Table of Contents
K.P.K, PUBLIC SERVICE COMMISSION, PESHAWAR
COMPETATIVE EXAMINATION FOR PROVINCIAL MANAGEMENT SERVICE, 2010
APPLIED MATHEMATICS, PAPER-II
| TIME: 3 hours | Max Marks: 100 |
Note: Attempt only FIVE questions, selecting at least ONE question from each section.
Each part carries 10 marks.
SECTION A
| Q.1 |
Solve any two of the following differential equations.
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| Q.2 (a) | Under certain conditions, cane sugar is converted into dextrose at a rate which is proportional to the amount unconverted at any time. If out of 75 grams of sugar at t=0, 8 grams are converted during the first 3 minutes. Find the amount converted in 90 minutes. | ||
| (b) |
Apply the method of variation of parameters to solve
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| Q.3 (a) |
Solve the following partial differential equation.
y2p – xyq = x(z – 2y) where
p = ∂z∂x , q = ∂z∂y |
| (b) |
The variation of an elastic string is governed by the P.D.E. ∂2U∂t2 = ∂2U∂x2. The length of the string is π and the ends are fixed. The initial velocity is zero and the initial deflection is U(x, 0) = 2(Sinx + Sin3x). Find the deflection u(x,t) of the vibrating string for t > 0. |
SECTION B
| Q.4 (a) | A covariant tensor has components xy , 2y – z2 , xz in rectangular coordinates. Find its covariant components in spherical coordinates. |
| (b) | If (ds)2 = r2(dθ)2 + r2Sin2θ(dφ)2. Find the value of. |
| Q.5 (a) | If Aij are the cofactors of aij in a determinant Δ of order 3, then show that aij Akj = Δδik | ||||||
| (b) |
Prove any two of the following.
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SECTION C
| Q.6 (a) |
Starting with x0 = 3. Use Newton Raphson method to find a root of x3 – 3x – 5 = 0 , correct to 3 decimal places. |
| (b) |
Find by the method of Regula falsi a root of the equation.
x3 + x2 – 3x – 3 = 0 , lying between 1 and 2.
|
| Q.7 (a) |
Use the method of iteration to solve the equation x = e-x starting with x = 1. Perform 4 iterations up to 4 decimal places. |
| (b) |
Evaluate ∫0π3 √1 – 13Sin2θ dθ , using Simpsons rule with 6 intervals, correct to 3 decimal places. |
| Q.8 (a) |
Solve the following system of equations by Jacobi’s method.
4x + y + 3z = 17
x + 5y + z = 14 2x – y + 8z = 12 |
| (b) |
Solve the following system of equations by using Gauss-Seidel method.
2x – y + 2z = 3
x + 3y + 3z = -1 x + 2y + 5z = 1 |
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