v2)2
y
y2
, 1 – 2 = c2x2
Putting v =
x
x

y2 
 1 + 2 
x 

2
2
2 2
x 2 − y2
x 2 − y2
c 2 ( x 2 + y2 ) 2
2 2 (x + y )
=
c
x
or
=
x2
x2
x4
x2
or
x2 – y2 = C(x2 + y2)2 where c2 = C
which is the required general solution.
5. Form the differential equation of the family of circles in the
first quadrant which touch the coordinate axes.
Sol. We know that the circle in the first
Y
quadrant which touches the coordinates axes has centre (a, a)
where a is the radius of the circle.
(See adjoining figure)
∴ Equation of the circle is
a
C(a, a)
(x – a)2 + ( y – a)2 = a2 ...(i)
a
or x2 + y2 – 2ax – 2ay + a2 = 0
Differentiating w.r.t x, we get,
X
or
O
MathonGo
68
Class 12
Chapter 9 - Differential Equations
2x + 2yy′ – 2a – 2ay′ = 0
Dividing by 2, x + yy′ = a(1 + y′)
or
x + yy′
1 + y′
a =
Substituting the value of a in (i), to eliminate a, we get

x + yy′ 
x −

1 + y′ 

2

x + yy′ 
+ y−

1 + y′ 

2
 x + yy′ 
= 

 1 + y′ 
2
2
2
2
 x + xy′ − x − yy′ 
 y + yy′ − x − yy′ 
 x + yy′ 

 + 
 = 

′
′
1+ y
1+ y




 1 + y′ 
Multiplying by L.C.M. = (1 + y′)2
(xy′ – yy′)2 + ( y – x)2 = (x + yy′)2
or
y′2(x – y)2 + (x – y)2 = (x + yy′)2
or
(x – y)2(1 + y′2) = (x + yy′)2
which is the required differential equation.
6. Find the general solution of the differential equation
or
dy
+
dx
1 – y2
= 0.
1 – x2
Sol. The given differential equation is
dy
+
dx
1 − y2
=0
1 − x2
1 − y2 dx
dy
Separating Variables,
=
1 − y2
⇒
− 1 − y2
dy
=
dx
1 − x2
⇒
1 − x2 dy = –
Integrating both sides,
∫
− dx
1 − x2
1
1
∫
dx
1 − x2
⇒
sin y = – sin x + c
⇒
sin– 1 x + sin– 1 y = c
which is the required general solution.
7. Show that the general solution of the differential equation
1− y
dy = –
2
–1
– 1
2
y + y +1
dy
= 0 is given by
+ 2
dx
x + x +1
(x + y + 1) = A(1 – x – y – 2xy), where A is parameter.
Sol. The given differential equation is
dy
dy
 y2 + y + 1 
y2 + y + 1
+ 2
= 0 ⇒
= –  2
 x + x + 1 
dx
dx
x + x+1


Multiplying by dx and dividing by y2 + y + 1, we have
− dx
dy
= 2
x + x +1
y2 + y + 1
MathonGo
69
Class 12
Chapter 9 - Differential Equations
dy
dx
+ 2
= 0
(Variables separated)
y2 + y + 1
x + x +1
Integrating both sides,
1
1
...(i)
∫ y2 + y + 1 dy + ∫ x2 + x + 1 dx = 0
⇒
Now, y2 + y + 1 = y2 + y +
1
1
–
+ 1
4
4
2
2

1
1

1
 To complete squares, add and subtract  coeff. of y  =   = 
4 
2

2

1

= y+ 
2

∫
∴
1
 3


 2 
2
3
1

= y+ 
+
4
2

1
dy =
y2 + y + 1
tan– 1
∫
1
2 =
3
2
y+
2
 3

+ 
 2 
1
2
 3
1

 y + 2  +  2 




∫
2
dy
2
2y + 1
tan– 1
3
3
1
dx =
x + x +1
Putting these values in eqn. (i),
Changing y to x,
2
2
2
2x + 1
tan– 1
3
3
2x + 1
2
2y + 1
2
tan– 1
+
tan– 1
=c
3
3
3
3
2y + 1
2x + 1
3
3
Multiplying by
,
tan– 1
+ tan– 1
=
c
3
3
2
2
or
tan– 1
2x + 1 2 y + 1
+
3
3
–1
2 x + 1 2 y + 1 = tan c′
1−
.
3
3


3
–1
–1
–1 a + b
c by tan – 1 c′
and replacing
∵ tan a + tan b = tan
ab
1
–
2


Multiplying every term in the numerator and denominator of
L.H.S. by 3, we have
3 (2 x + 2 y + 2)
= c′
3 − (4 xy + 2 x + 2 y + 1)
or
⇒
3 (2x + 2y + 2) = c′ (2 – 2x – 2y – 4xy)
2 3 (x + y + 1) = 2c′(1 – x – y – 2xy)
MathonGo
70
Class 12
Chapter 9 - Differential Equations
c′
(1 – x – y – 2xy)
3
c′
(x + y + 1) = A(1 – x – y – 2xy) where A =
.
3
Dividing every term by 2 3 , x + y + 1
or
=
8. Find the equation of the curve passing through the point
 π
 0, 4  whose differential equation is


sin x cos y dx + cos x sin y dy = 0.
Sol. The given differential equation is
sin x cos y dx + cos x sin y dy = 0
⇒
sin x cos y dx = – cos x sin y dy
− sin y
sin x
dx =
dy
cos x
cos y
Separating variables,
⇒
Integrating both sides,
tan x dx = – tan y dy
∫
tan x dx = – ∫ tan y dy
⇒
log | sec x | = – log | sec y | + log | c |
⇒ log | sec x | + log | sec y | = log | c |
⇒
log | sec x sec y | = log | c |
∴
sec x sec y = c
 π
To find c: Given: Curve (i) passes through  0,  .
 4
π
π
in (i), sec 0 sec
= c or
Putting x = 0 and y =
4
4
Putting c =
sec x
=
cos y
2
...(i)
2 = c.
2 in (i), equation of required curve is
⇒
2 cos y = sec x
⇒
cos y =
sec x
.
2
9. F i n d t h e p a r t i c u l a r s o l u t i o n o f t h e d i f f e r e n t i a l
equation (1 + e 2x ) dy + (1 + y2) ex dx = 0, given that y = 1
when x = 0.
Sol. The given differential equation is
(1 + e 2x) dy + (1 + y2) ex dx = 0
Dividing every term by (1 + y2)(1 + e2x), we have
x
e
dy
+
dx = 0
2
1 + e2 x
1+ y
Integrating both sides, we have
1
ex
∫ 1 + y2 dy + ∫ 1 + e2 x dx = c
or
tan– 1 y +
∫
ex
dx = c
1 + e2 x
MathonGo
...(i)
71
Class 12
Chapter 9 - Differential Equations
To evaluate
∫
ex
dx
1 + e2 x
∴ ex =
Put ex = t
ex dx
=
1 + e2 x
dt
dx
or ex dx = dt
dt
= tan– 1 t = tan– 1 ex
1 + t2
Putting this value in (i), tan– 1 y + tan– 1 ex = c
To find c: y = 1 when x = 0 (given)
∴
∫
∫
...(ii)
Putting x = 0 and y = 1 in (ii), tan– 1 1 + tan– 1 1 = c
π
π

π
+
= c
∵ tan 4 = 1
4

4
π
2π
or
c=
=
2
4
π
Putting c =
in (ii), the particular solution is
2
π
tan– 1 y + tan– 1 ex = .
2
10. Solve the differential equation
yex/y dx = (xex/y + y2) dy ( y ≠ 0).
Sol. The given differential equation is
y . ex/y dx = (x . ex/y + y2) dy, y ≠ 0
or
0
∵e = 1
∴ tan – 1 1 =
or
or
(
x ex / y + y2
x ex / y
y2
dx
=
=
+
x/ y
x/ y
dy
y. e
ye
y ex / y
dx
x
=
+ y e– x/y
dy
y
)
π
4 
...(i)
It is not a homogeneous differential equation (because of presence
of only y as a factor) yet it can be solved by putting
so that
dv
dx
= v + y
dy
dy
Putting these values of x and
v+y
or
dv
= v + ye– v
dy
dv
y
= y e– v
dy
or y
x
= v i.e., x = vy.
y
dx
in (i), we have
dy
dv
y
= v
dy
e
Cross-multiplying and dividing both sides by y,
ev dv = dy
Integrating ev = y + c or ex/y = y + c
which is the required general solution.
11. Find a particular solution of the differential equation
(x – y)(dx + dy) = dx – dy given that y = – 1 when x = 0.
MathonGo
72
Class 12
Chapter 9 - Differential Equations
Sol. The given differential equation is
(x – y)(dx + dy) = dx – dy
or
(x – y) dx + (x – y) dy = dx – dy
or (x – y) dx – dx + (x – y) dy + dy = 0
or
(x – y – 1) dx + (x – y + 1) dy = 0
⇒
(x – y + 1) dy = – (x – y – 1) dx
( x − y − 1)
dy
= –
x− y+1
dx
Put x – y = t
∴
...(i)
dy
dt
=
dx
dx
− dt
dy
dt
dy
⇒ –
=
+ 1
– 1
⇒
=
dx
dx
dx
dx
 t −1
− dt
Putting these values in (i),
+ 1 = – 

dx
 t + 1
 t −1
dt

⇒
–
= – 1 – 
dx
 t + 1
Differentiating w.r.t. x, 1 –
Multiplying by – 1,
t −1
dt
t +1+ t −1
= 1 +
=
t +1
dx
t +1
⇒
2t
dt
=
t +1
dx
⇒ (t + 1) dt = 2t dx
⇒
t +1
dt = 2 dx
t
t +1
dt = 2 ∫ 1 dx
t 
1
t 1

or ∫  +  dt = 2x + c
or ∫  1 +  dt = 2x + c
t
t
t



⇒ t + log | t | = 2x + c
Putting t = x – y, x – y + log | x – y | = 2x + c
⇒ log | x – y | = x + y + c
...(ii)
To find c: y = – 1 when x = 0
Putting x = 0, y = – 1 in (ii),
log 1 = 0 – 1 + c or 0 = – 1 + c
∴
c = 1
Putting c = 1 in (ii), required particular solution is
log | x – y | = x + y + 1.
 e– 2 x
y  dx

–
12. Solve the differential equation 
x
x  dy = 1 (x ≠ 0)

Sol. The given differential equation is
Integrating both sides,
∫ 
 e− 2 x
y  dx


−

x
x  dy = 1

MathonGo
73
Class 12
Chapter 9 - Differential Equations
dy
,
dx
Multiplying both sides by
e− 2 x
–
x
y
dy
=
dx
x
or
dy
+
dx
y
e− 2 x
=
x
x
dy
+ Py = Q.
dx
It is of the form
e− 2 x
1
and Q =
x
x
1
x1 / 2
− 1/ 2
P dx = ∫
dx = ∫ x
dx =
= 2 x
x
1/2
P dx
I.F. = e∫
= e2 x
Comparing, P =
∫
The general solution is
y(I.F.) =
∫ Q(I.F.)
dx + c
or
ye2
x
=
∫
e− 2 x
2
x e
or
y . e2
x
=
∫
x − 1 / 2 dx + c =
x
Multiplying both sides by e− 2
dx + c =
x
∫
1
dx + c
x
x1 / 2
+ c= 2 x + c
1
2
, we have
y = e− 2 x (2 x + c) is the required general solution.
13. Find a particular solution of the differential equation
dx
≠
dy + y cot x = 4x cosec x (x 0) given that y = 0, when
π
.
2
Sol. The given differential equation is
dy
+ y cot x = 4x cosec x
dx
(It is standard form of linear differential equation.)
dy
Comparing with
+ Py = Q, we have
dx
P = cot x and Q = 4x cosec x
x =
∫
P dx =
∫ cot x
I.F. = e∫
Solution is
y(I.F.) =
P dx
= log sin x
= elog sin x = sin x
∫ Q (I.F.) dx + c
∫ 4x cosec x sin x dx + c
⇒
y(sin x) =
⇒
y(sin x) = 4
∫
x.
1
sin x dx + c
sin x
MathonGo
74
Class 12
Chapter 9 - Differential Equations
∫
x dx + c = 4 .
or
y sin x = 4
or
y sin x = 2x2 + c
x2
+ c
2
...(i)
π
To find c: Given that y = 0, when x = .
2
π
and y = 0 in (i),
2
Putting
x =
or 0 =
π2
+ c
2
Putting c = –
⇒ c =
0 = 2 .
π2
+ c
4
− π2
2
π2
in (i), the required particular solution is
2
π2
.
2
14. Find a particular solution of the differential equation
dy
(x + 1)
= 2e– y – 1 given that y = 0 when x = 0.
dx
Sol. The given differential equation is
dy
(x + 1)
= 2e– y – 1
dx
y sin x = 2x2 –
dy
2
2 − ey
= y – 1 =
dx
e
ey
y
Cross-multiplying, (x + 1) e dy = (2 – ey) dx
or
(x + 1)
e y dy
dx
=
x+1
2 − ey
ey
∫ 2 − e y dy =
Separating variables,
Integrating both sides,
Put ey = t. ∴ ey =
∴
or
dt
dy
∫
1
dx
x+1
⇒ ey dy = dt
dt
2 − t = log | x + 1 |
log|2 − t|
= log | x + 1 | + c
−1
∫
Putting t = ev, – log | 2 – ey | = log | x + 1 | + c
or log | x + 1 | + log | 2 – ey | = – c
or
log | (x + 1)(2 – ey) | = – c
or
| (x + 1)(2 – ey) | = e– c
or
(x + 1)(2 – ey) = ± e– c
or
(x + 1)(2 – ey) = C where C = ± e– c ...(i)
When x = 0, y = 0 (given)
∴ From (i), (1)(2 – 1) = C
or
C = 1
Putting C = 1 in (i) the required particular solution is
(x + 1)(2 – ey) = 1.
Note. The particular solution may be written as
MathonGo
75
Class 12
Chapter 9 - Differential Equations
2 – ey =
1
x+1
or
 2x + 1 
or log ey = log 

 x+1 
ey = 2 –
2x + 1
1
=
x+1
x+1
 2x + 1 
y = log 

 x+1 
.
.
( . log ey = y log e = y as log e = 1)
which expresses y as an explicit function of x.
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 20,000 in 1999
and 25,000 in the year 2004, what will be the population of
the village in 2009?
Sol. Let P be the population of the village at time t.
According to the question, Rate of increase of population of the
village is proportional to the number of inhabitants.
dP
⇒
= kP (where k > 0 because of increase, is the constant of
dt
proportionality)
dP
⇒ dP = kP dt
⇒
= k dt
P
1
Integrating both sides, ∫
dP = k ∫ 1 dt
P
⇒
log P = kt + c
...(i)
To find c: Given: Population of the village was P = 20,000 in the
year 1999.
Let us take the base year 1999 as t = 0.
Putting t = 0 and P = 20000 in (i), log 20000 = c
Putting c = log 20000 in (i), log P = kt + log 20000
∴ log P – log 20000 = kt
or
P
= kt
20000
To find k: Given: P = 25000 in the year 2004
i.e., when t = 2004 – 1999 = 5
Putting P = 25000 and t = 5 in (ii),
⇒
log
log
25000
= 5k
20000
Putting k =
⇒ 5k = log
5
4
⇒ k =
...(ii)
1
5
log
5
4
5
1
5
P
1
log
in (ii), log
=  log  t
4
5
4
20000
5
...(iii)
To find the population in the year 2009,
i.e., when t = 2009 – 1999 = 10,
Putting t = 10 in (iii),
log
5
P
1
=  log  × 10
4
20000
5
MathonGo
76
Class 12
Chapter 9 - Differential Equations
= 2 log
5
5
= log  
4
4
2
= log
25
16
P
25
=
16
20000
25
⇒
P =
× 20000 = 25 × 1250 = 31250.
16
16. Choose the correct answer:
The general solution of the differential equation
∴
y dx – x dy
= 0 is
y
(B) x = Cy2
(A) xy = C
y dx − x dy
= 0
y
Sol. The given differential equation is
Cross-multiplying,
⇒
(D) y = Cx2
(C) y = Cx
y dx – x dy = 0
y dx = x dy
dy
dx
=
y
x
Integrating both sides, log | x | = log | y | + log | c |
⇒ log | x | = log | cy |
⇒ | x | = | cy |
Separating variables,
⇒
x = ± cy
⇒
y = ±
1
x
c
1
c
which is the required solution.
∴ Option (C) is the correct answer.
17. The general solution of a differential equation of the type
or y = Cx where C = ±
dx
dy + P1x = Q1 is
(A) ye ∫ P1 dy =
∫
 Q e∫
 1

(B) y . e ∫
=
∫
P1 dx
(C) xe∫ P1 dy =

P1 dy 
 dy + C

 Q e∫
 1

∫  Q1e∫
P1 dx

 dx + C

P1 dy 
 dy + C


∫ P1 dx 
 dx + C
(D) xe∫ P1 dx = ∫  Q1e


Sol. We know that general solution of differential equation of the type
dx
+ P1x = Q1 is
dy
x . (I.F.) =
∫
Q1 (I.F.) dy + c where I.F. = e∫
MathonGo
P1 dy
77
Class 12
Chapter 9 - Differential Equations
∴ x e∫ P1 dy = ∫ (Q1 e∫ 1 ) dy + c
∴ Option (C) is the correct answer.
18. The general solution of the differential equation
ex dy + (y ex + 2x) dx = 0 is
y
(A) x e + x2 = C
(B) x ey + y2 = C
x
2
(C) y e + x = C
(D) y ey + x2 = C
Sol. The given differential equation is
ex dy + ( y ex + 2x) dx = 0
Dividing every term by dx,
P dy
ex
or
dy
+ y ex + 2x = 0
dx
dy
+ y ex = – 2x
ex
dx
Dividing every term by ex to make coefficient of
dy
unity,
dx
− 2x
dy
(Standard form of linear differential equation)
+ y =
dx
ex
dy
Comparing with
+ Py = Q, we have
dx
− 2x
P = 1 and Q =
ex
∫
P dx =
∫1
I.F. = e∫
dx = x
P dx
∫ Q (I.F.) dx + C
Solution is y (I.F.) =
or
y ex =
∫
= ex
– 2x x
e dx + C
ex
x2
+ C
2
x
2
x
2
or
y e = – x +C
or y e + x = C
∴ Option (C) is the correct answer.
or
y ex = – 2
∫
x dx + C
MathonGo
or
y ex = – 2
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