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SUMMER 2013, ASSIGNMENT
| DRIVE | SUMMER 2015 |
| PROGRAM | BACHELOR OF COMPUTER APPLICATION |
| SUBJECT CODE & NAME | BCA3010 -COMPUTER ORIENTED NUMERICAL METHODS |
| SEMESTER | THIRD |
| CREDITS | 4 |
| MAX. MARKS | 60 |
| BK ID | B |
Answer all questions
- 1 Solve the system of equation by matrix inversion method
x +y +z = 1
x +2y + 3z = 6
x + 3y +4z = 6
Solution:-Write the given as a single matrix equation:
- 1 1 x 1
1 2 3 y = 6
1 3 4 z 6
- 2. Find all eigen values and the corresponding eigen vectors of the matrix.
A = I 8 – 6 2 I
I 6 7 4 I
I 2 4 3 I
- 3. Find the cubic polynomial which takes the following values y(0) = 1, y(1) = 0, y(2) = 1 and y(3) = 10. Hence or otherwise, obtain y (0.5).
Solution: –
y = ax^3 + bx^2 = cx +d.
y(0) = 1 = 0+0+0+d, d = 1
y(1) = 0 = a+b+c+1, a+b+c = -1
y(2) = 1 = 8a + 4b + 2c + 1, 8a+4b + 2c = 0
y(3) = 10 = 27a + 9a + 3c + 1, 27a + 9a + 3c = 9 –à 9a + 3b + c
a + b +c = -1
8a + 4b + 2c = 0
9a + 3b + c = 3
Eq. 1: a – 9a + b – 3b + c – c = -1 -3
-8a -2a = -4
-4a –b = -2
4a +b = 2
Eq.2 : 2 – 2*eq.3: 8a – 18a +4b -6b +2c – 2c = 0-6
- 4. Find the approximate value of ò p/2,0 √ cos q dq by Simpson’s 1/3rd rule by dividing [0, p/2] into 6 equal parts.
Solution:- A method for approximating the value of a function near a known value. The method uses the tangent line at the known value of the function to approximate the function’s graph. In this method Δx and Δy represent the changes in x and y for the function, and dx and dy represent the changes in x and y for the tangent line.
- 5. Use Picard’s method of successive approximations to find y1,y2, y3 to the solution of the initial value problem
Solution:-
Y ‘ = y
- 6. Solve x y + y = 0, y (1) = 0, y(2) = 1, h = 2/1
Answer : To solve an equation is to find what values (numbers, functions, sets, etc.) fulfill a condition stated in the form of an equation (two expressions related by equality). When searching a solution, one or more free variables are designated as unknowns. A solution is an assignment of expressions to the unknown variables that makes the equality in the equation true. In other words, a solution is an
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