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1. UNIT I PARTIAL DIFFERENTIAL EQUATIONS TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS Department of Applied Mathematics

2. Formation of partial differential equations Lagrange’s linear equation Solutions of standard types of first order partial differential equations Linear partial differential equations of second and higher order with constant coefficients Syllabus Department of Applied Mathematics

3. Department of Applied Mathematics Partial Differential Equation Partial differential equation is one which involves partial derivatives. The order of PDE is the order of highest derivative occurring in it.

4. Department of Applied Mathematics Formation of PDE by eliminating arbitrary constant Let us consider the functional relation f(x, y, z, a, b) = 0 -------- (1) Where a and b are arbitrary constant to be eliminated Differentiating (1) partially with respect to x and y, we get Equation (2) and (3) will contain a and b. If we eliminate a and b from (1), (2) and (3) we get the PDE (involving p and q) of the first order.

5. Department of Applied Mathematics Remarks: If the number of constants to be eliminated is equal to number of independent variables, the PDE got after elimination will be of first order. If the number of constants to be eliminated is more than the number independent variables, the resulting PDE will be of second or higher order. Answer is not unique .

6. Problem 1 Form the partial differential equation by eliminating a and b from Solution: Differentiating (1) partially w.r. t x and y we get Department of Applied Mathematics

7. Department of Applied Mathematics Differentiating (1) partially w.r. t ‘y’ we get Substitute (2) and (3) in equation (1), we have

8. Problem 2 Form the partial differential equation by eliminating the arbitrary constants a and b from Solution: Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y , we get

9. Department of Applied Mathematics Substitute (2) and (3) in equation (1), we have

10. Problem 3 Form the partial differential equation by eliminating the arbitrary constants a and b from Solution: Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y , we get

11. Problem 4 Form the partial differential equation by eliminating the arbitrary constants a and b from Solution: Department of Applied Mathematics Substitute (3) in equation (2), we have Diff. eqn. (1) p.w.r.t. x , we get

12. Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. y , we get

13. Problem 5 Find the partial differential equation of all planes cutting equal intercepts from the x and y axes . Solution: The equation of the plane cutting equal intercept from x and y axes is Department of Applied Mathematics Substitute (2) and (3) in equation (1), we have

14. Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y , we get Divide (2) by (3), we get

15. Problem 6 Find the partial differential equation of all planes passing through the origin Solution: The equation of the plane passing through the origin is Department of Applied Mathematics a x + by + cz = 0 where A and B are arbitrary constants

16. Department of Applied Mathematics Problem 7 Find the PDE of all planes which are at a constant distance ‘k’ from the origin . Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y , we get Substitute (2) and (3) in equation (1), we have

17. Department of Applied Mathematics Solution: The equation of the plane having constant distance ‘k’ from the origin is Diff. eqn. (1) p.w.r.t. x , we get Substitute (2) and (3) in equation (1), we have Diff. eqn. (1) p.w.r.t. y , we get

18. Department of Applied Mathematics Problem 8 Form the partial differential equation of all spheres whose centre lies on the z-axis . Solution: Any point on the z-axis is of the form (0, 0, a) Then the equation of the sphere with centre (0, 0, a) and radius k (say) is where ‘a’ is the arbitrary constant.

19. Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y , we get Divide (2) by (3), we get

20. Department of Applied Mathematics Problem 9 Find the partial differential equation of the family of spheres having their centres on the line x = y = z. Since the centre (a, b, c) lies on the line x = y = z, we have a = b = c Hence the equation of the sphere is Solution: ( x – a) 2 + (y – a) 2 + (z – a) 2 = r 2 ---------------- (1) where ‘a’ is the arbitrary constants . Diff. eqn. (1) p.w.r.t. x , we get

21. Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. y , we get Divide (2) by (3), we get

22. Department of Applied Mathematics Formation of PDE by eliminating arbitrary functions Let us consider the relation f (u, v) =0 --------(1) where u and v are functions of x, y ,z and f is an arbitrary function to be eliminated. Differentiating (1) partially with respect to x and y we get

23. Department of Applied Mathematics Remarks: Equation (4) is called Lagrange’s linear PDE whose solution will be discussed later. The order of PDE formed depends only on the number of arbitrary functions eliminated

24. Department of Applied Mathematics Problem 1 Form the partial differential equation by eliminating an arbitrary function from Solution: Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y , we get Divide (2) by (3), we get

25. Department of Applied Mathematics Problem 2 Form the partial differential equation by eliminating the arbitrary functions from Solution: Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y, we get Diff. eqn. (2) p.w.r.t. x , we get Diff. eqn. (2) p.w.r.t. y, we get

26. Department of Applied Mathematics Problem 3 Form the partial differential equation by eliminating an arbitrary function from Solution: Diff. eqn. (3) p.w.r.t. y, we get From (2) and (3) we have Diff. eqn. (1) p.w.r.t. x , we get

27. Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. y, we get Divide (2) by (3), we get

28. Department of Applied Mathematics Problem 4 Eliminate the arbitrary function ‘ f ’ from the relation Solution: Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y, we get

29. Department of Applied Mathematics Dividing (2) by (3), we have

30. Department of Applied Mathematics Problem 5 Form the partial differential equation by eliminating the arbitrary function from Solution: The given equation can be written as Diff. eqn. (1) p.w.r.t. x , we get

31. Department of Applied Mathematics Diff. eqn. (1) p.w.r.t. y, we get Divide (2) by (3), we get

32. Department of Applied Mathematics Problem 6 Eliminate the arbitrary function ‘ f ’ from the relation Solution: The given equation can be written as Diff. eqn. (1) p.w.r.t. x , we get Diff. eqn. (1) p.w.r.t. y, we get

33. Department of Applied Mathematics Dividing (2) by (3), we have

34. Department of Applied Mathematics Lagrange’s linear PDE:(Linear first order PDE) The linear PDE of first order is known as Lagrange’s linear equation is of the form Pp + Qq = R where P,Q, R are functions of x, y, z This is got by eliminating arbitrary function f (u, v)=0 or u=F(v) To solve Pp + Qq = R 1. Form the auxiliary equation of the form 2. Solve these auxiliary simultaneous equation, giving two independent solution u=C 1 and v= C 2 3. The general solution is f (u, v)=0 or u=F(v)

35. Department of Applied Mathematics Problem 1 Find the solution of Solution: This is Lagrange’s linear PDE of the form Pp + Qq =R Take 1 st and 2 nd ratio, we have Integrating, we get

36. Department of Applied Mathematics Take 2 nd and 3 rd ratio, we have Integrating, we get Hence the required solution is

37. Department of Applied Mathematics Problem 2 Using multiplier 1/x, 1/y, 1/z and then add, each ratio is Solution: Solve: This is Lagrange’s linear PDE of the form Pp + Qq =R

38. Department of Applied Mathematics Using multiplier 1, 1, 1 and then add, each ratio is Hence the required solution is

39. Department of Applied Mathematics Problem 3 Solution: Solve: This is Lagrange’s linear PDE of the form Pp + Qq =R Take 1 st and 2 nd ratio, we have Integrating, we get

40. Department of Applied Mathematics Using multiplier x,y,z and then add, each ratio is Integrating, we get Hence the required solution is

41. Department of Applied Mathematics Problem 4 Solution: Using multiplier 1/x, 1/y, 1/z and then add, each ratio is Solve: This is Lagrange’s linear PDE of the form Pp + Qq =R

42. Department of Applied Mathematics Using multiplier x,y,z and then add, each ratio is

43. Department of Applied Mathematics Problem 5 Solution: Integrating, we get Hence the required solution is Solve: This is Lagrange’s linear PDE of the form Pp + Qq =R

44. Department of Applied Mathematics Using multiplier l,m,n and then add, each ratio is

45. Department of Applied Mathematics Using multiplier x,y,z and then add, each ratio is Integrating, we get Hence the required solution is

46. Department of Applied Mathematics Problem 6 Solution: Using multiplier 1/x, (-1/y), 1/z and then add, each ratio is Solve : This is Lagrange’s linear PDE of the form Pp + Qq =R

47. Department of Applied Mathematics Using multiplier x,(-y),-1 and then add, each ratio is

48. Department of Applied Mathematics Integrating, we get Hence the required solution is

49. Department of Applied Mathematics Problem 7 Solution: Solve: This is Lagrange’s linear PDE of the form Pp + Qq =R Take 2 nd and 3 rd ratio, we have

50. Department of Applied Mathematics Using multiplier x,y,z and then add, each ratio is Integrating, we get

51. Department of Applied Mathematics Equate this to 2 nd ratio, we have

52. Department of Applied Mathematics Problem 8 Solution: Hence the required solution is Solve: This is Lagrange’s linear PDE of the form Pp + Qq =R

53. Department of Applied Mathematics (Subtracting 1 st and 2 nd ratio) (Subtracting 2 nd and 3 rd ratio) -----------(A)

54. Department of Applied Mathematics ---------(B) From (A) and (B) we have Integrating we get

55. Department of Applied Mathematics Using multiplier 1,1,1 and then add, each ratio is Using multiplier x,y,z and then add, each ratio is -------(C) ---------(D)

56. Department of Applied Mathematics From (C) and (D) Integrating we get

57. Department of Applied Mathematics Hence the required solution is

58. Department of Applied Mathematics Non-Linear first order PDE (Standard types) Those equations in which p and q occur other than the first degree and product of p, q terms are called non linear first order PDE Ex: p 2 +q 2 +pq = 4 Types of Solutions: A solution in which the number of arbitrary constants is equal to number of independent variable is called complete integral or complete solution . 2. In the complete integral if we give particular value to arbitrary constant we get particular integral

59. Department of Applied Mathematics Eliminate a and b from (1),(2) and (3), the resulting one is called singular integral

60. Department of Applied Mathematics Standard types of first order PDE f (p ,q)=0 ( i.e. equations containing p and q only) Type I Its complete integral is given by z = a x + b y + c -------- (1) where a and b are connected by f (a, b)=0----------(2) From (2) express b as function of a

61. Department of Applied Mathematics Hence there is no singular integral Eliminate ‘a’ from (A) and (B) we get general integral or general solution.

62. Department of Applied Mathematics Problem 1 Find the complete integral of Solution: The complete solution of equation (1) is where Hence the complete integral is This is of the form f(p,q)=0

63. Department of Applied Mathematics Problem 2 Find the complete integral of p-q=0 Solution: The complete solution of equation (1) is Given p – q = 0 ------------ (1) Where a-b=0 => b=a Hence the complete integral is This is of the form f(p,q)=0

64. Department of Applied Mathematics Problem 3 Find the complete integral of Solution: The complete solution of equation (1) is where This is of the form f(p,q)=0

65. Department of Applied Mathematics Hence the complete integral is

66. Department of Applied Mathematics f (p, q, z)=0 ( i.e. equations containing p ,q and z only) Type II The given PDE is f (p, q, z)=0--------(1) Put q=ap in equation (1) and find p and q as function of z Substitute p and q in dz = p dx + q dy (keep z terms in LHS and remaining in RHS) On integrating we get the complete integral Procedure for obtaining Singular integral and general solution are same as explained in type I

67. Department of Applied Mathematics Problem 1 Solve: Solution: This is of the form f(z , p, q) = 0 Let q = ap Then equation (1) becomes p (1 + a p ) = a p z 1 + a p = az

68. Department of Applied Mathematics Substitute p and q in the relation dz = p dx + q dy which is the complete integral

69. Department of Applied Mathematics To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, in turn, we get The last equation is absurd and shows that there is no singular integral. To find general integral, assume b = f(a) Then equation (2) becomes Diff. eqn. (3) p.w.r.t. ‘a’, we get The eliminant of ‘a’ between equations (3) and (4) gives the general integral.

70. Department of Applied Mathematics (i.e. equations containing p , q , x and y) Type III f 1 (x, p) = f 2 (y, q) Let f 1 (x, p) = f 2 (y, q) = a f 1 (x, p) = a and f 2 (y, q) = a solve for p and q (write p as function of x and q as function of y) p=f(x) and q=g(y) Substitute p and q in dz = pdx + qdy => dz= f(x) dx +g(y )dy

71. Department of Applied Mathematics On integrating we get Which is the complete integral contains two arbitrary constant a and b Procedure for obtaining Singular integral and general solution are same as explained in type I

72. Department of Applied Mathematics Problem 1 Solve: Solution: This is of the form f(x , p) = g( y , q )

73. Department of Applied Mathematics Substitute p and q in the relation dz = p dx + q dy which is the complete integral

74. Department of Applied Mathematics To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, in turn, we get The last equation is absurd and shows that there is no singular integral To find general integral, assume b = f(a)

75. Department of Applied Mathematics Then equation (2) becomes Diff. eqn. (3) p.w.r.t. ‘a’, we get The eliminant of ‘a’ between equations (3) and (4) gives the general integral .

76. Department of Applied Mathematics Problem 2 Find the complete integral of Solution: This is of the form f(x , p) = g( y , q ) Let q = a Then equation (1) becomes Substitute p and q in the relation

77. Department of Applied Mathematics which is the complete integral.

78. Department of Applied Mathematics Type IV (Clairaut’s form) An equation of the form z = p x + q y + f (p, q) is known as Clairaut’s equation Its complete integral is z = a x + b y + f (a, b) ------(1) (by replacing p by a and q by b) To find singular integral diff. (1) partially with respect to a, b we get Eliminate a and b from (1), (2) and (3) we get singular integral. Procedure for obtaining general solution are same as explained in type I

79. Department of Applied Mathematics Problem 1 Find the complete integral of Solution: The complete integral of equation (1) is (replacing p by a and q by b ) This is in Clairaut’s form

80. Department of Applied Mathematics Problem 2 Find the singular integral of z=p x +q y +p q Solution: Given z=p x +q y +p q This is in Clairaut’s form The complete integral of equation is z= a x + b y + a b -------(1 ) To find the singular integral, diff. (1) partially w.r.to a and b 0 = x + b => b = -x 0 = y + a => a = -y (1)=> z= -x y – x y + x y => z= -x y (replacing p by a and q by b )

81. Department of Applied Mathematics Problem 3 Find the singular solution of Solution: This is in Clairaut’s form The complete integral of equation (1) is (replacing p by a and q by b ) To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, we get

82. Department of Applied Mathematics Solving (3) and (4) we get Substitute the values of a and b in equation (2) we have

83. Department of Applied Mathematics Problem 4 Find the singular integral of the partial differential equation Solution: This is in Clairaut’s form

84. Department of Applied Mathematics The complete integral of equation (1) is (replacing p by a and q by b ) To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, we get Substitute the values of a and b in equation (2) we have

85. Department of Applied Mathematics Problem 5 Find the singular integral of Solution: This is in Clairaut’s form The complete integral of equation (1) is (replacing p by a and q by b )

86. Department of Applied Mathematics To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, we get Multiplying (3) and (4) we get x y=1, which is the singular integral.

87. Department of Applied Mathematics Problem 6 Solve: Solution: This is in Clairaut’s form The complete integral of equation (1) is To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, we get (replacing p by a and q by b )

88. Department of Applied Mathematics Substitute (3) and (4) in equation (2), we get

89. Department of Applied Mathematics Squaring and adding (3) and (4), we have

90. Department of Applied Mathematics which is the singular integral . To find general integral, assume b = f(a ) Then equation (2) becomes Diff. eqn. (6) p.w.r.t. ‘a’, we get The eliminant of ‘a’ between equations (6) and (7) gives the general integral.

91. Department of Applied Mathematics Problem 7 Find the complete and singular solutions of Solution: This is in Clairaut’s form The complete integral of equation (1) is (replacing p by a and q by b )

92. Department of Applied Mathematics To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, we get Substitute (4) in (3) , we get

93. Department of Applied Mathematics Substitute (4) and (5) in equation (2), we have which is the singular integral

94. Department of Applied Mathematics Problem 1 Find the complete solution of Solution:

95. Department of Applied Mathematics Equation (1) becomes Let Q = aP Then equation (2) becomes

96. Department of Applied Mathematics Substitute P and Q in the relation dz = P dX + Q dY which is the complete solution .

97. Department of Applied Mathematics Problem 2 Solution: Find the general solution of Equation (1) becomes

98. Department of Applied Mathematics Substitute p and q in the relation dz = p dx + q dy

99. Department of Applied Mathematics which is the complete integral. To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, in turn, we get The last equation is absurd and shows that there is no singular integral .

100. Department of Applied Mathematics To find general integral, assume b = f(a) Then equation (2) becomes Diff. eqn. (3) p.w.r.t. ‘a’, we get The eliminant of ‘a’ between equations (3) and (4) gives the general integral .

101. Department of Applied Mathematics Linear PDE of second and higher order with constant coefficients: Homogeneous PDE Non-Homogeneous PDE Homogeneous linear PDE An equation in which the partial derivatives occurring are all of same order and coefficients are constant is called homogeneous PDE Ex:

102. Department of Applied Mathematics Standard form : The standard form of homogeneous PDE of nth order with constant coefficients is of the form The general solution of equation (1) is z=C.F+P.I

103. Department of Applied Mathematics Procedure to find C.F: C.F is the solution of equation The above equation is called auxiliary equation and solving the A.E we get n roots m 1 , m 2 , m 3 , ......... m n , Case I (All the roots are distinct)

104. Department of Applied Mathematics Case II (All the roots are equal) Note : roots may be real or complex Particular Integral: (1) R.H.S is an exponential function then

105. Department of Applied Mathematics (2) R.H.S is an Trigonometric function then (3) R.H.S is an Polynomial function then

106. Department of Applied Mathematics (4) Exponential shift rule (5) General rule

107. Department of Applied Mathematics Problem 1 Solve Solution: A.E. is m 3 – 2m 2 = 0 [Put D = m and D′ = 1] m 2 (m – 2) = 0 m 2 = 0 (or) m – 2 = 0 m = 0, 0, 2

108. Department of Applied Mathematics Problem 2 Solve Solution: A.E. is (m – 1) 3 = 0 [Put D = m and D′ = 1] (m – 1)(m – 1)(m – 1) = 0 m = 1, 1, 1

109. Department of Applied Mathematics Problem 3 Solve Solution: A.E. is m 2 – 2m + 1 = 0 [Put D = m and D′ = 1] (m – 1)(m – 1) = 0 m = 1, 1

110. Department of Applied Mathematics Problem 4 Solve Solution: A.E. is m 3 + m 2 – m – 1 = 0 [Put D = m and D′ = 1] m 2 (m + 1) –1(m + 1) = 0 (m + 1)(m 2 – 1) = 0 m = –1, m 2 = 1 m = 1, –1, –1

111. Department of Applied Mathematics Problem 5 Solve Solution: The given equation can be written as A.E. is m 3 – 2m 2 – 4m + 8 = 0 [Put D = m and D′ = 1] m 2 (m – 2) – 4(m – 2) = 0 (m – 2)(m 2 – 4) = 0 m = 2, m 2 = 4 => m = 2, 2, –2

112. Department of Applied Mathematics Problem 6 Solve Solution: A.E. is m 3 – 7m – 6 = 0 [Put D = m and D′ = 1] m = –1 is a root The other roots are m 2 – m – 6 = 0 ( m – 3)(m + 2) = 0 m = 3, –2 m = –1, –2, 3

113. Department of Applied Mathematics

114. Department of Applied Mathematics

115. Department of Applied Mathematics z = C.F + P.I 1 + P.I 2

116. Department of Applied Mathematics Problem 7 Solve Solution: A.E. is m 3 + m 2 – m – 1 = 0 [Put D = m and D′ = 1] m 2 (m + 1) –1(m + 1) = 0 (m + 1)(m 2 – 1) = 0 m = –1, m 2 = 1 m = 1, –1, –1

117. Department of Applied Mathematics Since the denominator = 0, we have to multiply x on Nr. and Diff. Dr. w.r.t.‘D’

118. Department of Applied Mathematics z = C.F + P.I 1 + P.I 2

119. Department of Applied Mathematics Problem 8 Solve Solution: A.E. is 4m 2 – 4m + 1 = 0 [Put D = m and D′ = 1] (2m – 1) 2 = 0 P.I 1 =

120. Department of Applied Mathematics P.I 2 = z = C.F + P.I 1 + P.I 2

121. Department of Applied Mathematics Problem 9 Solve Solution: A.E. is m 2 + 2m + 1 = 0 [Put D = m and D′ = 1] (m + 1)(m + 1) = 0 m = –1, –1 P.I 1 = Since the denominator = 0, we have to multiply x on Nr. and Diff. Dr. w.r.t.‘D’

122. Department of Applied Mathematics P.I 2 =

123. Department of Applied Mathematics

124. Department of Applied Mathematics z = C.F + P.I 1 + P.I 2 Problem 10 Solve Solution: A.E. is m 2 + 3m – 4 = 0 [Put D = m and D′ = 1] (m – 1)(m + 4) = 0 m = 1, – 4

125. Department of Applied Mathematics P.I 1 =

126. Department of Applied Mathematics P.I 2 = z = C.F + P.I 1 + P.I 2

127. Department of Applied Mathematics Problem 11 Solve Solution: The given equation can be written as A.E. is m 2 + m – 2 = 0 [Put D = m and D′ = 1] (m + 2)(m – 1) = 0 m = –2, 1

128. Department of Applied Mathematics P.I 1 = Since the denominator = 0, we have to multiply x on Nr. and Diff. Dr. w.r.t. ‘D’ we get

129. Department of Applied Mathematics P.I 2 =

130. Department of Applied Mathematics

131. Department of Applied Mathematics z = C.F + P.I 1 + P.I 2

132. Department of Applied Mathematics Problem 12 Solve Solution: The given equation can be written as A.E. is m 2 – 5m + 6 = 0 [Put D = m and D′ = 1] (m – 2)(m – 3) = 0 m = 2, 3

133. Department of Applied Mathematics P.I = where y = c – 3 x

134. Department of Applied Mathematics z = C.F + P.I where y = c – 2 x

135. Department of Applied Mathematics Problem 13 Solve Solution: A.E. is m 2 – 1 = 0 [Put D = m and D′ = 1] m 2 = 1 P.I =

136. Department of Applied Mathematics

137. Department of Applied Mathematics

138. Department of Applied Mathematics z = C.F + P.I

139. Department of Applied Mathematics Non-Homogeneous PDE The general solution of equation (1) is z=C.F+P.I The method to find P.I. is same as those for homogeneous PDE To find C.F.

140. Department of Applied Mathematics

141. Department of Applied Mathematics Problem 1 Solve Solution: The given equation is non-homogeneous.

142. Department of Applied Mathematics Problem 2 Solve Solution: The given equation is non-homogeneous and it can be written as P.I 1 =

143. Department of Applied Mathematics P.I 2 = P.I 3 = z = C.F + P.I 1 + P.I 2 + P.I 3

144. Department of Applied Mathematics Problem 3 Solve Solution: The given equation is non-homogeneous and it can be written as P.I =

145. Department of Applied Mathematics z = C.F + P.I Problem 4 Solve Solution:

146. Department of Applied Mathematics P.I =

147. Department of Applied Mathematics

148. Department of Applied Mathematics z = C.F + P.I Problem 5 Solve Solution:

149. Department of Applied Mathematics P.I 1 =

150. Department of Applied Mathematics

151. Department of Applied Mathematics P.I 2 =

152. Department of Applied Mathematics

153. Department of Applied Mathematics z = C.F + P.I 1 + P.I 2