Model Answer Solution
(SPPU, Dec 2014 Exam)
Engineering Mathematics – III
S.E( Mech /Auto), (2012 Pattern)
Q1)(a) Solve any two of the following
i)

(D 2 -4D+4)y=e x .cos 2 x
A.E. is
D 2 - 4D+4=0
∴ D = 2,2
y c = (c 1 x+c 2 )e 2 x
yp =
1
1
= ex
= ex
=
=
=
=
=
e x cos 2 x
(D−2)
ex
2
ex
2
ex
2
ex
2
ex
2
cos 2 x
(D−1)2
1
(D−1)2
[
1
(D−1)2
1+cos2x
(
2
eox +
1
2
(D −2D+1)
cos2x]
1
[1 − (2D+3) cos2x]
[1 −
[1 +
[1 −
(2D−3)
4D2−9
1
25
1
25
cos2x]
(2D − 3) cos2x]
(4sin2x + 3cos2x)]
y c = (c 1 x+c 2 )e 2 x +
ii)
)
𝐞𝐱
𝟐
[𝟏 −
𝟏
𝟐𝟓
(𝟒𝐬𝐢𝐧𝟐𝐱 + 𝟑𝐜𝐨𝐬𝟐𝐱)]
(D 2 +5D+6)y = e - 2 x .sec 2 x (1+2tanx)
 ( D 2 +5D+6)y = e - 2 x .sec 2 x (1+2tanx)
A.E.is D 2 +5D+6 = 0
(D+2) (D+3) = 0

D= -2, -3
C.F. = c 1 e - 2 x +c 2 e - 3 x
P.I. =
=
1
(D+3)(D+2)
1
(D+3)
[e − 2x. sec2x (1 + 2tanx)]
[e − 2x ∫ e2x. e − 2x sec2x (1 + 2tanx)dx]
= (D+3) [e − 2x ∫ sec2x (1 + 2tanx)dx]
1
Put tanx = t

sec 2 xdx=dt
= (D+3) [e − 2x ∫(1 + 2t) dt]
1
= (D+3) [e − 2x (t + t2)]
1
= (D+3) [e − 2x (tanx + tan2x)]
1
=e - 3 x ∫ e3x. e − 2x[(tanx − 1) + sec2x]dx
=e - 3 x ∫ ex[(tanx − 1) + sec2x]dx
=e - 3 x [ex[(tanx − 1)]]
(∴ ∫ ex(f(x) + f ′ (x))dx = e x f(x)
=e
-2x
(tanx-1)
Hence the complete solution is
y=c 2 e - 3 x +e - 2 x (c1 + tanx − 1)
y=c 2 e - 3 x +e - 2x (𝐂𝟑 + 𝐭𝐚𝐧𝐱)
iii)
𝐝𝟐 𝐲
𝐝𝐱𝟐
+ 𝐲 = 𝐜𝐨𝐬𝐞𝐜𝐱,
By variation of parameters.

A.E. is
D 2 +1=0
∴ D = ±i
c.f = Acosx + Bsinx
= Ay 1 +By 2
Here y 1 = cosx & y 2 = sinx
Let P.I = uy 1 +vy 2
y
y2
cosx sinx
w= | 1
| =1
|=|
y′1 y′2
−sinx cosx
u=∫
=∫
dx
w
−sinxcosecx
1
and v = ∫
=∫
−y2x
−y1x
D
dx = ∫ −dx = −x
dx
cosxcosecx
dx = cotxdx
1
= log sinx
∴ P. I. = (−x)cosx + {log(sinx)} sinx
Hence the general solution is
y = Acosx + Bsinx – xcosx + sinx log (sinx)
b) Find the fourier transform of:
𝐱 ≤
f(x) = 1
= 0 x >a
 The fourier transform of f(x)
∞
F(λ) = ∫−∞ f(u)e
a
= ∫−∞ f(u)e
a
= ∫−∞(1)e
− i λu
− i λu
− i λu
du
∞
du + ∫a f(u)e
du +
∞
∫a (0)e
− i λu
− i λu
du
−
du = [
e
iλu −a
−iλ
]
a
=
F(λ)=
e+iλa−e−iλa
iλ
2sinλa
λ
λ ≠0
For λ = 0,
We obtain F(𝛌) = 2a
Q2) a) Weight of 1N, stretches a spring 5 cms, A weight of 3N is
attached to the spring & weight W is pilled 10 cm below the
equilibrium position & released. Determine the position & the
velocity as a function of time.
 Since a weight
W = 1N, stretches a spring
5cm = 0.05m,
By Hooke’s law
Fo = kSo
1 = K X 0.05
∴k=
1
100 20N
=
(or kg/s 2
0.05
5 m
By Newton′ ssecond law,
The eq n of motion of the body is,
mX
or
d2 x
dt2
w d2 x
9 dt2
= −kx
= −kx
1 d2 x
= −20x
9.8 dt 2
d2 x
≠ 196x = 0
dt 2
Which is L. D. E. with cont. coeff ′ s
A.E. is
D2 + 196 = 0
∴ D2 = −196
D = ± √−196
= ± 2√49 i
G. S. is
x=C 1 cos(2√49) t + C2 sin(2√49 )t
………(1)
Differentiating (1) w.r.t t, we get
dx
dt
. . … … (2)
= C 1 sin 2√49 t. 2√49 + C2 cos 2√49 t .2√49
Using x = 0.01 at t = 0 in eq n (1), we get 0.01 = C 1 (1) + C 2 (0)
 C 1 = 0.01
Using
dx
dt
= 0 at t = 0 in eqn (2), we get 0 = 0 + C2. 2√49
 C2 = 0
1) X(t) = (0.01) cos 2 √𝟒𝟗𝐭
2) V =
𝐝𝐱
𝐝𝐭
= - (0.01) 2 √𝟒𝟗 sin 2 √𝟒𝟗 t
b) Solve any one of the following
i) Find Laplace transform of the function f(t) =
We have
L[e−at − e−bt ] =
∴ L[
e−at −e−bt
t
1
s+a
-
∞ 1
] = ∫s
s+a
1
s+a
−
1 ds
s+a
= [log(s + a) − log(s + b]∞
s
= [log
a
s
b
1+
s
1+
∞
]
s
𝐞−𝐚𝐭 −𝐞−𝐛𝐭
𝐭
= log 1 = 0-
a
s
b
log 1+
s
log 1+
log s+a
log s+b
𝐋[
𝐞−𝐚𝐭 −𝐞−𝐛𝐭
]=
𝐭
𝐥𝐨𝐠 𝐬+𝐛
𝐥𝐨𝐠 𝐬+𝐚
ii) Find inverse Laplace transform of the function
F(s) =
Let
𝟑𝐬+𝟏
(𝐬−𝟏)(𝐬 𝟐 +𝟏)
3s+1
(s−1)(s2 +1)
=
A
(s−1)
+
Bs+C
……(1)
(s2 +1)
Multiplying both sides (i) by (s − 1)(s 2 + 1),
We obtain
3s + 1 = A (s 2 + 1) + (Bs + C) (s − 1)
……(2)
Putting s= 1, we get
4 = A(2)
∴A = 2
To determine B & C, equate coefficients of like powers of s 2
& constant terms in (ii) we get,
1=A–C
0 = A +B &
B = -A = -2
&
C =A – 1 = 1
Thus,
L−1 [
3s+1
(s−1)(s2 +1)
] = L−1 [
2
s−1
= 2 L−1 [
𝐋−𝟏 [
−
1
s−1
−2s+1
s2 +1
]
]- 2L−1 [
s
s2 +1
]
𝟑𝐬 + 𝟏
] = 𝟐𝐞𝐭 − 𝟐𝐜𝐨𝐬𝐭 + 𝐬𝐢𝐧𝐭
𝟐
(𝐬 − 𝟏)(𝐬 + 𝟏)
c) Solve the following equation using Laplace transform method
𝐲 𝟏𝟏 (t) - 3𝐲 𝟏 (t) + 2y(t) = 12. 𝐞−𝟐𝐭 ;
y(0) = 2, 𝐲 𝟏 (0)= 6
L{y11 (t)– 3y1 (t) + 2y(t)} = L {12. e−2t }
L{y11 (t)} – 3L{y1 (t)} + 2L { y(t)} = 12.
1
(s+2)
s 2 Y(s) – sy(0) – y1 (0) − 3 {sY(s) − y(0)} + 2Y(s) =
s2 Y(s) – sy(0) – y1 (0) − 3sY(s) + 3Y(0) + 2Y(s) =
(s 2 - 3s + 2) Y(s) – S(2) – 6+3(2) =
(s 2 - 3s + 2) Y(s) – 2s =
(s 2 - 3s + 2) Y(s) =
(s 2 - 3s + 2) Y(s) =
Y(s) =
12
s+2
12
s+2
12
s+2
12
s+2
12
s+2
+ 2s
12+2s2 +4s
(s+2)
(2s2 +4s+12)
(s+2)((s2 −3s+2)
(2s2 +4s+12)
=
(s−1)((s−2) (s+2)
Using method of partial fractions, we can express Y(s) in the
form
Y(s) =
−6
(s−1)
+
7
(s−2)
+
1
(s+2)
Taking the inverse Laplace transform of both slides, we get
y(t) = -6𝐞𝐭 + 𝟕𝐞𝟐𝐭 + 𝐞−𝟐𝐭
Which is required solution.
Q3) a) If the directional derivative of ∅ = 𝐚𝐱𝐲 + 𝐛𝐲𝐳 +
𝐜𝐱𝐳 𝐚𝐭 (𝟏, 𝟏, 𝟏) has maximum 4 in a direction parallel to x axis.
Find the values of a,b,c

∂∅
∂x
= ay + cz,
∂∅
∂y
= ax + bz,
∂∅
∂z
= by + cx,
∇∅ = i̅(ay + cz) + j̅(ax + bz) + k̅(by + cx)
∇∅(1,1,1) = (a + c)i̅ + (a + b)j̅ + (b + c)k̅
Now,
(a + c)i̅ + (a + b)j̅ + (b + c)k̅ = 4i̅
(given)
∴a+c= 4
a+b=0
b+C=0
Which gives,
a = 2, b = -2, c = 2
b) Show that the vector field,
𝐅̅ = (𝟐𝐱𝐳 𝟑 + 𝟔𝐲)𝐢̅ + (𝟔𝐱 − 𝟐𝐲𝐳)𝐣̅ + (𝟑𝐱 𝟐 𝐳 𝟐 − 𝐲 𝟐 )𝐤̅
is irrotational & find scalar function ∅ such that 𝐅̅ = 𝛁∅
i̅
 ∇ X F̅ =
k̅
j̅
∂⁄
∂⁄
∂⁄
∂x
∂y
∂z
(2xz 3 + 6y) (6x − 2yz) (3x 2 z 2 − y 2 )
= i̅ [
+ j̅ [
∂
∂y
∂
∂y
(3x2 z 2 − y 2 ) −
(3x 2 z 2 − y 2 ) −
∂
+ k̅ [ (6x − 2yz) −
∂y
∂
∂z
∂
∂z
∂
∂y
(∂x − 2yz)]
(2xz 3 + 6y)]
(2xz 3 + 6y)]
= i̅ (- 2y + 2y) – j̅ (6xz 2 − 6xz 2 ) + k̅(6 − 6)
=0
Which shows that F̅ is irrotational to find corresponding scalar ∅,
Consider the relation
̅̅̅
d∅ ≡ ∇∅ . dr
but
F̅ ≡ ∇∅
∴
d∅ ≡ F̅. ̅̅̅
dr
≡ {(2xz 3 + 6y)i̅ + (6x − 2yz)j̅ + (3x 2 z 2 − y 2 )k̅ } .
(i̅ dx + j̅dy + k̅dz)
≡ (2xz 3 + 6y)dx + (6x − 2yz)dy + (3x 2 z 2 − y 2 )dz
≡ d(x 2 z 3 ) + (6d (xy) + d (−y 2 z)
Integrating we get,
∅ = 𝐱 𝟐 𝐳 𝟑 + 𝟔𝐱𝐲 − 𝐲 𝟐 𝐳 + 𝐂
c) Find the coefficient of correlation for the following data:
x
y
10
18
14
12
18
24
22
6
26
30
30
36
 Preparing table as follows calculation can be made simple
x
y
x2
y2
xy
10
18
100
324
180
14
12
196
144
168
18
24
324
576
432
22
6
484
36
132
26
30
676
900
780
30
36
900
1,296
1080
T o tal 120
126
2680
3,276
2,772
Here n =6
x̅ =
∑x
120
=
= 20
N
6
Y̅ =
∑Y
126
=
= 21
N
6
r =
=
=
=
∑ xy−nx̅y
̅
̅2 )
√(∑ x2 −nx̅2 ) X(∑ y2 −ny
2772−6 X 20 X 21
√(2680−6 X 20 X 20)X (3276−6 X 21 X 21)
252
√(280)X (630)
252
420
= 0.6
r = 0.6
Q4) a) Porte the following (any one)
1) 𝛁 𝟒 𝐞𝐫 = 𝐞𝐫 +

𝟒
𝐞𝐫
𝐫
∇4 er = ∇2 ∇2 (er )
Let f (r) = er
 f 1 (r) = er , f 11 (r) = er , ∇2 f(r) = f 11 (r) +
∇2 (er )
2
=
r
2
er + er = ( + 1) er
r

f 1 (r) = er ( + 1) + er ( 2 )
r
r
2
−2
2
−2
2
4
r
r2
= er ( + 1 −
&
r
F1 (r) =
4
r2
er +
2 r
e
r
∇4 er = ∇2 (∇2 er )
−
4
+ 3)
r
4
r3
er
4
r3
… … … (1)
… … … . by (1)
r
f 11 (r) = er ( + 1) + er ( 2 ) +
r
r
2
f 1 (r)
2
f(r)
∴
r
= er ( + 1)
Let
∴
2
er -
2
r2
er
2
= ∇2 {er ( + 1)}
r
2
4
r
r2
= er { + 1 −
∇4 er =
+
4
r3
+
4
r2
2
4
r
r3
+ −
}
……..by(1)
4 r
e + er
r
𝟏
𝐧(𝐧−𝟐)
ii) 𝛁. [𝐫 𝛁 ( 𝐧 )] =
𝐫
𝐫 𝐧+𝟏
1
1
= ∇r . ∇ ( n) +r.∇2 ( n)
r
r
= ∇r . ∇(r −n ) +r.∇2 (r −n )
−nr−n−1
1
= ( ) r̅. (
r
r̅
−n
r
rn+2
= .
=
=
=
=
=
=
=
−nr2
rrn+2
−n
rn+ 1
r
) . r̅ + r. ∇. (∇r −n )
−nr̅
+ r. ∇ ( n+2)
r
+ r.∇ (
−nr̅
rn+2
)
+ r.[−n ∇ (r −n−2 ). r̅ + (−n). r −n−2 (∇. r̅)]
r̅
−n
rn+ 1
−n
rn+ 1
−n
rn+ 1
−n
rn+ 1
+ r.[−n. (−n − 2)r −n−3 . r̅ + (−n)r −n−2 (3)]
r
+ r. [
+ r.[
+
n(n+2)r−n−3
r
n(n+2)
rn+2
n(n+2)
rn+1
n2 +n+3(−n)
rn+1
+
=
+
3(−n)
+
rn+2
]
3(−n)
rn+2
]
3(−n)
rn+1
n2 −2n
rn+1
=
n(n−2)
rn+1
𝟏
𝛁. [𝐫 𝛁 ( 𝐧 )] =
𝐫
𝐧(𝐧−𝟐)
𝐫 𝐧+𝟏
b) Between 2p.m. & 3p.m. the average number of phone calls per
minute coming into company are 2. Find the probability that
during one particular minute there will be 2 or less calls.
 The p.m.f. is given by
P(r) =
P(r) =
zr e−2
R = 0,1,2
r!
e−2 2r
r!
p[r≤2] = p [r = 0] + p[r = 1] + p[r = 2]
=
e−2 20
o!
+
e−2 21
1!
e−2 22
+
2!
= e−2 + 2e−2 + 2e−2
= 5e−2
p[r≤2] = 5 X 0.1353
p[r≤2] = 0.6767
c) Calculate the first four moments about the mean of given
distribution also find 𝛃𝟏 & 𝛃𝟐
x
f
2
5
2.5
38
3
65
3.5
92
4
70
4.5
40
5
10
 Taking A = 3.5, h = 0.5
&u=
X−A
h
=
X−3.5
0.5
We prepare the table for calculating μ11 , μ12 , μ13 , & μ14
x−3.5
fu
fu2
fu3
fu4
-3
-15
45
-135
405
38
-2
76
152
-304
608
3
65
-1
-65
65
-65
65
3.5
92
0
0
0
0
0
x
f
2
5
2.5
u=
0.5
4
70
1
70
70
70
70
4.5
40
2
80
160
320
640
5
10
3
30
90
270
810
176
582
156
2598
Total
∑ f = 320
For moments about arbitrary mean A = 3.5,
∑ fur
μ′r = h
=
∑f
r
μ′1 = h
∑ fu
176
= (0.5)
= 0.275
∑f
320
μ′2 = h2
∑ fu2
582
= (0.5)2 =
= 0.4547
∑f
320
∑ fu3
156
μ′3 = h
= (0.5)3 =
= 0.0609
∑f
320
3
∑ fu4
2598
μ′4 = h
= (0.5)4 =
= 0.5074
∑f
320
4
μ1 = 0
μ2 = μ′ 2 − μ′1
= (0.4547) − (0.275)2
= 0.3791
μ3 = μ′3 − 3μ′ 2 μ′1 + 2 (μ′1 )3
= (0.0609) – 3(0.4547).(0.275) + 2(0.275)3
= 0.0609 – 1.3641 . 0.275 + 0.0416
= 0.0609 – 0.3751 + 0.0416
= - 0.2726
2
μ4 = μ′4 − 4μ′ 3 μ′1 + 6μ′ 2 (μ′1 ) − 3(μ′1 )
4
= 0.5074 – 4 X 0.0609 X 0.275 + 6 X 0.4547
X (0.272)2 – 3(0.275)4
= 0.5074 – 0.06699 + 0.2063 – 0.01716
= 0.62955
β2 =
β2 =
μ23
μ32
μ4
μ22
=
=
(−0.2726)2
(0.3791)3
0.62955
(0.3791)2
0.07431
=
=
0.05448
0.62955
0.1437
= 1.3640
= 4.3807
̅̅̅ for:
Q5) (a) Evaluate ∫𝐜 𝐅̅. 𝐝𝐫
𝐅̅ = 3𝐱 𝟐 𝐢̅ + (𝟐𝐱𝐳 − 𝐲)𝐣̅ + 𝐳𝐤̅
Along the straight line joining (1,2,1) & (2,1,3)
̅̅̅)
Ler I = ∫c F̅. ̅̅̅
dr = ∫C (3x 2 dx + (2xz − y)dy + zdz
Equation of line joining the points (1,2,1) to (2,1,3)
x−1
−2+1
=
y−2
−1+2
z−1
=

−3+1
x−1
−1
y−2
=
1
=
z−1
Multiplying each ratio by -1
−2
x - 1 = +t,
-y + z = t,
z – 1= +2t
x = +t+1,
y = -t + 2,
z = +2t +1
dx = dt
dy = - dt
dz = 2dt
1
∴ W = ∫ {3(t + 1)2 . dt + [2. (t + 1)(2t + 1) − (2 − t)]. (−dt)
0
+ (2t + 1). 2dt}
1
= ∫0 {3(t + 1)2 + [2(2t 2 + t + 2t + 1) − 2 + t]. (−dt) +
(4t + 2)dt}
1
1
1
W = ∫0 {3(t + 1)2 . dt + ∫0 (4t 2 + 7t)(−dt) + ∫0 (4t + 2)dt }
=[
=[
=
3(t+1)3
3
3(t+1)3
3
3(1+1)3
3
−
−
−
4t3
3
4t3
3
4(1)3
3
−
−
−
7t2
2
3t2
2
+
1
4t2
+ 2t]
2
1
+ 2t]
3(1)2
2
0
+ 2(1)
0
=
24
3
−
4
3
8+9
= 10 – (
= 10 =
60−17
6
3
−
6
2
+2
)
17
6
43
=
6
b) Evaluate
𝐝𝐬 where 𝐅̅ = (𝐱 + 𝐲 𝟐 )𝐢̅ − 𝟐𝐱𝐣̅ + 𝟐𝐲𝐳 𝐤̅
∬ 𝐅̅. ̅̅̅
& s is the surface of the tetrahedron bounded by the coordinate
planes & the plane. : 2x + y + 2a = 6
 The surface s of tetrahedron has four faces
Let us first evaluate the volume integral. Given plane cut – off
intercept 3,6,3 on x,y& z axes res p.
∇. F̅ =
∂
∂
∂
(x + y 2 ) +
(−2x) +
(2yz)
∂x
∂y
∂z
z
= 1 + 0 +2y = 1+2y
C ( 0,0,3 )
z
3
6−2x
∭v ∇. F̅ dv = ∫0 ∫0
3
6−2x
= ∫0 ∫0
=
(6−2x−y)/2
(1 + 2y)dxdydz
∫0
(1 + 2y) [z](6−2x−y)/2
dxdy
0
3 6−2x
(1
∫0 ∫0
+ 2y)
( 0,6,0 )
z
0
(6−2x−y)
dxdy
2
B y
z
A
2x+y=6
( 3,0,0 )
z
=
=
3
{(6
∫
2 0
1
1
2
y2
− 2x) (y + 2. ) −
2
y2
2
y3
−2 }
3
3
∫0 [(6 − 2x){6 − 2x + (6 − 2x)2 } ∫
x)3 ] dx
∴ volume integral
=
1
2
[
(6−2x)3
(−6)
+
(6−2x)4
(−8)
+
3
12
+
6−2x
(6−2x)4
12
3
]
0
dx
0
(6−2x)2
2
x
−
2
3
(6 −
1
= [36 + 162 − 18 − 108] = 36
2
To evaluate the surface integral’s consider the four surfaces
s 1 [plane ABC (s 2 )plane z = 0], s 3 (plane y = 0)
First consider the surface s 1 whose eq n is
2x +y +2z – 6 = 0
Let ∅ = 2x +y +2z – 6
∂∅
∂x
∂∅
= 2,
∂y
∂∅
= 1,
∂z
=2
∇∅ = 2i̅ + j̅ + 2k̅
n̂ =
̅
2i̅+ j̅+2k
√4+1+4
=
̅
2i̅+ j̅+2k
3
̅
2i̅+j̅+2k
f.̅ n̂ = {(x + y 2 )i̅ − 2xj̅ + 2yzk̅} . {
}
3
=
1
3
{2(x + y)2 − 2x + 4yz}
Let ds be an element of area in plane ABC
Taking its projection in xoy plane, we get dscos θ = dxdy,
Where θ is angle between normals to the surfaces s1 & xoy plane resp.
Unit normal to the xoy plane is k̅
∴ cosθ = n̂ . k̅
or ds =
dxdy
̅|
|n̂ .k
In the problem,
n̂
=
n̂ . k̅ =
ds
=
̅
2i̅+ j̅+2k
3
̅)
(2i̅+ j̅+2k
3
dxdy
2/3
=
1
3
2
. k̅ =
2
3
dxdy
3
I1 = ∬s1 {2(x + y)2 − 2x + 4yz} . dxdy
3
2
Putting
6−2x−y
z=
2
3
1
6−2x
I1 = ∫0 ∫0
2
=
3
y3
{2
∫
2 0
3
1
6−2x−y
{2(x + y)2 − 2x + 4y (
+ (6 − 2x). 2
1
3
1
(6−2x)4
y2
2
−2
y3
2
)} dxdy
6−2x
}
3 0
dx
= ∫0 (6 − 2x)3 dx
2
=
2
{
−8
3
}
0
= 81
Next consider the surface S 2 ,
(plane z =0)
n̂ = −k̅
F̅. n̂ = F̅. (− k̅) = −2yz
Ds = dxdy
I = ∬ −2yzdxdy = 0
as z = 0
Now consider the surface S 3
(plane y = 0)
F̅. (− j̅) = 2x
n̂ = −j̅
3
3−x
I 3 = ∫0 ∫0
=
3
∫0 2x(3
1
2x dxdz = ∫0 2x [z]3−x
0 dx
− x)dx = {6
x2
2
x3
−2 }
3
3
0
= 27 - 18= 9
Lastly consider the surface S 4
(plane x = 0)
F̅. (− i̅) = −(x + y 2 )
n̂ = −i̅
I 4 = ∬s −(x + y 2 )dydz
4
6
(6−y)/2
= ∫0 ∫0
=-
1
2
6
6
(6−y)/2
−y 2 dz = − ∫0 y 2 [z]0
∫0 y 2 (6 − y)dy
dy
==-
1
2
1
2
[
6y3
3
−
y4
6
]
4 0
[2X216 − 324] =
−108
2
= −54
Surface integral = I 1 +I 2 +I 3 +I 4
Volume inte. = 36
Div Theorem verified
c) Evaluate
∫𝐜 (𝐱𝐲𝐝𝐱 + 𝐱𝐲 𝟐 𝐝𝐲)
By stoke’s theorem, where C is the square in x -y plane with
vertices (0,0), (1,0), (1,1), (0,1).
 To verify stoke’s theorem, we prove that,
∮ F̅. ̅̅̅
dr = ∬(∇ X F̅)n̂. ds
c
s
∬s (∇ X F̅)n̂. ds
(∇ X F̅)
i̅
j̅
k̅
= ∂/ ∂x ∂/ ∂y ∂/ ∂z
xy
xy 2
0
= i̅( 0 – 0)-j̅ (0 − 0) + k̅(y 2 − x)
n̂ = k̅
∴ (∇ X F̅)n̂ = k̅(y 2 − x) . k̅
= y2 – x
ds = dxdy
1
1
∬(∇ X F̅). n̂ds = ∫ ∫(y 2 − x). dxdy
s
x=0 y=0
1
y3
1
= ∫x=0 [ − xy] dx
3
0
x=1 1
= ∫x=0 [ − x] dx
3
x2
3
2 0
= [ x−
1
=
1
1
3
−
]
1
2
−1
=
………(1)
6
For L.H.S (from fig.)
dr = ∫OA F̅. ̅̅̅
dr + ∫AB F̅. ̅̅̅
dr + ∫BC F̅. ̅̅̅
dr + ∫CO F̅. ̅̅̅
dr
∮c F̅. ̅̅̅
Along OA
y=0 dy = 0
Along AB
x=1 dx = 0
Along BC
y=1 dy = 0
Along CO
x=0 dx = 0
dr = ∫OA xy dx + ∫AB xy 2 dy + ∫BC xy dx + ∫CO xy 2 dy
∮c F̅. ̅̅̅
y=1
x=0
y=0
= 0 + ∫y=0 xy 2 dy + ∫x=1 xydx + ∫y=1 xy 2 dy
1
y3
x2
0
= (1)| | + (1) | | +0
3 0
2 1
1
1
= ( + 0) + (0 − )
3
2
=
1
3
−
1
2
=
−1
…………(2)
6
From eq n (1) & (2)
̂𝐝𝐬 = ∮ 𝐅̅. ̅̅̅
∬(𝛁 𝐗 𝐅̅). 𝐧
𝐝𝐫
𝐬
Hence stoke’s theorem verified.
𝐜
Q6) a) Using Green’s theorem, show that the area bounded by a
simple closed curve C is given by,
𝟏
𝟐
∫ 𝐱 𝐝𝐲 − 𝐲 𝐝𝐱
Hence find the area of circle
x= acos𝛉, y =asin𝛉
I =
1
2
∫c ∫ x dy − y dx
Since C is simple closed curve, by Green’s theorem
I =
=
1
∂
∂
∬R (∂x x − ∂y (−y)) dxdy
2
1
∬R 2dxdy = ∬R dxdy
2
= Area of region bounded by given closed curve.
We know that,
Area = ∬R dxdy
Area =
1
2
∮c xdy − ydx
Here circle c is x 2 + y 2 =a2 , z = 0
x = acosθ, y = asinθ
dx = - asinθdθ, dy = acosθdθ, dz = 0
∴I
=
=
∫0 (acosθ) (acosθdθ) − (asinθ)(−asinθdθ)
2
2π
1
∫0 a2 dθ
2
a2
=
=
2π
1
=
2
a2
[θ]2π
0
2
a2
2
2π
∫0 dθ
(2 π)
Area = πa2
b) Evaluate
𝐝𝐬
∬𝐬 (𝐱 𝟑 𝐢̅ + 𝐲 𝟑 𝐣̅ + 𝐳 𝟑 𝐤̅) ̅̅̅
Where s is the surface of the sphere 𝐱 𝟐 + 𝐲 𝟐 + 𝐳 𝟐 = 𝐚𝟐
 Since s is closed surface,
We apply Gauss divergence theorem.
∴ ∬s F̅ . n̂ds = ∭v ∇. F̅ dv
F̅ = x 3 i̅ + y 3 j̅ + z 3 k̅
∇. F̅ = 3x 2 + 3y 2 + 3z 2 = 3(x 2 + y 2 + z 2 )
∇. F̅ = 3(x 2 + y 2 + z 2 )
∴ ∬ F̅ . n̂ds = ∭ 3(x 2 + y 2 + z 2 ) dxdydz
s
Transforming the variables to spherical polar co -ordinates
x = rsinθcos∅, y = rsinθsin∅, z = rcosθ
x2 + y2 + z2 = r2
dxdydz = r 2 sinθdrdθd∅
2π
π
2π
π
a
= 3 ∫0 ∫0 ∫0 r 2 (r 2 sinθ) drdθd∅
= 3 ∫0 ∫0 sinθ r 4 dr dθd∅
2π
π
a
= 3[∫0 d∅] [∫0 sinθ dθ][∫0 r 4 dr]
r5
π
= 3[∅]2π
0 [−cos∅]0 [ ]
a
5 0
= 3(2π) (2)
a5
5
̂𝐝𝐬 =
∬ 𝐅̅ . 𝐧
𝐬
𝟏𝟐𝛑𝐚𝟓
𝟓
c) Evaluate
∬𝐬 (𝛁 𝐗 𝐅̅) . 𝐧̂ 𝐝𝐬 where
𝐅̅ = (𝐱 − 𝐲)𝐢̅ + (𝐱 𝟐 + yz) 𝐣̅ – 3x𝐲 𝟐 𝐤̅ & s is the surface of the cone.
Z = 4 - √𝐱 𝟐 + 𝐲 𝟐 above xoy plane.
Since, surface of cone is above XY plane, given surface is open
surface.
∴ By stokes theorem
dr
∬s (∇ X F̅) . n̂ ds = ∮c F̅. ̅̅̅
Where, C is projection of cone on XY plane, which is circle
x 2 + y 2 = 16
Consider,
̅̅̅ = ∮ (x − y)dx + (x 2 + yz)dy − 3xy 2 dz
∮c F̅. dr
c
y
C: x 2 + y 2 = 16
Z=0
x = 4cost,
y = 4sint,
z=0
dx = -4sintdt,
dy = 4costdt,
dz = 0
x
𝐱 𝟐 + 𝐲 𝟐 = 𝟏𝟔
zt
:0
y
2π
y
2π
I = ∫0 (4cost -4sint) (-4sintdt) + (16 cos 2 t)(4cost dt)
2π
= ∫0 ( − 16 cost sint) +16sin2 t + 64 cos 3 t) dt
π/2
= 0 + 16X4 ∫0
1
π
2
2
sin2 t. dt + 0
= 64 [ X ]
̅̅̅̅ = 𝟏𝟔𝝅
̅ . 𝒅𝒓
∮𝑭
𝒄
Q7) a) If
𝛛𝟐 𝐲
𝛛𝐭 𝟐
= 𝐜𝟐
𝛛𝟐 𝐲
𝛛𝐱 𝟐
Represents the vibrations of a string of length & fixed at both
ends, find solution with cond n s
i)
ii)
y(0,t) = 0
y(i,t) = 0
iii)
( 𝛛𝐭 )
iv)
𝐲(𝐱, 𝟎) = 𝐤(𝐭𝐱 − 𝐱 𝟐 ), 0≤ 𝐱 ≤ 𝐥
𝛛𝐲
𝐭=𝟎
=𝟎
 y(x,t) be the displacement at any time t
The differential equation vibration of stretched string is
∂2 y
∂t2
= c2
∂2 y
∂x2
& the given conditions are,
i)
ii)
y(0,t) = 0, ∀ t
y(i,t) = 0, ∀ t
iii)
( ∂t )
iv)
y(x, 0) = k(tx − x 2 ), 0≤ x ≤ l
∂y
t=0
= 0, ∀ x
The most suitable solution is.
y(x,t) = (c1 cosmx + c2 sinmx) (c3 coscmt + 4 sincmt )
……..(1)
By conditional (1),y(0,t) = 0
∴ put x =0, y = 0 in eq n (1)
∴ 0 = (𝐜𝟏 + 𝟎) (c3 coscmt + 4 sincmt )
 𝐜𝟏 = 𝟎
Eq n (1) becomes,
y(x,1) = c2 sinmx (c3 coscmt + 4 sincmt )
For condition (iii),
∂y
∂t
………(2)
= 0 at t = 0
∴ 0 = c2 sinmx ( 0 + mc c4 )

𝐜𝟒 = 𝟎
(∴ If we take c2 = 0 then eq n (2) complete becomes zero)
Eq n (2) becomes
……….(3)
Y(x,t) = c2 c3 sinmx cosmct
By condition (ii), x=l, y=0
From eq n (3)
∴ 0 =c2 c3 sinml cosmct
c2 c3 = 0 or sinml = 0 or cosmct = 0
Only sinml = 0
ml = nπ
m=
nπ
n= 0,1,2,3……….
l
Eq n (3) becomes
nπ
nπ
y(x,t) = c2 c3 sin ( ) x cos (
l
l
c) t;
nπ
nπ
y(x,t) = ∑∞
n=0 bn sin ( l ) x cos (
l
n = 0,1,2,3 … … …
……….(4)
c) t
Now, by using cond n (iv);
y(x,0) = k(lx – x 2 )
0≤ x ≤ l
∴ 𝑝𝑢𝑡 𝑡 = 0 𝑖𝑛 𝑒𝑞 𝑛 (4)
𝑛𝜋
k(lx – 𝑥 2 ) = ∑∞
𝑛=0 𝑏𝑛 𝑠𝑖𝑛 ( 𝑙 ) 𝑥
0≤ 𝑥 ≤ 𝑙
which is half range sine series in 0 ≤ 𝑥 ≤ 𝑙
𝑙
2
𝑛𝜋
∴ 𝑏𝑛 = ∫ 𝑓(𝑥) 𝑠𝑖𝑛 ( ) 𝑥𝑑𝑥
𝑙
𝑙
0
=
2
𝑙
2
𝑙
𝑛𝜋
∫0 𝑘(𝑙𝑥 − 𝑥 2 ) 𝑠𝑖𝑛 ( 𝑙 ) 𝑥𝑑𝑥
= [𝑘(𝑙𝑥 − 𝑥
𝑙
2)
(
𝑛𝜋
)𝑥
𝑙
𝑛𝜋
𝑙
−𝑐𝑜𝑠(
(−2𝑘) (
) − (𝑘(𝑙 − 2𝑥)). (
𝑛𝜋
)𝑥
𝑙
3
𝑛𝜋
( )
𝑙
−𝑐𝑜𝑠(
𝑛𝜋
)𝑥
𝑙
2
𝑛𝜋
( )
𝑙
−𝑠𝑖𝑛(
𝑙
)]
0
)+
=
2
𝑙
[{0 − 0 −
(2𝑘)(−1)𝑛
𝑛𝜋 3
( )
𝑙
} − {0 − 0 −
(2𝑘)
(
𝑛𝜋 3
)
𝑙
}]
(∴ 𝑠𝑖𝑛 𝑛𝜋 = 0 , 𝑐𝑜𝑠𝑛𝜋 = (−1)𝑛 )
=
2
𝑙
2
𝑙3
𝑙
𝑛3 𝜋 3
= x2k
𝑏𝑛 =
3
𝑙
3
𝑙
[−2𝑘 (𝑛𝜋) (−1)𝑛 + 2𝑘 (𝑛𝜋) ]
8𝑘𝑙 3
𝑛3 𝜋 3
;
(sin0=0,cos0=1)
[1 − (−1)𝑛 ]
if n is an odd no.
0;
if n is an even no.
∴ 𝑜𝑑𝑑 𝑛𝑜. = 2𝑛 + 1;
(
= 2𝑛 − 1;
𝑛 = 0,1,2,3 … … … .
)
𝑛 = 1,2,3 … … …
Put the value of b n in eq n (4), we get
y(x,t) =∑∞
𝑛=0
y(x,t) =
8𝑘𝑙 2
𝜋3
8𝑘𝑙 2
(2𝑛+1)𝜋
𝜋3 (2𝑛+1)2
∑∞
𝑛=0
. sin(
1
(2𝑛+1)3
𝑙
) 𝑥. 𝑐𝑜𝑠 (
(2𝑛+1)𝜋
sin (
𝑙
(2𝑛+1)𝜋𝑐
) 𝑥. 𝑐𝑜𝑠 (
𝑙
)
(2𝑛+1)𝜋𝑐
𝑙
)
which is required deflection of the string, at any time t & at any
distance x.
b) A homogenous rod of conducting material of length 100cm
has ends kept at zero temp. & the temp. initially is
u(x,0) = x 0≤ 𝒙 ≤ 𝟓𝟎
= 100 – x, 50≤ 𝒙 ≤ 𝟏𝟎𝟎
Find the temp. u(x,t) at any time t.
Let u(x,t) be the temp at any time t. The partial differential eq n of
one dimensional heat flow is
Given conditions are,
i)
ii)
u(0,t) = 0 ∀ 𝑡
u(100,t) = 0∀ 𝑡
𝜕𝑢
𝜕𝑡
𝑐2
𝜕2 𝑢
𝜕𝑥 2
u(x,0)= {
iii)
𝑥
100 − 𝑥
0 ≤ 𝑥 ≤ 50
}
50 ≤ 𝑥 ≤ 100
The most suitable sol n is.
u(x,t) = [𝑐1 𝑐𝑜𝑠𝑚𝑥 + 𝑐2 𝑠𝑖𝑛𝑚𝑥] 𝑒 −𝑚
2𝑐 2𝑡
……..(1)
By condition (i), u = 0, x = 0
∴ 0 = (𝑐1 + 0 ) 𝑒 −𝑚
2𝑐 2𝑡
 𝑐1 = 0
Eq n (1) becomes.
U(x,t) = 𝑐2 𝑠𝑖𝑛𝑚𝑥𝑒 −𝑚
2 𝑐 2𝑡
………(2)
By condition (ii) u = 0 as x = 100
∴ 0 = 𝑐2 𝑠𝑖𝑛𝑚 100𝑒 −𝑚
But 𝑐2 ≠ 0 or 𝑒 −𝑚
2 𝑐 2𝑡
2𝑐 2𝑡
≠ 0, only sinm100=0
 m100 = n𝜋
m=
𝑛𝜋
n = 1,2,3…….
100
Eq n (2) becomes,
u(x,t) = c 2 sin
𝑛𝜋
100
x𝑒
−(
𝑛𝜋 2 2
) .𝑐 𝑡
𝐿
n=1,2,3……
2
u(x,t)
𝑛𝜋
𝑛𝜋
−( ) .𝑐 2 𝑡
𝐿
=∑∞
𝑏
𝑠𝑖𝑛
.
𝑒
𝑛=1 𝑛
100
By condition (iii) at t = 0,
u = {x
0≤ 𝑥 ≤ 50}
………..(3)
u = {100-x
50≤ 𝑥 ≤ 100}
…………(3)
∴ 𝑥;
0≤ 𝑥 ≤ 50
=
100 − 𝑥; 50≤ 𝑥 ≤ 100
∞
2
b n sin(
n=1
Which is half range sone series in 0 ≤ 𝑥 ≤ 50
∴ 𝑏𝑛 =
=
2
𝐿
2
100
𝐿
n𝜋
∫0 𝑓(𝑥) 𝑠𝑖𝑛 ( 100) 𝑥 𝑑𝑥
100
∫0
𝑓(𝑥) 𝑠𝑖𝑛 (
𝑛𝜋
100
) 𝑥 𝑑𝑥
𝑛𝜋
100
)x
=
=
50
1
50
𝑛𝜋
[∫0 𝑓(𝑥) 𝑠𝑖𝑛 (
𝑏𝑛 =
𝑥) (
1
100
𝑛𝜋
)𝑥
100
𝑛𝜋
( )
100
𝑛𝜋
−𝑠𝑖𝑛(
)𝑥
100
𝑛𝜋 2
( )
100
) − (−1) (
𝑏𝑛 = {[−(50)
50
𝑛𝜋
. 𝑐𝑜𝑠
2
100
[−(50) ( 𝑛𝜋 ) 𝑐𝑜𝑠
𝑛𝜋
2
𝑛𝜋
2
+(
𝜋
𝑛𝜋
)𝑥
100
𝑛𝜋 2
( )
100
−𝑠𝑖𝑛(
) − (1) (
50
+ [(100 −
)]
0
100
)]
}
0
100 2
𝑛𝜋
) 𝑠𝑖𝑛 ( 2 ) − 0] + [0 − 0] −
𝜋
100 2
−(
𝑛𝜋
) 𝑥 𝑑𝑥 + ∫50 (100 − 𝑥) 𝑠𝑖𝑛 ( 100) 𝑥 𝑑𝑥 ]
{[𝑥 (
50
𝑛𝜋
−𝑐𝑜𝑠(
)𝑥
100
𝑛𝜋
(100)
100
−𝑐𝑜𝑠(
1
𝑛𝜋
) 𝑥 𝑑𝑥 + ∫50 𝑓(𝑥) 𝑠𝑖𝑛 ( 100) 𝑥 𝑑𝑥 ]
100
50
1
50
100
𝑛𝜋
[∫0 𝑓(𝑥) 𝑠𝑖𝑛 (
) 𝑠𝑖𝑛
𝑛𝜋
2
]}
1
100 2
𝑛𝜋
=
[2. (
) 𝑠𝑖𝑛 ( )]
50
𝑛𝜋
2
=
𝑏𝑛 =
10000
25(𝑛𝜋)2
400
𝑛𝜋
sin ( )
2
𝑛𝜋
(𝑛𝜋)2
sin( )
2
Eqn (3)becomes
𝟒𝟎𝟎
u(x,t)=∑∞
𝒏=𝟏 (𝒏𝝅)𝟐
𝒏𝝅
𝒏𝝅
−(
sin( ) sin( ) 𝒙. 𝒆
𝟐
𝟏𝟎𝟎
𝒏𝝅 𝟐 𝟐
) .𝒄 𝒕
𝑳
is required temp. at any time
Q8)a) An infinitely long uniform metal plate is enclosed bet n
lines y = 0 & y = l for x>0. The temperature is zero along the
edges.
y=0&y=l
& at infinity. If the edge x = 0 is kept at a constant temp. 𝒖𝟎 ,
find the temp. distribution u(x,y).
 Let u(x,y) be the temp. at any point x,y
y
The given conditions are
i)
y=L
u=0
(∞, 𝑦) = 0; ∀𝑦
x=0
u=0
(x,0) = 0; ∀𝑥
(x,L) = =; ∀𝑥
(0,y) = u o , 0≤ 𝑦 ≤ 𝐿
ii)
iii)
iv)
u=0
y=0
x ∞
x
The most suitable solution is .
𝑢(𝑥, 𝑦) = [𝐶1 𝑒 𝑚𝑥 + 𝐶2 𝑒 −𝑚𝑥 ][𝐶3 𝑐𝑜𝑠𝑚𝑦 + 𝐶4 𝑠𝑖𝑛𝑚𝑦] ……..(1)
By condition (i) as x∞, u0
∴ 0 = [𝐶1 + 0] [𝐶3 𝑐𝑜𝑠𝑚𝑦 + 𝐶4 𝑠𝑖𝑛𝑚𝑦]
u0 iff 𝐶1 = 0
∴ eqn (1)becomes
u(x, y) = C2 e−mx [C3 cosmy + C4 sinmy]
……….(2)
𝐵𝑦 𝑐𝑜𝑛𝑑 𝑛 (𝑖i)𝑎𝑠 𝑦 = 0, 𝑢 = 0
∴ 𝑜 = 𝐶2 𝑒 −𝑚𝑥 [𝐶3 + 0]
 𝐶3 = 0
∴ 𝑒𝑞 𝑛 (2)𝑏𝑒𝑐𝑜𝑚𝑒𝑠
𝑢(𝑥, 𝑦) = 𝐶2 𝑒 −𝑚𝑥 [ 𝐶4 𝑠𝑖𝑛𝑚𝑦]
………(3)
By condition (iii), as y = 1, u0
∴ 0 = 𝐶2 𝐶4 𝑒 −𝑚𝑥 𝑠𝑖𝑛𝑚𝐿
𝑒 −𝑚𝑥 ≠ 0 𝑜𝑛𝑙𝑦,
𝐶2 𝐶4 ≠ 0,
𝑠𝑖𝑛𝑚𝐿 = 0
 𝑚𝐿 = 𝑛𝜋, 𝑛 = 1,2,3 … …
 𝑚𝐿 =
Put m =
𝑛𝜋
𝐿
𝑛𝜋
𝑛 = 1,2,3 … …
𝐿
in eq n (3)
𝑛𝜋
𝑢(𝑥, 𝑦) = 𝐶2 𝐶4 𝑒 ( 𝐿 )𝑥 . 𝑠𝑖𝑛 (
∞
𝑢(𝑥, 𝑦) =
𝑛𝜋
−( )𝑥
∑ 𝑏𝑛 𝑒 𝐿
𝑛=1
𝑛𝜋
)𝑦
𝐿
𝑛𝜋
. 𝑠𝑖𝑛 ( ) 𝑦
𝐿
Now, by condition (iv)
As x = 0, u = u 0
in 0≤ 𝑦 ≤ 𝐿
𝑛 = 1,2,3 … …
𝑛𝜋
∴ u 0 = ∑∞
𝑛=1 𝑠𝑖𝑛 ( 𝐿 ) 𝑦
𝑖𝑛 0 ≤ 𝑦 ≤ 𝐿
Which is half range sine series
𝐿
2
𝑛𝜋
= ∫ 𝑓(𝑦) 𝑠𝑖𝑛 ( ) 𝑦 𝑑𝑦
𝐿
𝐿
𝑏𝑛
0
=
𝑏𝑛
=
2
𝐿
2
𝐿
𝐿
𝑛𝜋
∫0 𝑢0 𝑠𝑖𝑛 ( 𝐿 ) 𝑦 𝑑𝑦
𝑢0 [
−𝑐𝑜𝑠(
𝑛𝜋
)𝑦
𝐿
𝑛𝜋
( 𝐿 )𝑦
2
𝐿
]
0
𝐿
= − 𝑢0 ( ) (𝑐𝑜𝑠𝑛𝜋 − 1)
𝐿
𝑛𝜋
=−
2𝑢0
𝑛𝜋
[(−1)𝑛 − 1]
Eq n (4) becomes,
u(x,y) =
𝟐𝒖𝟎
𝒏𝝅
∑∞
𝒏=𝟏 [
𝟏−(−𝟏)𝒏
𝒏
𝒏𝝅
−(
] 𝒔𝒊𝒏 ( ) 𝒚 . 𝒆
𝑳
𝒏𝝅
)𝒙
𝒍
Which is required temp. at any point(x,y)
b) Use fourier transform to solve
𝝏𝒖
𝝏𝒕
=
𝝏𝟐 𝒖
𝝏𝒙𝟐
, 0<x<∞, 𝒕 > 𝟎
Subject to the following conditions.
i)
ii)
u(0,t) = 0, t>0
u(x,0) = 1, 0<x<1
= 0, x>1
u(x,t) is bounded
iii)
Since the interval is 0<x<∞ & u(0,t) is given. Taking fourier
cosine transform of both sides of the given eq n , we get.
𝜕𝑢
𝜕2 𝑢
𝐹𝑐 ( ) = 𝐹𝑐 ( 2 )
𝜕𝑡
𝜕𝑥
∞
∞
𝜕u
𝜕2𝑢
∫
𝑐𝑜𝑠𝜆𝑥𝑑𝑥 = ∫ 2 𝑐𝑜𝑠𝜆𝑥𝑑𝑥
𝜕𝑡
𝜕𝑥
0
0
̅̅̅𝑐̅
𝑑𝑢
𝑑𝑡
̅̅̅𝑐̅
𝑑𝑢
𝑑𝑡
𝜕𝑢
- 𝜆2 ̅̅̅(𝜆,
𝑢𝑐 𝑡)
= −[ ]
𝜕𝑥 𝑥=0
= - 𝜆2 ̅̅̅
𝑢𝑐
(by variable separable form)
Solving this ordinary D.E.,we get
2𝑡
𝑢𝑐 (λ,t) = A𝑒 −𝜆
̅̅̅
To determine arbitrary constant A, we take fourier cosine transform
of initial condition (iii). i.e.
𝑢𝑐 (λ,0) = 𝐹̅𝑐 [𝑢(𝑥, 0)]
̅̅̅
∞
= ∫0 𝑢(𝑥, 0)𝑐𝑜𝑠 λxdx
1
1
= ∫0 𝑥𝑐𝑜𝑠 λxdx + ∫0 (0)𝑐𝑜𝑠 λxdx
= [𝑥.
=
=
𝑠𝑖𝑛𝜆𝑥
𝜆
𝑠𝑖𝑛𝜆
+
𝜆
−𝑐𝑜𝑠𝜆𝑥
− (1) (
𝜆2
1
)]
0
𝑐𝑜𝑠𝜆−1
𝜆2
𝜆𝑠𝑖𝑛𝜆+𝑐𝑜𝑠𝜆−1
𝜆2
Substituting t = 0 in (vi) & using (vii), we get
𝑢𝑐 (λ,0) = A =
̅̅̅
𝜆s𝑖𝑛𝜆+𝑐𝑜𝑠𝜆−1
… … … … (𝑣𝑖ii)
𝜆2
Putting this value of A in (vi),
uc (λ,t) =
̅̅̅
λsinλ+cosλ−1
λ2
e−λ
2t
……….(ix)
Taking inverse fourier cosine transform of i(x), we obtain
u(x,t) =
𝟐
𝛑
∞ 𝛌𝐬𝐢𝐧𝛌+𝐜𝐨𝐬𝛌−𝟏
∫𝐨
Which is required solution.
𝛌𝟐
𝟐
𝐞−𝛌 𝐭 𝐜𝐨𝐬 𝛌𝐱𝐝𝛌