MM

JNTU ONLINE EXAMINATIONS

[Mid 3 - MM]

1. Solving the differential equation 1 + y

, y(0) = 0 by Taylor's series method, y(0.2)

2

=

1. 0.10266

2. 0.00266

3. 0.30266

4. 0.20266

2. Solving the differential equation x + y, y(1) = 0 h = 0.1, by Taylor's series method,

y(1.1) =

1. 0.110338

2. 1.110338

3. 1.210338

4. 1.310338

3. Solving y

+ x, y(0) = 1 by Taylor's series method y(0.1) =

2

1. 0.1164

2. 1.1164

3. 2.1164

4. 3.1164

4. Solving , y(4) = 4 by Taylor's series method y(4.2) =

1. 3.0098

2. 2.0098

3. 1.0098

4. 4.0098

5. Solving the differential equating x

+ y

, y(0) = 1 h = 0.1 by Taylor's series

2

2

method. y(0.1) =

1. 0.11145

2. 2.11145

3. 1.11145

4. 3.11145

6. To find the solution of the x

y - 1, y(0) = 1 by Taylor series y (0) =

2

1. -6

2. 6

3. 8

4. -8

7. Solving x - y

, y(0) = 1 by Taylors series method y(0.1) =

2

1. 0.7138

2. 0.8138

3. 0.9138

4. 0.6138

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8. For the differentia equation y(x0), y(x0)= y0 the solution by Taylor's series

is given by yn+1 =

1.

2.

3.

4.

9. To solve the differential equation y(0) = 1 by Taylor's series method y

(0) =

1. 3

2. 4

3. 2

4. 1

10. To find the solution of the , y(1) = 1 by Taylor's series method y (o) =

1.

2.

3.

4. 1

11. Solving the differential equation 1 + xy, y(0) = 1 by Picard's method y(0.1) =

1. 0.1053

2. 0.2053

3. 1.1053

4. 1.2053

12. To solve the differential equation , y(0) = 0 by Picard's method, the

approximation =y

1.

2.

3.

4.

13. The solution of the differential equation , y(0) = 0 by Picard method is

1.

2.

3.

4.

14. The formula to find yn by Picard's method is given by y

=

n

1.

2.

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3.

4.

15. If x + y, y(0) = 1, the first approximation by Picard's method in given by y1 =

1.

2.

3.

4.

16. Using Picards method, the solution of the equation y, y(0) = 1 is given by y4 =

1.

2.

3.

4.

17. The solution of xy + 1, y(0) = 1 by Picard's method is given by y

=

2

1.

2.

3.

4.

18. Using Picard's method, the solution of the equation y

+x

. y(0) = 1 is given by y

2

2

1

=

1.

2.

3.

4.

19. To solve the differential equation , y(0) = 1 by Picard method, y

=

1

1. x + x log (1 + x)

2. 1 - x + 2 log (1 + x)

3. 1 - x log (1 + x)

4. 1 + x + 2 log (1 + x)

20. To solve the differential equation f(x, y), y(x

) = y

the formula by Euler's method

0

0

is given by y =

1. y

- hf(x

, y

)

n

n

n

2. y

= hf(x

, y

)

n

n

n

3. y

= hf (x , y )

n

4. y

+ hf(x , y )

n

21. Using modified Euler's method, the second approximation of the differential equation

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1 - y , y (0) =0, h = 0.1, y_1

=

2

1. 1.09525

2. 2.09525

3. 3.09525

4. 0.09525

22. Using Euler's method, the solution of the differential equation with y(0) = 1

when x = 0.1 is given by y(0.1) =

1. 0.0200

2. 1.0200

3. 2.0200

4. 3.0200

23. Using modified Euler's method, the solution of the differential equation 2xy, y(0)

=1, h = 0.25 is given by y(0.25) =

1. 1.0645

2. 0.0645

3. 2.0645

4. 1.2645

24. Solving the differential equation 1-2xy, y(0) = 0, h =0.2 by Euler's method y(0.6)

=

1. 1.4748

2. 0.4748

3. 2.4748

4. 3.4748

25. Solving the differential equation by modified Euler's method y(2)

=

1. 4.051

2. 5.051

3. 6.051

4. 3.051

26. Solving the differential equation x+ y + xy, y(0) = 1, h = 0.025 by Euler's method,

y(0.1) =

1. 0.1448

2. 2.1448

3. 1.1448

4. 1.2448

27. Solving the differential equation -xy

, y(0) = 2, h = 0.1 by modified Euler's method

2

y(0.2) =

1. 1.9227

2. 0.9227

3. 2.9227

4. 3.9227

28. Solving the differential equation y + e

, y(0) = 0 by modified Eule's method y(0.2)

x

=

1. 0.2468

2. 0.4268

3. 1.2468

4. 1.4268

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29. To solve the differential equation = f(x, y) y(x

) = y

, the formula by modified

0

0

Euler's method

1.

2.

3.

4.

30. To solve the differential equation f(x, y), y(x

) = y

for the formula by Runge Kutta

0

0

method is gives by K =

1.

2.

3.

4.

31. Give x

+ y

, y(1) = 0, he solution of y when x = 1.1 by Runge-Kutta method is

2

2

given by

1. 0.11072

2. 1.11072

3. 0.21072

4. 1.21072

32. The solution of differential equation , y(0)=0 by Runge Kuta method, y(0.2) =

1. 0.2027

2. 1.2027

3. 1.0227

4. 0.0227

33. The solution of differential equation - y, y(0) = 1 by Runge Kutta method, y(0.1) =

1. 1.9048

2. 0.9048

3. 0.0948

4. 1.0948

34. The solution of differential equation = x + y, y(0) = 1, h = 0.1 by Runge Kutta

method y(0.1) =

1. 0.11034

2. 2.11034

3. 1.11034

4. 3.11034

35. Using Runge-Kutta method, to solve the differential equation x + y, y(0) = 1, h =

0.1, K

=

4

1. 1.12105

2. 0.12105

3. 1.22105

4. 0.22105

36. To solve the differential equation + y + xy

= 0, y(0) = 1, h = 0.1 by Runge Kutta

2

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method, K

=

2

1. 0.0995

2. 1.0995

3. - 1.0995

4. - 0.0995

37. To solve the differential equation , y(0) = 1 by Runge Kutta method K

=

2

1. 1.19672

2. 0.19672

3. 0.16972

4. 1.16972.

38. Which is the best for solving initial value problems

1. Runge Kutta method

2. Euler's method

3. Modified Euler's method

4. Taylor's method.

39. Milne's Preductor formula

1.

2.

3.

4.

40. Milne's corrector formula

1.

2.

3.

4.

41. Solving differential equation , y(0) = 1, y(0.1)=1.01, y(0.2) = 1.022, y(0.3) =

1.023 y_0

= 0 y_1

= 0.0505, y_2

= 0.1022, y_3

= 0.1535 by Adams - Moulton

1

1

1

1

formula, y_4

= y(0.4) =

c

1. 1.0410

2. 1.0410

3. 2.0408

4. 1.2408

42. Solving differential equation 1 + y

, y(0) = 0, y(0.2) = 0.2027, y(0.4) = 0.4228, y

2

(0.6) = 0.6841, y_0

= 1.000, y_1

= 1.0411, y_2

= 1.1787, y_3

= 1.4681 by Milne's

1

1

1

1

formula, y_4

= y(0.4) =

c

1. 0.0294

2. 1.0294

3. 0.1294

4. 1.1294

43. If y(0.1) = 1.06, y(0.2) = 1.12, y(0.3)=1.21, y_1

= 0.567,

1

y_2

= 0.6522, y_3

= 0.7980 by Milne's formula, = y(0.4) =

1

1

1. 1.2797

2. 0.2797

3. 1.7297

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4. 0.7297

44. Given x - y

, y(0) = 0, y(0.2) = 0.200, y(0.4) = 0.0795, y(0.6) = 0.1762 y_1

=

2

1

0.1996, y_2

= 0.3937, y_3

= 0.5989 by Milne's formula, = y(0.8) =

1

1

1. 0.3046

2. 0.4046

3. 1.3046

4. 1.4046

45. Solving differential equation , y(0) = 1, y(0.1)=1.01, y(0.2) = 1.022, y(0.3) =

1.023 y_0

= 0 y_1

= 0.0505, y_2

= 0.1022, y_3

= 0.1535 by Adams - Moulton

1

1

1

1

formula, y_4

= y(0.4) =

p

1. 1.0408

2. 0.0408

3. 2.0408

4. 1.2408

46. Solving differential equation 1 + y2, y(0) = 0, y(0.2) = 0.2027, y(0.4) = 0.4228, y

(0.6) = 0.6841, y_0

= 1.000, y_1

= 1.0411, y_2

= 1.1787, y_3

= 1.4681 by Milne's

1

1

1

1

formula, y_4

= y(0.4) =

P

1. 1.0239

2. 0.0239

3. 1.1239

4. 0.1239

47. If y(0.1) = 1.06, y(0.2), y(0.3)=121, y_1

= 0.567, y_2

=

1

1

0.6522, y_3

= 0.7980 by Milne's formula, = y(0.4) =

1

1. 0.2772

2. 1.2772

3. 0.7225

4. 1.7225

48. Given x - y

, y(0) = 0, y(0.2) = 0.200, y(0.4) = 0.0795, y(0.6) = 0.1762 y_1

=

2

1

0.1996, y_2

= 0.3937, y_3

= 0.5689 by Milne's formula, = y(0.8) =

1

1

1. 0.0346

2. 1.0346

3. 1.3046

4. 0.3046

49. Adams Moluton predictor formula

1.

2.

3.

4.

50. Adams-Moulton corrector formula

1.

2.

3.

4.

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51. Given x

+ y

, y(1) = 0, y(1.1) = 1.1107, y(1.2) = 0.24631, y(1.3) = 0.41357, y_1

2

2

1

= 0.11072, y_2

= 1.50067, y_3

= 1.86104 by Adams-Moulton method, y_4

= y(1.4)

1

1

1

=

1. 1.62208

2. 1.26208

3. 0.62208

4. 0.25208

52. Given x

+ y

, y(1) = 0, y(1) = 0, y(1.1) = 1.11073, y(1.2) = 0.24631, y(1.3) =

2

2

0.4157, y_1

= 0.11072, y_2

= 1.50067, y_3

= 1.86104 by Adams-Moulton method,

1

1

1

= y(1.4) =

1. 0.62274

2. 1.62274

3. 0.26274

4. 1.26274

53. Which is a step by step method

1. Taylor's Series

2. Adams Bashfortion

3. McGauran's Series

4. Modified Euler's method

54. Given 1 + y

, y(0) = 0, y(0.2) = 0.2027, y(0.4) = 0.4228, y(0.6) = 0.6841, y_0

=

2

1

1.0, y_1

= 1.041099 y_2

= 1.17876, y_3

= 1.46799 by Adams-Moulton method, y_4

1

1

1

c

= y(0.8) =

1. 0.0296

2. 2.0296

3. 1.0296

4. 3.0296

55. Given y(0.2) = 2027, y(0.4) = 0.4228, y(0.6) = 0.6841, ,

y_1

= 1.041099, y_2

= 1.17876, y_3

= 1.46799 by Adam=Moulton method, y_4

= y

1

1

1

P

(0.8) =

1. 0.02337

2. 1.02337

3. 0.3237

4. 1.3237

56. The period of sin2x is

1. 2

p

2.

p

3.

p

/2

4.

p

/4

57. In the Fourier expansion of f(x) defined in the internal ( + 2

p

), a

=

n

1.

2.

3.

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4.

58. If a function f(x) in defined in the internal ( + 2

p

) the Fourier expansion is

1.

2.

3.

4.

59. If x = a is a point of discontinuity of the function f(x), then the Fourier series of f(x) is

convergent to

1. point itself

2. f(a)

3.

4.

60. If x = a is a point of continuity of the function f(x) then the Fourier series of f(x) is

convergent to

1. f(a)

2. point itself

3.

4.

61. In the Fourier expansion of f(x) = 2x - x

in (0, 3) a

=

2

n

1.

2.

3.

4.

62. In the Fourier expansion of f(x) = in (-

p

,

p

), b

=

n

1.

2.

p

3.

4. 0

63. In the Fourier expansion of = in -

p

x

p

, a

=

0

1.

p

e

p

2. 1

3.

4. 0

64. In the Fourier expansion of f(x) = cos

a

x in (0. 2

p

)

a

being non-integral a

=

0

1.

2.

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3.

4.

65. The Fourier series of f(x) = 0, -

p

x 0

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = sinx, 0 x

p

at x =

p

/2 is convergent to

1. 0

2.

p

3. 1

4. -

p

66. If

x then the series at x = 0 becomes

1.

2.

3.

4.

67. In the Fourier expansion of f(x) = e defined in 0 x 2

p

, a

=

n

1. )

2. )

3. )

4.

68. The Fourier series of f(x) = -

p

, -

p

x 0

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = 0, 0 x

p

at x = 0 is convergent to

1. -

p

/2

2.

p

/2

3. 0

4.

p

69. In the Fourier expansion of f(x) in (c, c + 2l) b

=

n

1.

2.

3.

4.

70. In the Fourier expansion of f(x) = l - x, 0 x

=

l

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = 0, l x

=

2l

a

=

0

1. /2

2.

3. -

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4. - /2

71. The Fourier expansion f(x) = -k, -

p

x 0

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = k, 0 x x

is given by

1.

2.

3.

4.

72. In the Fourier expansion of =

1. 0

2.

3.

4.

p

73. In the Fourier expansion of =

1.

2.

3.

p

4. 0

74. The function

1. odd

2. even

3. constant function

4. neither even nor odd

75. The function is

1. odd

2. even

3. constant function

4. neither even nor odd

76. Expanding the function f(x) = x sin x as a fourier series in the interval

a

=

1

1.

2.

3.

4.

77. The Fourier expansion of an even function f(x), - x contains

1. Sine terms only

2. Both sine and cosine terms

3. Cosine terms only

4. constant terms

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78. The Fourier expansion of an odd function f(x), in - x contains

1. cosine terms only

2. both sine and cosine terms only

3. sine terms only

4. constant terms

79. The even function f(x) in the interval - x is symmetrical about

1. y - axis

2. x - axis

3. y = x

4. both the axis

80. The odd function f(x) in the interval - x is symmetrical about

1. y - axis

2. y = x

3. x - axis

4. in opposite quadrants

81. For half range cosine expansion of f(x) in (- , ) an =

1.

2.

3.

4.

82. The half-range cosine expansion of x in 0 x 2 is given by

1.

2.

3.

4.

83. The half range sine series for e

in 0 x 1 in given by

x

1.

2.

3.

4.

84. In the half range cosine expansion of f(x) = x

in 0, 1) a

=

3

n

1.

2.

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3.

4.

85. If f(x) = (x - 1)

in 0 x 1, the half range cosine series is given by

2

1.

2.

3.

4.

86. Fourier integral of f(x) is

1.

2.

3.

4.

87. Fourier sine integral is

1.

2.

3.

4.

88. Fourier cosine integral is

1.

2.

3.

4.

89. For half range sine expansion of f(x) in (- , ), b

=

n

1.

2.

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3.

4.

90. Fourier Transform of f(x) is

1.

2.

3.

4.

91. Fourier sine transform of f(x) is

1.

2.

3.

4.

92. If F

(s) and F

(s) are the Fourier sine and cosine transform of f(x) respectively then F

[f

s

c

s

(x)]=

1

1. sF

(s)

c

2. -sF

(s)

c

3. -sF

(s)

s

4. sF

(s)

s

93. Fourier cosine transform of f(x) is

1.

2.

3.

4.

94. Inverse Fourier sine transform of F

(s) is

s

1.

2.

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3.

4.

95. Inverse Fourier cosine transform of F

(s) is

c

1.

2.

3.

4.

96. In F

(s) is the Fourier sine transform of f(x),

s

1.

2.

3.

4.

97. If F

(s) and F

(s) are the Fourier sine and cosine transform of f(x) respectively then Fc

s

c

[x f(x)] =

1.

2.

3.

4.

98. If F

(s) and F

(s) are the Fourier sine and cosine transform of f(x) respectively then Fs

s

c

[x f(x)]

1.

2.

3.

4.

99. If Fc (s) is the Fourier Cosine transform of f(x) F

[f(x) sin ax].

c

1.

2.

3.

4.

100. If F

(s) and F

(s) are the Fourier sine and Cosine transform of f(x) respectively, then F

s

c

s

[f(x) sin ax]

1.

2.

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3.

4.

101.

1.

2.

3.

4.

p

102. F

(e ) =

c

1.

2.

3.

4.

103. The Fourier transform of is

1.

2.

3.

4.

104. Fourier transform of is

1.

2.

3.

4.

105. Fourier cosine transform of f(x) where is

1.

2.

3.

4.

106. Fourier cosine transform of is

1.

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2.

3.

4.

107. Fourier transform of is

1.

2.

3.

4.

108. If F(s) is the Fourier transform of f(x), then F[f (ax)]

1.

2.

3.

4.

109. If F

(s) is the Fourier cosine transform of f(x) then

c

1. F

(a s)

c

2. - F (as)

3.

4.

110. =

1.

2.

3.

4.

111.

1.

2.

3.

4.

112. Fourier sine transform of is

1.

2.

3.

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4.

113. F

(e ) =

s

1.

2.

3.

4.

114. F

[f(x)] where f(x) = 2e + 5e is

s

1.

2.

3.

4.

115. F

[f(x)] where f(x) = 2e + 5e is

c

1.

2.

3.

4.

116. If then =

1.

2.

3.

4.

117. If

1.

2.

3.

4.

118. F

[f(x)] where is

s

1.

2.

3.

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4.

119. If , then f(x) =

1. e

x

2. e

3. e

4. e

120. If inverse Fourier sine transform of as then

1.

2.

3. 1

4. 0

121. If , then F[f(x)] =

1.

2.

3.

4.

122. If

then F[f(x)]

1.

2.

3.

4.

123. Finite Fourier sine transform of f(x) = x

, 0 x 4 is

2

1.

2.

3.

4.

124. Finite Fourier sine transform of f(x) = x, 0 x 4 is

1.

2.

3.

4.

125. Finite Fourier cosine transform of f(x) x, 0 x 4 is

1.

2.

3.

4.

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126. Finite Fourier cosine transform of in (0,

p

) is

1.

2.

3.

4.

127. Finite Fourier sine transform of , 0 x

p

is

1.

2.

3.

4.

128. Finite Fourier sine transform of f(x) in 0 x l is

1.

2.

3.

4.

129. Finite Fourier cosine transform of f(x) in 0 x l is

1.

2.

3.

4.

130. Finite cosine transform of f(x) = x

, 0 x 4

2

1.

2.

3.

4.

131. Finite sine transform of , 0 x

p

is

1.

2.

3.

4.

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132. The partial differential equation formed by eliminating the arbitrary constants a and b

from z = (x

+ a) (y

+ b)

2

2

1. xyz = pg

2. yz = p

3. 4xyz = pq

4. xyz = q

133. The partial differential equation formed by eliminating the arbitrary constants a and b

from (x - a)

+ (y - b)

+ z

= r

2

2

2

2

1. z

(p

- q

+ 1) = r

2

2

2

2

2. z

(p

+ q

+1) = r

2

2

2

2

3. z

(p

+ q

- 1 ) = r

2

2

2

2

4. z

(p

q

1) = r

2

2

2

2

134. Partial differential equation of all planes passing though the origin is

1. z = py + qx

2. z = px - qy

3. z = px + qy

4. z = py - qx

135. Partial differential equation of all spheres whose centers lie on z-axis given radius r is

1. xp - yq

2. xq - yp = 0

3. xp + yq = 0

4. xq + yp = 0

136. Partial differential equation formed from z = f(x

+ y

+ z

) is

2

2

2

1. py = qx

2. px = qy

3. xy = pq

4. xyp = q

137. The partial differential equation formed by eliminating the arbitrary constant a and b

from z = ax + by + ab is

1. z = p + q + pq

2. Z = px + qy + pq

3. z = py + qx

4. z = py + qx + pq

138. Partial differential equation formed from z = f(x

+ y

) is

2

2

1. py - xq = 0

2. py - qx = 0

3. py + xq = 0

4. px + qy = 0

139. Partial differential equation formed from z = is

1. px - qy = 0

2. px - qy = 0

3. py + qx = 0

4. px + qy = 0

140. Partial differential equation formed from z = f

(x) f

(y) is

1

2

1. pq = zp

2. pq = zq

3. pq = zs

4. pq = zr

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141. Partial differential equation forme from z = f(x + it) + g( x- it) is

1.

2.

3.

4.

142. Partial differential equation formed from Z = is

1. px + qy = 0

2. py - qx = 0

3. px - qy = 0

4. py + qx = 0

143. The partial differential equation formed from xyz = Ø (x + y + z ) is

1. x(y + z) + y(z + x) = z(x + y)

2. x(y - z) + y(z - x) = z(x - y)

3. y(y z) + x (z- x) = (x y)

4. z(y z) + z (z- x) = (x y)

144. The solution of px + qy = z is

1.

2.

3.

4.

145. One of the solutions of the differentia equation

x(y

- z

)p + y (z

- x

)q = z (x

- y

) taking x, y, z as multipliers

2

2

2

2

2

2

1. x

+y

-z

= a

2

2

2

2. x

+y

+z

=a

2

2

2

3. x

-y

-z

= a

2

2

2

4. x

-y

+z

= a

2

2

2

146. The solution of (y

+ z

)p - xyq = -zx is

2

2

1. Ø(xd+y+z, xyz) = 0

2. Ø(x+y+z, ) = 0

3. Ø(x

+y

+z

, ) = 0

2

2

2

4. Ø(x

+y

+z

, ) = 0

2

2

2

147. One of the solutions of the differential equation

x(y - x)p + y(z - x)q = z(x- y) taking 1, 1, 1 as multiplexes is

1. x+y+z =a

2. x-y-z = a

3. x + y - z = a

4. x - y + z = a

148. One of the solution of the differential equation

x(y - z)p + y(z - x) q = z (x - y) taking as multipliers is

1. xy = za

2. xz = ay

3. yz = ax

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4. xyz = a

149. One of the solutions of the differential equation

x(y

- z

)p + y (z

- x

)q = z (x

- y

) taking as multipliers is

2

2

2

2

2

2

1. xy = za

2. xz = ay

3. xyz = a

4. yz = ax

150. The solution is

1.

2.

3.

4.

151. The solution px + qy = nz is

1.

2.

3.

4.

152. The solution of xzp + y = q = xy

1.

2.

3.

4.

153. The solution of z = pq is

1. (x + ay + c)

= 4 az

2

2. (x + ay + c)

= 4 ax

2

3. (x + ay + c)

= 4 ay

2

4. (x + ay + c)

= 4 axy

2

154. The solution of pq = y is

1.

2.

3.

4.

155. The solution of p + q = 1 is

1. Ø (x - z, y - z) = 0

2. Ø (x+z, y-z) = 0

3. Ø (x + z, y + z) = 0

4. Ø (x - z, y + z) = 0

156. Complete integral of p

+ q

= 1 is

2

2

1.

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2.

3.

4.

157. The singular integral of pq = k is

1.

2.

3. z = ax + y

4. no singular integral

158. The complete solution of is

1.

2.

3.

4.

159. The complete solution of p = q

is

2

1. z = bzx + by + c

2. z = -bzx + by + c

3. z = -bzx - by+c

4. z = bzx - by + c

160. The solution of pe

= qe

is

y

x

1. z = ae

+ c

x

2. z = ae

+ c

y

3. z = ae + c

4. z = a(e

+ e

) + c

x

y

161. The solution of x

p

+ y

q

= z

2

2

2

2

2

1. y = a logx

2.

3. x

z

+ y

b

= z

2

2

2

2

2

4. y = a logx + logy + c

162. Solution of p

+ q

= x + y is

2

2

1.

2.

3.

4. x + y = a

+ b

2

2

163. The solution of px = qy is

1. z = logx + logy + c

2. z = logx

+ logy

+ c

a

a

3. z = xyz b + c

4. z = xy + c

164. The solution of z = px + qy - 2 is

1.

2.

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3.

4.

165. The solution of z

= pqxy is

2

1.

2.

3. axy

4. abxy

166. The solution is

1.

2.

3.

4.

167. Solution by p

- q

= x - y is

2

2

1.

2.

3.

4.

168. The general solution of dx = dy = dz is

1. f(x - y, y - z) = 0

2. f(x + y, y +z) = 0

3. f(x + y z, x+ y) = 0

4. f(x + y +2z, x - y ) = 0

169. The general solution of p

= qz is

3

1. 4z=a(x+ay-b)

2

2. 4z=a(x-ay-b)

2

3. 4z=a(ay-x-b)

2

4. 4z=a(b-ay-x)

2

170. If the solution of the equation by the method of separation of variables

is z = X(x) Y(y) then X(x)

1.

2.

3.

4.

171. z(n

) =

2

1.

2.

3.

4.

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172. z(cosn

) =

1.

2.

3.

4.

173. z(sin n

) =

1.

2.

3.

4.

174. The general solution of p cot x + q cot y = cot z is

1.

2.

3.

4.

175. If the solution of the equation = 0 by the method of separation of variables

is z = X(x) Y(y) the Y(y) =

1.

2.

3.

4.

176. If the solution of the equation by the method of separation of variables

is u = X(x) Y(y) then X(x)

1.

2.

3.

4.

177. If z(4

) = there z(a u

) =

n

n

1.

2.

3.

4.

178. If z(4

) = then z(a

u

) =

n

n

n

1.

2.

3.

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4.

179. z(-1)

=

n

1.

2.

3.

4.

180. z[4z

+ 5(-1)

]

n

n

1.

2.

3.

4.

181. z(coshn

) =

1.

2.

3.

4.

182. z(n

) =

4

1.

2.

3.

4.

183. Z(sinh n

) =

1.

2.

3.

4.

184.

1.

2.

3.

4.

185.

1.

2.

3.

4.

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186.

1. (n - 1)a

n

2. (n + 1)a

n

3. na

n

4. a

n

187.

1.

2.

3.

4.

188.

1. 1

2. 4

3. 2

4. 0

189. 1 * 1 * 1 =

1. 3

2. 2

3. 1

4. 0

190. for 2 > > 3 is given by

1. and

2. and

3. and

4. and

191. for > 3 is given by

1.

2.

3. and 0 , n < 0 and 0, n< 0 and 0, n < 0

4. 0, n 1

192.

is given by

1.

2.

3.

4.

193. If

1.

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2.

3.

4.

194. Solution of difference equation = 1, y0 = 0 is

1.

2. - 2(-1)

3.

4. 2(-1)

195. Solution of difference equation y_{n + 2} - 4y_{n - 1} + 3y_n = 5

is

n

1.

2.

3.

4.

196. Solution

by z-transform is

1.

2.

3.

4.

197. The solution of by z-transform is

1.

2.

3.

4.

198. where

1.

2.

3.

4.

199. =

1. 3

2. 4

3. 2

4. 1

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