UNIT- IV
APPLICATION OF INTEGRATION AND
FIRST ORDER DIFFERENTIAL
EQUATION
AREA AND VOLUME
4.1 Area – Area of circle, volume – volume of cone and sphere –
simple problems
FIRST ORDER DIFFERENTIAL EQUATION
4.2 Definition of order and degree of differential equation – solution
of first order variable separable type differential equation –
simple problems
LINEAR TYPE DIFFERENTIAL EQUATION
4.3 Solution of linear differential equation –simple problems
Introduction
In mathematics – iii, we discussed the basic concepts of
integration. In mathematics – ii, we studied the differential equation
and formation of differential equation. In this unit, we shall study the
application of integration and first order differential equation.
4.1. AREA AND VOLUME
W e apply the concept of definite integral to find the area and
volume.
Area
The area under the curve y = f(x) between the x-axis and the
b
ordinates x=a and x=b is the definite integral
³
a
266
b
f( x )dx or
³ ydx
a
b
b
³
Area(A ) = f ( x )dx or
a
³ ydx
a
Similarly, the area between the curve x=g(y) and the y=axis and
d
the lines y=c and y=d is the definite integral
³ g(y )dy
c
X=g(y)
d
³
Area (A ) = g( y )dy or
c
d
³ xdy
c
267
d
or
³ xdy
c
Volume
The volume of solid obtained by rotating the area bounded by the
curve y=f(x) and x-axis between x=a and x=b about the x-axis is the
b
b
definite integral π [f( x )] dx or π y 2 dx
³
³
2
a
a
b
b
Volume (v)= π [f( x )] dx or π y 2 dx
³
³
2
a
a
Similarly, the volume of solid obtained by rotating the area
bounded by the curve x = g(y) and y-axis between y=c and y=d about
d
d
the y-axis is the definite integral π [g(y )] dy or π x 2dy
³
2
c
d
c
d
Volume (v)= π [g(y )] dy or π x 2 dy
³
c
2
³
³
c
268
4.1 WORKED EXAMPLES
PART - A
3
1) Find the area bounded by the curve y=4x , the x-axis and the
ordinates x=0 and x=1
Solution:
1
b
1
ª x4 º
A = ³ ydx = ³ 4x 3dx = «4 »
a
0
¬« 4 ¼» 0
[ ]
= x 4 0 = (1) − 0 = 1 − 0 = 1 Sq. units
x
2) Find the area bounded by the curve y=e , the x-axis and the
ordinates x=0 and x=6.
Solution:
1
b
6
a
0
4
A = ³ ydx = ³ e x dx
[ ]
(
6
)
= e x 0 = e6 − e0 = e6 − 1 Sq. units
3) Find the area bounded by the curve y=cosx, x-axis and between
π
x=0 and x=
2
Solution:
π
2
b
A = ³ ydx = ³ cos xdx
a
0
π
2
0
π
− sin 0 = 1Sq. units
2
2
4) Find the area bounded by the curve x=4y-y , the y-axis and the
lines y=0 and y=3
Solution:
= [sin x ] = sin
d
A = ³ xdy
c
3
3
(
)
= ³ 4y − y 2 dy
0
3
ª 4y 2 y 3 º
ª
y3 º
=«
− » = « 2y 2 − »
3 ¼»
3 »¼
¬« 2
¬«
0
0
= 18 − 9 = 9 Sq. units
269
5)
Find the volume of the solid formed when the area bounded by
2
the curve y =4x between x=1 and x=2 is rotated about x-axis.
Solution:
b
2
³
v = π y 2 dx
³
= π 4xdx
a
1
ª x2 º
= 4π « »
«¬ 2 »¼
2
ª (2)2 (1)2 º
= 4π «
−
»
2 »¼
«¬ 2
3
= 4π.
2
= 6π Cubic units
6)
Find the volume of the solid formed when the area bounded by
the curve y= 10 + x between x=0 and x=5 is rotated about xaxis.
Solution:
Given y = 10 + x
Squaring on both sides, we get
2
y =10+x
b
³
v = π y 2 dx
5
= π (10 + x )dx
³
a
0
5
ª
x2 º
= π«10 x +
»
2 ¼»
¬«
0
25 º
ª
= π«50 +
2 »¼
¬
=
125
π Cubic units.
2
270
7)
Find the volume of the solid formed when the area bounded by
π
2
2
the curve x = sec y between y=0 and y = is rotated about y4
axis.
Solution:
d
³
v = π x 2 dy
c
π
4
³
= π sec 2 ydy
0
π
= π[tan y ]04
π
ª
º
= π«tan − tan 0» = π[1 − 0]
4
¬
¼
= π Cubic units.
PART - B
1.
Find the area of the circle whose radius is ‘a’ units.
Solution:
2
2
2
Area of AOB bounded by x +y =a , the x-axis between x=0 and
x=a is
271
b
Area of
AOB = ³ ydx
a
a
=
³
a 2 − x 2 dx
0
a
ªx
a2
§ x ·º
=«
a2 − x 2 +
sin −1¨ ¸»
2
© a ¹»¼ 0
«¬ 2
ª§ a 2
º
a2
§ a ··
a − a2 +
sin −1¨ ¸ ¸ − (0 + 0)»
= Ǭ
¸
¨
2
© a ¹¹
«¬© 2
»¼
=
a2
.sin −1(1)
2
a 2 π πa 2
. =
2 2
4
∴ Re quired area = 4 × Area of AOB
=
= 4×
πa 2
4
∴ Area of circle = πa 2 Squnits.
Aliter
a
I
=
³
a 2 − x 2 dx
0
Let I
=
π
2
³
a2 − x 2 dx
0
=
π
2
³
a 2 − a 2 sin2 θ (a cos θ)dθ
Put x = a sin θ
0
=
π
2
³
(
)
a 2 1 − sin2 θ (a cos θ)dθ
dx = a cos θ dθ
0
272
=
π
2
³
(
)
a 2 1 − sin2 θ (a cos θ)dθ
when x = 0, θ = 0
0
π
2
= ³ a 2 cos θ cos θdθ
and x = a, θ =
0
π
2
π
2
= a 2 ³ cos2 θdθ
0
π
2
1 + cos 2A ·
§1 1
·
§
= a 2 ¨ + cos 2θ ¸dθ ¨∴ cos 2 A =
¸
2 2
2
¹
©
¹
0©
³
π
1 sin 2θ º 2
=a « θ+
2 2 »¼ 0
¬2
2ª1
a2
=
2
π
1
ª
º2
«θ + 2 sin 2θ»
¬
¼0
ªπ 1
1
§π· §
·º
« + sin 2/ ¨ / ¸ − ¨ 0 + sin 2(0)¸»
2
2
2
2
© ¹ ©
¹¼
¬
=
a2
2
=
a2 ª π 1
1 º πa 2
0
0
+
(
)
−
−
(0) =
2 «¬ 2 2
2 »¼
4
πa 2
= πa 2
4
2
Find the area bounded by the curve y=x +x+2, the x-axis and the
ordinate x=1 and x=2.
∴ Area of the circle = 4
2.
Solution:
b
³
a
³ (x
2
A = ydx =
2
)
+ x + 2 dx
1
º
ª x3 x2
=«
+
+ 2x »
2
¼»
¬« 3
2
1
273
ª 23 2 2
º ª 13 12
º
=«
+
+ 2(2)» − « +
+ 2(1)»
2
2
¬« 3
¼» ¬« 3
¼»
§8
· §1 1
·
= ¨ + 2 + 4¸ − ¨ + + 2¸
3
3
2
©
¹ ©
¹
§ 8 6 · § 2 + 3 + 12 ·
= ¨ + ¸−¨
¸
6
©3 1¹ ©
¹
26 17 52 − 17
−
=
3
6
6
35
=
Sq units
6
2
Find the area bounded by the curve y=x -6x+8 and the x-axis.
=
3.
Solution:
The curve meets the x-axis at y=0
2
y = 0 Ÿx -6x+8=0
Ÿ(x-2)(x-4)=0
Ÿx=2,4
b
³
A = ydx =
a
³ (x
4
2
2
)
4
§ x3
·
x2
− 6 x + 8 dx = ¨
−6
+ 8x ¸
¨ 3
¸
2
©
¹2
4
§ x3
·
=¨
− 3x 2 + 8x ¸
¨ 3
¸
©
¹2
274
ª§ 64
· §8
·º
= Ǭ
− 48 + 32 ¸ − ¨ − 12 + 16 ¸»
¹ ©3
¹¼
© 3
64
8
64 8
64 − 8 − 60
=
− 16 − − 4 =
− − 20 =
3
3
3 3
3
−4 4
Sq units.
=
=
3
3
4. Find the volume of a right circular cone of base radius r and
altitude h by integration.
Solution:
Let y=mx be rotated about the x-axis to get the cone.
r
Then m = tan θ = .
h
r
Now y = mx Ÿ y = x
h
b
h
2
§r ·
Volume V = π ³ y dx = π³ ¨ x ¸ dx
a
0©h ¹
2
=
πr 2
h2
h
³x
2
dx
0
h
=
V=
πr 2 ª x 3 º
πr 2
» = 2
2 «
h ¬« 3 ¼» 0 h
πr 2h3
h
2
=
ª h3
º
« − 0»
»¼
¬« 3
1 2
πr h Cubic units.
3
275
5. Find the volume of the sphere of radius r by integration.
Solution:
2
2
2
When the semi-circle x + y = r is rotated about X-axis solid
sphere is obtained.
b
Volume of the sphere
V = π³ y 2dx
a
(
)
(
)
r
= π ³ r 2 − x 2 dx
−r
r
= 2π³ r 2 − x 2 dx
( function is even )
0
r
ª
x3 º
= 2π«r 2 x −
»
3 »¼
«¬
0
ª
º
r3
= 2π«r 3 −
− (0 − 0)»
3
«¬
»¼
= 2π.
2r 3
4
v = πr 3cubic units.
3
3
276
6.
Find the volume of the solid generated when the region enclosed
2
3
2
by y =4x +3x +2x between x=1 and x=2 is revolved about x-axis.
Solution:
b
= π³ y 2 dx
v
a
³(
2
)
ª x4
x2 º
x3
= π 4x + 3 x + 2x dx = «4
+3
+2 »
2 »¼
3
«¬ 4
1
1
2
[
= π[(2
3
2
]
+ 2 ) − (1
= π x4 + x3 + x2
2
1
2
4
+ 23
+ 13 + 12
= π[(16 + 8 + 4) − (1 + 1 + 1)]
= π[(28 − 3)]
= 25π Cubic units.
7.
4
)]
Find the volume generated by the area enclosed by the curve
2
2
y =x(x-1) and the x-axis, when rotated about x-axis.
Solution:
The curve meets x-axis
∴y=0
2
Ÿy =0
2
Ÿx(x-1) =0
Ÿx=0,1
277
b
v
= π³ y 2 dx
a
1
= π³ x (x − 1) dx
2
0
(
1
)
= π³ x x 2 − 2x + 1dx
0
1
(
)
= π³ x 3 − 2x 2 + x dx
0
1
ª x 4 2x 3 x 2 º
= π«
−
+
»
3
2 »¼
¬« 4
0
ª§ 1 2 1 ·
º
= 𫨠− + ¸ − (0 − 0 + 0)»
¬© 4 3 2 ¹
¼
ª3 − 8 + 6º
ª1º π
= π«
» = π «12 » = 12 Cubic units.
12
¬
¼
¬ ¼
4.2. FIRST ORDER DIFFERENTIAL EQUATION
Introduction:
Since the time of Newton, physical problems have been
investigated by formulating them mathematically as differential
equations Many mathematical models in engineering employ
differential equations extensively.
Order and degree of Differential Equation:
The order of a differential equation is the order of the highest
differential coefficient appearing in the equation.
The degree of an equation is the degree of the highest
differential coefficient is free from radicals and fractional exponents.
278
Solution of First Order Differential Equation:
Recalling that we formed differential equation by differentiating
algebraic equations involving x,y etc and constants. Now, we will
consider the reverse process.
Consider the differential equation
dy
= 3x 2
dx
…(1)
3
A solution of this equation is y=x , since this satisfy (1) All the
Possibilities which satisfy are of the form
y = x3 + c
…(2)
This is called the general solution of (1), here c may be any
constant. W e call c an arbitrary constant. The general solution of (1)
has one arbitrary constant. For second order differential equation, two
arbitrary constants will be there.
A particular solution is one where a value is given to c. Particular
solutions arise when we are required to find a solution fitting certain
conditions.
If we give y=1 when x=0 in (2), we get c=1.
3
Hence the particular solution is y=x +1.
Solution of the variable separable differential equation:
dy
= f (x, y ) , if the
dx
function of x can be grouped with dx on one side and the function of y
can be grouped with dy on the other side, then this type of equation is
called variable separable differential equation. The solution can be
obtained by integrating both sides after separating the variables.
In the first order differential equation, say
279
1.
4.2 WORKED EXAMPLES
PART - A
Write the order and degree of the following differential equations
i
y=2
dy
d2 y
+3 2 +5
dx
dx
2
ii y’+y = 0
2
Iii (D +5D+4)y=e
iv
d2 y
dx
2
x
dy
+ 2y
dx
= 3x =
Solution:
i.
y=2
dy
d2 y
+3 2 +5
dx
dx
Here, Highest order=2
Degree of highest order=1
∴Order=2, degree=1
ii. y ' + y 2 = 0 Ÿ
dy
+ y2 = 0
dx
Here, Highest order=1
Degree of Highest order=1
∴Order=1, degree=1
iii
(D
2
Ÿ
)
+ 5D + 4 y = e x
d2 y
dx
2
+5
dy
+ 4y = e x
dx
Here, Highest order=2
Degree of Highest order=1
∴Order=2, degree=1
280
iv
d2 y
dx 2
+ 3x =
dy
+ 2y
dx
To eliminate the radical in the above equation, raising to the
2
§ d2 y
·
dy
+ 2y
power 2 on both sides, we get ¨ 2 + 3x ¸ =
¨ dx
¸
dx
©
¹
Here, Highest order=2
Degree of Highest order=2
∴Order=2, degree=2
2.
8
8
8
8
Solve x dx+y dy=0
Solution:
Given x dx+y dy=0
Integrating, we get,
³x
8
³
dx + y 8 dy = 0
x 9 y9
+
=c
9
9
3.
Solve
dy
2
=
dx 1 + x 2
Solution:
Given
dy
2
=
dx 1 + x 2
Ÿ dy =
Integrating, we get,
2
1+ x 2
dx
2dx
³ dy = ³ 1+ x 2
y = 2 tan −1 x + c
281
1
4.
dy §¨ 1 − y 2 ·¸ 2
Solve
=
dx ¨© 1 − x 2 ¸¹
Solution:
1
dy §¨ 1 − y 2 ·¸ 2
=
dx ¨© 1 − x 2 ¸¹
Given
1− y 2
dy
=
dx
1− x 2
dy
dx
=
Ÿ
2
1− y
1− x 2
Ÿ
Integrating, we get
-1
³
dy
1− y
2
=
³
dx
1− x2
-1
Sin y=sin x+c
5.
Solve
dy 3 + x
=
dx 3 + y
Solution:
Given
dy 3 + x
=
dx 3 + y
Ÿ (3 + y )dy = (3 + x )dx
Integrating, we get,
3y +
³ (3 + y )dy = ³ (3 + x )dx
x2
y2
= 3x +
+c
2
2
282
6.
Solve
dy
= e x − 5y
dx
Solution:
Given
dy
= e x − 5y
dx
dy
= e x e − 5y
dx
dy
Ÿ − 5 y = e x dx
e
Ÿ
Ÿ e 5 y dy = e x dx
³e
Integrating, we get
5y
³
dy = e x dx
e5 y
= ex + c
5
7.
Solve x
dy
=y
dx
Solution:
x
Given
Ÿ
Integrating, we get,
dy
=y
dx
dy dx
=
y
x
³
dy
=
y
³
dx
x
log y = log x + log c
Ÿ log y = log cx
Ÿ y = cx
283
8.
dy
− y cos x = 0
dx
Find the solution of Ÿ
Solution:
dy
− y cos x = 0
dx
Given
dy
= y cos x
dx
dy
Ÿ
= cos xdx
y
Ÿ
Integrating, we get,
³
dy
= cos xdx
y
³
Logy = sinx + c
PART - B
1)
2
2
Solve (x - y) dx + (y - x) dy = 0
Solution:
Given
2
2
(x - y) dx + (y - x) dy = 0
2
2
Ÿx dx –ydx + y dy –xdy = 0
2
2
2
2
Ÿx dx + y dy = xdy + ydx
Ÿx dx + y dy = d(xy)
Integrating, we get
³x
2
dx + y 2 dy = d(xy )
³
x3 y3
+
= xy + c
3
3
284
³
2)
Solve
dy 1 + cos 2y
+
=0
dx 1 + cos 2x
Solution:
Given
dy 1 + cos 2y
+
=0
dx 1 + cos 2x
Ÿ
dy
1 + cos 2y
=−
dx
1 + cos 2x
dy
2 cos 2 y
=−
dx
2 cos 2 x
dy
dx
Ÿ
=−
2
cos y
cos 2 x
Ÿ
Ÿ sec 2 ydy = − sec 2 xdx
Integrating, we get,
³ sec
2
³
ydy = − sec 2 xdx
tany = -tanx + c
Ÿtanx+tany=c
3)
Solve
dy
= e x − y + 3x 2e − y
dx
Solution:
dy
= e x − y + 3x 2 e − y
dx
Given
dy
= e x e − y + 3x 2 e − y
dx
dy
Ÿ
= e − y e x + 3x 2
dx
Ÿ
[
[
]
Integrating, we get. ³ e dy = ³ (e
]
Ÿ e y dy = e x + 3x 2 dx
y
y
x
3
e =e +x +c
285
x
)
+ 3x 2 dx
4)
Solve
dy y 2 + 4y + 5
=
dx x 2 − 2x + 2
Solution:
dy y 2 + 4y + 5
=
dx x 2 − 2x + 2
Given
Ÿ
Ÿ
dy
y + 4y + 5
2
dy
(y + 2)
2
Integrating, we get
+1
dx
=
=
x − 2x + 2
2
dx
(x − 1)2 + 12
dy
dx
³ (y + 2)2 + 12 = ³ (x − 1)2 + 12
-1
-1
tan (y+2)=tan (x - 1) + c
5)
x
2
x
x
2
x
Solve (1-e ) sec ydy + e tany dx = 0
Solution:
Given (1-e ) sec ydy + e tany dx = 0
x
2
x
Ÿ(1-e ) sec ydy = -e tany dx
Ÿ
sec 2 ydy − e x dx
=
tan y
1− ex
Integrating, we get
(
d(tan y )
d1− ex
=
³ tan y ³ 1 − e x
x
log (tany) = log (1-e ) + log c
x
Ÿlog (tany) = log c (1-e )
x
Ÿtany = c (1-e )
286
)
4.3. LINEAR TYPE DIFFERENTIAL EQUATION
A first order differential equation is said to be linear in y if the
dy
and y are unity.
power of the terms
dx
dy
+ Py = Q is a linear differential equation. Here P and Q are
dx
function of x. The solution of linear differential equation is given by
³
ye ³ pdx = Qe ³ pdx dx + c
y (IF) = ³Q(IF) dx + c
i.e.
Here IF = e ³ pdx is called an Integrating factor
Note:
e log f (x ) = f (x )
1.)
4.3 WORKED EXAMPLES
PART - A and B
dy 5
Find the integrating factor of
+ y=x
dx x
Solution:
Given
dy 5
+ y=x
dx x
Here P = 5
x
IF = e ³ pdx = e
5³
1
dx
x
= e 5 log x
= elog x
5
IF = x 5
287
2.)
Find the integrating factor of
dy
−x−y =0
dx
Solution:
Given
dy
−x−y =0
dx
Ÿ
dy
−y = x
dx
Here P=-1
IF = e ³ pdx = e ³ − dx = e − x
3.)
Find the integrating factor of
dy
− sin 2x = y cot x
dx
Solution:
Given
dy
− sin 2x = y cot x
dx
Ÿ
dy
− y cot x = sin 2x
dx
Here P = −cotx
IF = e ³ pdx
= e − ³ cot xd x
= e − log sin
x
= elog cos ecx
IF = cos ecx
288
PART - B
1.)
Solve
dy
1
+ 3y =
dx
3
Solution:
Given
dy
1
+ 3y =
dx
3
Here P = 3 and Q =
1
3
Now, IF= e ³ pdx
= e 3 ³ dx = e 3 x
The Required solution is
y × IF = ³ Q × IF dx + c
1
y e3 x = ³ e3 x dx + c
3
Ÿ y e3 x =
2.)
e3 x
+c
9
dy 3 x 2 y
2
+
=
3
dx 1 + x
1+ x3
2
3x
2
Here P =
and Q =
3
1+ x
1+ x3
Solve
Now, IF = e ³ pdx
=e
³
3x2
1+ x 3
dx
= eLog(1+ x
3
)
IF = 1 + x 3
The Required solution is
y (IF ) = ³ Q(IF )dx + c
y(1 + x 3 ) =
2
3
³ 1 + x 3 (1 + x )dx + c
( )
y (1 + x ) = 2x + c
y 1 + x 3 = 2³ dx
3
289
3.)
dy
+ y cot x = 2 cos x
dx
Solve
Solution:
dy
+ y cot x = 2 cos x
dx
Here P = Cotx , Q= 2 Cos x
Given
Now, If = e ³ pdx
= e ³ cot xdx = elog sin x = sin x
The Required Solution is
y (IF ) = Q (IF )dx + c
³
y sin x = ³ 2 cos x sin xdx + c
Ÿ y sin x = ³ sin 2xdx + c
Ÿ y sin x = −
4.)
cos 2x
+c
2
dy 3
− y = x 3 cos x
dx x
Solve
Solution:
dy 3
− y = x 3 cos x
dx x
Here P = − 3 ,Q = x 3 cos x
x
Given
Now, IF = e ³ pdx
=e
1
³ − 3 dx
x
= e − 3 log x = elog x
The Required solution is
y (IF ) = ³ Q (IF )dx + c
y
1
x3
= ³ x 3 cos x
Ÿ
Ÿ
y
x3
y
x3
1
x3
dx + c
= ³ cos xdx + c
= sin x + c
290
−3
= x −3 =
1
x3
5.)
(
Solve 1 + x 2
) dy
+ 2xy = 1
dx
Solution:
(
) dy
+ 2xy = 1
dx
Divide both sides by (1+ x ) , we get
Given 1 + x 2
2
dy
2xy
1
+
=
2
dx 1 + x
1+ x 2
2xy
1
,
Q=
Here P=
2
1+ x
1+ x2
Now If = e ³ pdx
2x
= e 1+ x
2
dx
= elog(1+ x
2
)
= 1+ x2
The Required solution is
(
y (IF ) = ³ Q (IF )dx + c
) ³ 1 +1x (1 + x )dx + c
Ÿ y (1 + x ) = ³ dx + c
Ÿ y (1 + x ) = x + c
y 1+ x2 =
2
2
2
2
EXERCISE
PART - A
1. Find the area bounded by the curve y= 2x the x-axis and the
ordinates x =0 and x=1.
2. Find the are bounded by the curve y=x² and x-axis between x=0
and x=2
3. Find the area bounded by the curve y =
x=1, and x=3.
291
x2
, x-axis, and between
2
4.
Find the area under the curve y =
1
1+ x 2
x- axis,
x = -1 and x = 1
5.
Find the area bounded by the curve y=sinx, x-axis and between
x=0 and x = π
6.
Find the area bounded by the curve y²=3x, the x-axis and line
x=3
7.
Find the area bounded by the curve xy=1 the y-axis and the lines
y=1 and y=5.
8.
Find the area bounded by the curve x=2y+5, the y-axis and the
lines y=1 and y=2.
9.
Find the volume of the solid formed when the area bounded by
the curve y²=25x³ between x =1 and x =3is rotated about x- axis.
10. Find the volume of the solid formed when the area bounded by
the curve y²=8x between x=0 and x=2 rotated about x-axis.
11. Find the volume generated by rotating the triangle with vertices
at (0,0),(3,0) and (3,3) about x-axis.
12. Find the volume generated when the area bounded by the curve
x²=3y² between y=0 and y=1 is rotated about y-axis.
13. Write the order and degree of the following differential equations.
(i)
4
§ d2 y ·
dy
+¨ 2 ¸ +
+ y = ex
3
¨
¸
dx
dx
© dx ¹
d3 y
(ii) y11 + y 2 = 0
(iii) y"=
(iv)
d2 y
dx 2
dy dx
+
dx dy
dy ·
§
= ¨1 +
¸
dx ¹
©
1
3
14. Solve xdx+ydy=0
292
15. Solve
dy x 2
=
dx y 2
dy
= 3x 2 y
dx
dy 1 + x
=
dx 1 + y
16. Solve
17.
dy
= xy + y + x + 1
dx
dy
= e 2x − y
19. Solve
dx
dy − y
=
20. Solve
dx
x
18.
dy
+ y sin x = 0
dx
dy 1
22. Find the integrating factor of
+ y=x
dx x
21. Find the solution of
23. Find the integrating factor of
dy
1
+
y =1
dx 1 + x 2
24. Find the integrating factor of
dy
− y tan x = e x sec x
dx
25. Find the integrating factor of 2 cos x
1)
dy
+ 4y sin x = sin 2x
dx
PART - B
Find the area of the circle whose radius is 4 units using
integration.
2)
Find the area of region bounded by the curve y= 3x 2 − 4x + 5 ,
the x-axis and the lines x=1 and x=2
3)
Find the area bounded by the curve y = x 2 + x + 1 and x-axis and
the ordinates x=1 and x=3
293
4)
Find the area bounded by the curve y = 4x − x 2 and the x-axis
5)
Find the area bounded by the curve y = 10 − 3x − x 2 and the xaxis
6)
Find the volume of the solid formed when the area bounded by
the curve y 2 = 2 + x − x 2 the x-axis and the lines x=-1 and x=2.
7)
Find the volume of the solid formed when the area bounded by
the curve x 2 =
(
)
a2
b2
(b
2
)
− y 2 , the y-axis and the lines y=-b and y=b
(
)
8)
Solve xy 2 + x dx + yx 2 + y dy = 0
9)
Solve
dy 1 + cos y
=
dx 1 + cos x
10) Solve
dy
= e x + y + xe y
dx
11) Solve
dy
= e 3 x − 2 y + x 3 e − 2y
dx
12) Solve
dy y 2 + 2y + 10
=
dx x 2 + 2x + 10
( )
Solve (1 + e )sec
13) Solve 1 + x 2 sec 2 ydy = 2x tan ydx
14)
x
2
ydy − e x tan ydx = 0
(
)
15) Solve 3e x tan ydx + 1 + e x sec 2 ydy = 0
(
)
16) Solve e x + 1 cos ydy + e x sin ydx = 0
17) Solve
dy
+ 3y = 6
dx
18) Solve
dy y
+ = x4
dx x
19) Solve
dy
2xy
1
+
=
dx 1 + x 2 1 + x 2
294
20) Solve
dy
+y = x
dx
21) Solve
dy
+ y cot x = e x cos ecx
dx
22) Solve
dy
+ y tan x = 4x cos x
dx
23) Solve
dy 2y
−
= x 2 sin x
dx
x
24) Solve x
(
dy
− 3y = x 4 e x
dx
25) Solve 1 + x 4
1
) dy
+ 4x y =
dx
1+ x
3
4
ANSWERS
PART - A
8
Sq unit
2)
3
3)
13
Sq unit
3
1)
1 Sq unit
4)
π
Sq unit
2
5) 2 Sq unit
6) 6 Sq unit
7)
log 5 Sq unit
8) 8 Squnit
9) 500 π cubic unit
10) 16 π Cubic units
11) 9 π cubic units
12) π cubic units
13) (i) order =3, degree=1
(iii) order =1, degree=2
14)
x2 y2
+
=c
2
2
17)
y+
y2
x2
=x+
+c
2
2
(ii) order =2, degree=1
(iv) order =2, degree=3
15)
y3 x3
−
=c
3
3
18) Log (1 + y ) =
295
16) log y = x 3 + c
x2
+x +c
2
19)
ey =
22)
X
e 2x
+c
2
20) xy=c
−1
(23) e tan x
21) logy=cosx+c
(25) sec 2 x
(24) Cos x
PART - B
(3)
44
Sq. units
3
(6)
9π
Cubic units
2
1)
16π Sq. units
(2) 6 Sq. units
4.)
32
Sq. units
3
(5)
7)
4 2
πa b cubic units
3
9)
tan
11)
e 2y e 3 x x 4
=
+
+c
2
3
4
13)
tan y = c 1 + x 2
15)
tan y 1 + e x
17)
ye3 x = 2e3 x + c
19)
y 1+ x 2 = x + c
)
(20)
ye x = xex − e x + c
21)
y sin x = e x + c
(22)
y sec x = 2x 2 + c
23)
y
+ cos x = c
x2
(24)
25)
y 1+ x 4 = x + c
y
x
= tan + c
2
2
(
(
(
(
)
3
343
Sq. units
6
(8)
x2 + 1 y2 + 1 = c
(10) e x + e − y +
x2
=c
2
§ y + 1·
−1§ x + 1 ·
(12) tan −1¨
¸ − tan ¨
¸=c
© 3 ¹
© 3 ¹
)
14) tan y = c 1 + e x
(
=c
16)
(e + 1)sin y = c
x
(18) xy =
)
296
)
y
x3
x6
+c
6
= ex + c