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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 UNIT–1 1. a) What is periodic function, gave an example of periodic function. 2 Solution: A function f (x) is said to be periodic if there exist a least positive integer T such that Suppose f ( x ) f ( x T ) f ( x 2T ) ...... f ( x nT ) , n I Where T is called period of f (x). Example: sin x, cos x of period 2 while period of tan x is b) What are the Dirichlet’s conditions for a Fourier series expansion? 2 Solution: Suppose f (x) is any finite function defined in (–L, L), then the Fourier series of f (x) is exists only if the following conditions are satisfied: (i). f (x) is periodic i.e. f (x) = f (x + 2L), where 2L is the period of f (x) (ii). f (x) and its integral are finite and single valued. (iii). f (x) has a finite number of discontinuities. (iv). f (x) has a finite number of maxima and minima. These conditions are known as Dirichlet’s conditions. c)
2 Find the Fourier transform of e a x , where a > 0
Solution:
3
2 Given the function: F ( x) e x
The Fourier transform of a function F(x) is given by f ( p)
f ( p)
1
1
F ( x) e 2
2
e
x2
ipx
e
ipx
dx
1
2
dx
f ( p)
1 2
e
x 2 ipx
dx
p2 p2 x 2 ipx 4 4 e dx
2
1 2
ip p2 x 2 4 e dx
[Add and subtract p2/4]
Putting
f ( p) e x
p2 4
1 2
ip x 2 e
2
dx
ip t , so that dx dt 2
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006
f ( p) e
f ( p) e
p2 4
p2 4
1
2
F e x
Thus
2
1
2
2
e t dt
2
e
2
e
p2 4
1
1
p2 4
since
2
e t dt
f ( p)
By change of scale property, we have
F e
a x2
F e
F e a x
Thus
d)
2
2
p f a a
1
e
2a
1 p F F (ax) a f a
1
ax
1 1 1 4 e a 2
p a
2
1 2a
p2 4a
e
p2 4a
Answer
Expand f (x) = x sin x, 0 < x < 2 in a Fourier series.
7
Solution: Here, 2L = 2 – 0 i.e. L = Suppose the Fourier series of f (x) with period 2L is, a0 n x n x f ( x) an cos bn sin 2 n 1 L n 1 L
f ( x)
a0 an cos nx bn sin nx 2 n 1 n 1
Now, a0
1
0
an
1
0
and
2
2
f ( x) dx
1
2
0
f ( x) cos nx dx
x sin x dx 1
2
0
---- (1)
[Since L = ]
1 x ( cos x) 1( sin x) 02 2
x sin x cos nx dx
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 1 2
1 0 x 2 cos nx sin x dx 2 0 x sin (n 1) x sin (n 1) x dx 2
2
…. (2) 2
cos(n 1) x cos( n 1) x sin( n 1) x sin( n 1) x x 1 2 n 1 n 1 ( n 1) (n 1)2 0
1 2
2
an
; n 1
2
n 1
For n = 1, from equation (2), we get
1 a1 2 and bn
1
2
0
2
f ( x) sin nx dx
0
1 2
2
0
2
cos 2 x sin 2 x 1 x 2 1 4 2 0
1 x sin 2 x dx 2 1
2
0
x 2 sin nx sin x dx
x sin x sin nx dx 1 2
2
0 x cos(n 1) x cos(n 1) x dx
…. (3) 2
1 sin(n 1) x sin( n 1) x cos( n 1) x cos( n 1) x x 1 2 n 1 n 1 (n 1) 2 (n 1)2 0 bn 0; n 1 For n = 1, from equation (3), we get
2
1 b1 2
2
0
1 x 1 cos 2 x dx 2
sin 2 x x 2 cos 2 x x x 1 2 2 4 0
b1
From equation (1), we have a0 f ( x) a1 cos x an cos nx b1 sin x bn sin nx 2 n2 n2
1 2 x sin x 1 cos x 2 cos nx sin x 2 n 2 n 1
Answer
Or
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 d)
Express f (x) = x as a (i). Half range cosine series in 0 < x < 2 (ii). Half range sine series in 0 < x < 2 Solution: (i). Suppose the half range cosine series is, f ( x)
a0 n x an cos 2 n 1 2
7
…. (1)
2
x2 2 2 2 Now, a0 f ( x) dx x dx 2 0 2 0 2 0
and
an
2 2 2 n x n x f ( x ) cos dx x cos dx 0 0 2 2 2
2 4 (1)n 2 4 4 n x n x x sin 1 2 2 cos 0 2 2 0 2 2 2 n 2 0 n n n
an
4
(1) n 1 n 2 2
Putting in equation (1) we get
(1)n 1 cos n x f ( x) 1 2 2 2 n 1 n 4
(ii).
Answer
suppose the half range sine series is,
f ( x)
n 1
Now, bn
n x 2
bn sin
…. (2)
2 2 2 n x n x f ( x) sin dx x sin dx 0 2 0 2 2
2 4 (1)n 2 4 n x n x x cos 1 sin 0 0 0 n 2 n 2 2 2 0 n
bn
4 n
Putting in equation (2), we get
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 f ( x)
4 (1) n n x sin n 1 n 2
Answer UNIT–2
2. a)
L 2sin t cos t L sin 2t
2
2
2
p 4
b) Write the Linearity property of Laplace Transform Solution: If a and b are any constants and F (t) and G(t) be any two function of t, then
L a F (t ) b G (t ) a L F (t ) b L G(t )
2
or L a F (t ) b G (t ) a f ( p) b g ( p)
c) Find the Laplace Transform of f (t),
3
2, 0 t 2 f (t ) t 1, 2 t 3 7, t 3 Solution: By definition of Laplace transform,
L f (t )
0
f (t ) e pt dt
2
3
0
2
3
2 e pt dt (t 1) e pt dt 0. e pt dt
e pt e pt 2 ( t 1) p 0 p
2
L f (t )
d)
e pt 1 2 p
3
0 2
2 2 2 p e3 p 1 e 2 p e 1 e 3 p 2 e2 p 2 p p p p p
e 2 p 2 e 2 p 2e 3 p e 3 p p p p p2 p2
Answer
6 s 2 22s 16 Evaluate: L 3 2 s 6 s 11s 6
Solution:
1
Suppose
6s 2 22 s 16
s 3 6s 2 11s 6 Where A, B and C determine by
7
6s 2 22 s 16 A B C ( s 1)( s 2)( s 3) s 1 s 2 s 3
…. (1)
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 6 s 2 22s 16 6 22 16 A 0 (1)(2) ( s 2)(s 3) s 1 6 s 2 22s 16 24 44 16 B 4 (1)(1) ( s 1)( s 3) s 2 and
6 s 2 22 s 16 54 66 16 C 2 ( s 1)( s 2) ( 2)( 1) s 3
Putting in equation (1) and taking Inverse Laplace Transform, we get 6 s 2 22s 16 1 1 1 1 L 3 4L 2L 2 s 2 s 3 s 6 s 11s 6 1
6 s 2 22s 16 2t 3t L1 3 4e 2e 2 s 6 s 11s 6
Answer
Or d)
s Using Convolution theorem, Evaluate L1 2 2 2 ( s a )
Solution:
Suppose f ( s )
s 2
and g ( s )
1
s a s a2 s L1 f ( s ) L1 2 cos at F (t ) s a2 sin at 1 And L1 g (s ) L1 2 G (t ) a s a2 By Convolution theorem of Inverse Laplace transform, we have
L1 f ( s ) g ( s )
2
2
t
0 F ( x) G(t x) dx
L 2 s2 a2
0
1 2a
1
s
t
cos ax.
t
sin[a (t x)] 1 dx a 2a
0 sin ax at ax
t
0 2 cos ax.sin (at ax) dx
sin [ ax ( at ax )] dx
2 cos A sin B sin( A B) sin( A B)
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 1 2a
t
0
sin at sin 2ax at dx
t 1 sin at 1 dx 0 2a
cos 2ax at 1 x sin at 2a 2a 0
1 cos at cos at t sin at 0 2a 2a 2a
1 t sin at 2a
t
0 sin 2ax at dx
t
Thus
t sin at L 2 2a s2 a2 1
s
UNIT–3 3. a) Explain the ordinary point and singular point of differential equation. Solution: Suppose the second order differential equation is d 2y
dy P2 ( x) y 0 dx dx Where P0(x), P1(x) and P2(x) be polynomial function of independent variable x. P0 ( x )
(i).
2
P1 ( x )
2
------ (1)
Ordinary Point: A point x = 0, is said to be an ordinary point of equation (1) if P0(0)
(ii).
Singular Point: A point x = 0, is said to be an singular point of equation (1) if P0(0)
b) Give the complete solution of differential equation when the roots of indicial equations are equal. Solution: When the roots of indicial equation are equal i.e. m1 = m2 y The Complete solution is y c1 y m m c2 1 m m m 1
c)
Solve the differential equation
3
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 d 2y dx Solution:
2
2 tan x
dy 5y e x sec x dx
Given the differential equation is, d 2y 2
2 tan x
dy 5y e x sec x dx
dx Here, P = –2 tan x, Q = 5 and R = ex sec x Now this problem solve by Removable of first derivative method. Suppose the complete solution is, y = v y1 Where v is a function of x only. Now we can find the value of y1 such as y1 e
and
1 Pdx 2
e
1 ( 2 tan x ) dx 2
….. (1)
….. (2)
elog sec x sec x
1 1 dP 1 1 Q1 Q P 2 5 (2 tan x ) 2 (2 sec 2 x) 5 tan 2 x sec2 x 6 4 2 dx 4 2 R1
R e x sec x ex y1 sec x
The normal form of Removable of first derivative is, d 2v dx 2
d 2v 2
Q1v R1 6v e x
…. (3)
dx This is LDR of higher order. The A.E. is m2 6 0
m 6i
Therefore C.F . c1 cos Now, P.I .
1
ex
6 x c2 sin 1
D2 6 12 6 The solution of equation (3) is,
ex
6x
ex 7
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 v C.F . P.I . c1 cos
6 x c2 sin
6x
ex 7
Putting in equation (2), which our complete solution
y c1 cos d)
Solve (1 x 2 )
Solution:
d 2y
6 x c2 sin
2x
ex sec x 7
6x
Answer
dy y 0 in series solution. dx
7
dx 2 Given the differential equation is, d 2y
(1 x 2 )
dx 2
2x
dy y0 dx
….. (1)
Here, P0(x) = 1– x2 Clearly at x = 0, P0(x) = 1– 0 = 1 ≠ 0 Therefore x = 0 is ordinary point. Suppose the complete solution of equation (1) by Power series method is
y = a0 a1x a2 x 2 a3 x3 + ........
ak x k
y=
….. (2) …. .(3)
k 0
Differentiating both sides w.r.t. x, we get
dy = dx
2
ak k (k 1) x k
2
k 0
d 2y dy and in equation (1), we get dx dx 2
ak k (k 1) x k
2
k 0
and
d 2y dx
k 0
Putting the value of y,
(1 x 2 )
ak k x
k 1
ak k (k 1) x k
2 x
ak k x k
1
k 0
2
k 0
k 0
ak k (k 1) x k 2
ak x k 0
k 0
ak k x k
k 0
ak x k 0
…. (4)
k 0
0
Equating the coefficient of x on both sides in equation (4), we get 2 (2 1) a2 0 0 a0 0
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 a0 2 1 Equating the coefficient of x on both sides in equation (4), we get 3(3 1)a3 0 2a1 a1 0
a2
a1 2 2 Equating the coefficient of x on both sides in equation (4), we get
a3
a2 1 a a 0 0 4 4 2 8 Equating the coefficient of x3 on both sides in equation (4), we get 5 (5 1)a5 3(3 1)a3 6a3 a3 0 a4
a5
a3 1 a a 1 1 20 20 2 40
Putting the values of a2, a3, a4 and a5 in equation (2), we get a a a a y a0 a1x 0 x 2 1 x 3 0 x 4 1 x5 ..... 2 2 8 40
d)
x2 x4 x3 x5 y a0 1 ..... a1 x ..... 2 8 2 40
Answer
Or Solve by the method of variation of parameter
7
D 2 1 y x Solution:
The given differential equation is, d 2y
yx dx 2 Here, P = 0, Q = 1 and R = x The A.E. is
….. (1)
m2 1 0
m i
Therefore C.F. is yc c1 cos x c2 sin x
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 Suppose
u cos x and v sin x
u sin x and v cos x
and
w
cos x sin x cos 2 x sin 2 x 1 0 sin x cos x
Suppose the complete solution of equation (1) is y Au Bv A cos x B sin x
….. (2)
Where A and B determine by the formula,
and
v.R A dx c1 x sin x dx c1 x( cos x) ( sin x) c1 w A x cos x sin x c1 u.R B dx c2 x cos x dx c2 x(sin x ) 1.( cos x ) c2 w B x sin x cos x c2
Putting the values of A and B in equation (2), we get y x cos x sin x c1 cos x x sin x cos x c2 sin x
y c1 cos x c2 sin x x
Answer
UNIT–4 4. a) Find the Partial differential equation by eliminating a and b from the relation
2
( x a )2 (y b) 2 z 2 c
Solution:
Given the equation is, ( x a )2 (y b) 2 z 2 c
Partially differentiating w.r.t., x we get z 2 ( x a) 0 2 z 0 x x a zp
…..(1)
…. (2)
Again equation (1) Partially differentiating w.r.t., y, we get z 0 2 (y b) 2 z 0 y
y b zq
…. (3)
Putting in equation (1), we get
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 z 2 p2 q 2 z 2 c
b).
Answer
Solve y z p z x q x y
Solution:
Given differential equation is y z p z xq x y
2 ….. (1)
This is Lagrange PDE. Here P = y z, Q = z x and R = x y The Lagrange A.E. is dx dy dz yz zx xy Taking first two ratios, we get dx dy x dx y dy yz zx Integrate both sides, we get
x 2 y2 x 2 y2 c1 c1 2 2 2 Taking Last two ratios, we get dy dz y dy z dz zx xy Integrate both sides, we get
y2 z 2 y2 z 2 c2 c2 2 2 2 The General solution of equation (1), we get x 2 y2 y2 z 2 , 0 2 2 c)
Answer
2z 2z Solve 2 2 2 e3x 2y x y y x
Solution:
2 z
3
The given Partial differential equation is
2 z x 2
2
2z 2z e3x 2y x y y 2
….. (1)
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 Suppose,
D
From (1), we have
, D x y
D 2 2 DD D2 z e3x 2y
The A.E. is, m2 2m 1 0 m 1, 1
The C.F. is,
P.I .
C.F . 1 (y x) x 2 (y x)
1 D 2 2 DD D2
e
3 x 2y
1
3 x 2y
(3)2 2(3)(2) (2)2
e
e3 x 2y 25
The Complete solution is,
z 1 (y x) x 2 (y x)
d)
e3 x 2y 25
Solve the Charpits method ( p 2 q 2 ) y q z
Solution:
Answer
7
Given Partial differential equation is, ( p2 q2 ) y q z
…. (1)
p2 y q 2 y q z 0
Suppose
f p2y q2 y q z
f f f f f 0, p2 q2 , q, 2 p y, 2q y z x y z p q
The Charpits A.E.is dp dq dz dx dy f f f f f f f f p q p q x z y z p q p q
dp dq dz dx dy 2 2 2 0 pq p(2 p y) q (2q y z ) 2 p y z 2q y p q q
Taking first two ratios, we get dp dq pq p2
p dp q dq
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 Integrate both sides, we get p2 q 2 a 2
…. (2)
where a2 = 2c1
Putting in equation (1), we get
a2 y q z q
a2 y z
Putting in equation (2), we get 2
a2 y p2 a2 z p2
a 2 z 2 a 4 y2 z
2
p
a 2 z a 2 y2 z
dz p dx q dy
Since
z dz a 2 y dy
2
2 2
a 2 a2 y z a 2 y2 dx dy z z
a dx
z a y
Integrating both sides, we get
z 2 a2 y2 a x c
Answer Or
d)
Solve
Solution:
u u 2 u by the method of separation of variables, where u x, 0 6 e3 x x t Given differential equation is, u u 2 u ….. (1) x t
7
With initial condition u x, 0 6 e3 x Suppose the complete solution is, u ( x, t ) X ( x).T (t ) X . T
…. (2)
Where X is function of x and T is function of t only. Now equation (2), partially differentiate w.r.t. x and t respectively
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 u dX u dT T. X T and X. = X .T x dx t dt Putting the values in equation (1), we get X T 2 X .T X .T X T (2.T T ) X
X 2.T T X 2T 1 k (Let) X T X T
X 2T k ….. (3) and 1 k …. (4) X T Now from (3) and integrate w.r.t. x, we get
X log X kx log c1 log kx c 1
X ( x) c1 e kx
And From equation (4), we have
2T T k 1 1 k T T 2 Integrate w.r.t. “t” on both sides, we get
T k 1 log T t log c log 2 c 2 2
T (t ) c 2
k 1 t 2
k 1 t e 2
Putting the values of T(t) and X(x) in equation (2), we get u ( x, t ) c1c 2
k 1 t k x 2 e e
….. (5)
By initial condition, u 6 e3 x at t = 0
6 e3 x c1 c 2 ek x
Equating both sides, we get c1 c 2 6 and k 3 Putting in equation (5), we get
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 u ( x, t ) 6 e 3 x 2t
Answer
UNIT–5 2 d 5. a) If u t i t j (2t 1)k and v (2t 3)i j t k , find u.v ? at t = 1 2 dt Solution: Now, u . v t 2 i t j (2t 1)k . (2t 3)i j t k 2t 3 3t 2 t 2t 2 t 2t 3 5t 2 2t d u.v 6t 2 10t 2 6 10 2 6 at t = 1 Answer dt
b)
Find the directional derivative of x y y z z x in the direction of the vector i 2 j 2k at the
point (1, 2, 0). Solution: Given x y y z z x grad i j k (x y y z z x ) y z x (y z ) i ( x z ) j (y x)k
Now
Suppose
2
grad 2 i j 3k at (1, 2, 0) a i 2 j 2k i 2 j 2k i 2 j 2k a a 3 1 4 4 a
Directional Derivative = grad . a
i 2 j 2k 1 10 D.D. 2 i j 3k (2 2 6) 3 3 3 c) If r xi y j zk , then show that grad r n n r n 2 r Solution: Given r xi y j zk
Answer 3
r 2 x2 y2 z 2
Partially differentiate w.r.t., x, y and z respectively, we get
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 2r
r r r 2y and 2r 2 z 2 x , 2r x y z
r y r x r z and , x r y r z r
grad r n i j k r n y z x
Now, d)
r r r n r n 1 i n r n 1 j n r n 1 k x z y z x y grad r n n r n 1 i j k n r n 2 r Proved r r r Verify Stoke’s theorem for F ( x 2 y 2 ) i 2 x y j taken round the rectangle bounded by
x = a, y = 0, y = b Solution: Given the function is, F ( x 2 y 2 ) i 2 x y j
7
By Stoke’s theorem we have, F . d r curl F .n ds -----(1) c
s
To evaluate Line integral: The curve C consists of four lines AB, BC, CD and DA.
c
F .d r
AB
F .d r
BC
F .d r
CD
F .d r
Now , F d r ( x 2 y 2 ) i 2 xy j dx i + dy j
DA
F .d r
------ (2)
( x 2 y 2 ) dx 2 xy dy
….. (3) Along line AB, Here x = a, dx = 0 and 0 y b
AB
F .d r
b
0
b
0
2ay dy a y 2
ab 2
Along the line BC, Here, y = b, dy = 0 and a x a
BC
F .d r
a
a
a
x3 ( x b ) dx b 2 x 3 a 2
2
2a3 2ab2 3
Along the line CD,
x = 0, dx = 0 and b y 0
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006
CD F .d r
since F d r 0
0
Along the line OA,
y = 0, dy = 0 and a x a
DA
F .d r
a
x3 2a3 a x dx 3 3 a a
2
Putting these values in equation (2), we get
2a3 2a 3 2 2 2 F . d r a b 2 ab a b 4a b 2 c 3 3
--- (3)
To Evaluate surface integral:
Now,
Curl F F
i
j
k
x
y
z
x2 + y2
2xy Curl F = ( 2y 2y) k 4yk
0
Since the surface on the xy-plane, then n k (Projection on XY-Plane)
and Curl F . n
S
4yk .k
curl F .n dS
4y
b a
0 a 4y dx dy b
y2 4 2 0 From (4) and (3), we get
c
F .d r
s
x a a
4ab 2
…… (4)
curl F .n ds 4ab 2
Hence Stoke’s theorem is verified.
Or Prove that div grad r m . r m m (m 1)r m 2 Solution: r xi y j zk d)
7
r 2 x2 y2 z 2
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 Partially differentiate w.r.t., x, y and z respectively, we get r r r 2y and 2r 2 z 2r 2 x , 2r x y z Now,
r y r x r z and , x r y r z r grad r m i j k r m y z x
r r r m r m 1 i m r m 1 j m r m 1 k x z y y z x grad r m m r m 1 i j k m r m 2 r r r r L.H.S = div grad r m div m r m 2 r m div r m 2 r m r m 2 div r r grad r m 2 m r m 2 (3) r (m 2) r m 4 r div .a div a a grad
m 3r m 2 (m 2) r m 2 m (m 1)r m 2 = R.H.S.
Proved
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006 Subjects available for Engineering students
1. Engineering Mathematics-I 2. Engineering Drawing 3. Engineering Mechanics and civil 4. Basic Electrical and Electronics
1. Engineering Mathematics-II 2. Network Analysis 3. Strength of Material 4. Machine Design & Drawing 5. Electro-magnetic theory 6. C, C++ and JAVA 1. Engineering Mathematics-III 2. Electro-magnetic theory 3. Electronic Device Circuit 4. Fluid Mechanics 5. Theory of Mechanism
1. Structural design and Drawing 2. Theory of structure 3. Control system 4. Digital signal processing 5. Theory of computation 1. Structural design and Drawing 2. Theory of structure-II 3. Control system 4. Heat Mass Transfer
FIRST YEAR M1 ED Civil EM BEEE
Sonendra Gupta Mohit Pandya Ajay Sir Shruti Jain
THIRD SEMESTER M2 Sonendra Gupta NA Rahul Arrora SOM Ishwar singh MDD Ankit Sir EMT Rahul Arrora Vishal Thakur FORTH SEMESTER M3 Sonendra Gupta EMT Rahul Arrora EDC Mohit Sir FM Ajay Sir TOM Ankit Sir FIFTH SEMESTER Steel Ishwar singh TOS-I Ishwar singh Control Rahul Arrora DSP Mohit Sir TOC Sunil Tripathi SIXTH SEMESTER Steel Ishwar singh TOS-II Ishwar singh Control Rahul Arrora HMT Ankit Sir
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Solution of Engineering Mathematics-II [June 2015] Subject: Engg. Mathematics-II Paper Code BE-301 Faculty: Sonendra Gupta Mobile: 9893455006
EKLAVYA ACADEMY
Announcing
GATE/NET st
FROM 1 Feb. 2016 [MATHEMATICS]
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