Numerical Methods Important Questions BCA 5th Sem CCSU

Here are the detailed answers to each question tailored for CCSU exams, categorized by their respective marks.

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1. Gauss’s Elimination Method (7 Marks)

Gauss’s elimination method is a systematic way of solving a system of linear equations. The method involves three steps:

1. Forward Elimination: Convert the system of equations into an upper triangular matrix form using elementary row operations.
2. Backward Substitution: Solve for the variables starting from the last equation.
3. Solution: Use back substitution to compute all variables.

Example: Solve:

$$2x + y + z = 3, \\ 4x + 3y + z = 7, \\ 6x + 5y + z = 11.$$

1. Convert the equations to an augmented matrix:

$$\begin{bmatrix} 2 & 1 & 1 & | & 3 \\ 4 & 3 & 1 & | & 7 \\ 6 & 5 & 1 & | & 11 \end{bmatrix}$$

2. Perform row operations to form an upper triangular matrix.
3. Back substitution gives $x = 1, y = 0, z = 1$.

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2. Bisection Method (Four Iterations) (7 Marks)

The bisection method divides the interval $[a, b]$ into two halves to find a root of $f(x) = 0$.

Problem: Solve $x^3 - 4x - 9 = 0$ using the bisection method for four iterations.

1. $f(a) = f(2) = -9$, $f(b) = f(3) = 6$ (root lies in [2, 3]).
2. Iterations:
   • Iteration 1: $c = \frac{2 + 3}{2} = 2.5$, $f(2.5) = -2.375$, root in [2.5, 3].
   • Iteration 2: $c = 2.75$, $f(2.75) = 2.234$, root in [2.5, 2.75].
   • Iteration 3: $c = 2.625$, $f(2.625) = -0.586$, root in [2.625, 2.75].
   • Iteration 4: $c = 2.6875$, $f(2.6875) = 0.797$.

Approximate root: $x \approx 2.6875$.

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3. Trapezoidal Rule and Euler’s Method (7 Marks)

Trapezoidal Rule:

Numerical integration method for approximating the value of a definite integral:

$$\int_a^b f(x) dx \approx \frac{h}{2} [f(a) + f(b)],$$

where $h = b - a$.

Euler’s Method:

A numerical method to solve first-order ODEs. It uses:

$$y_{n+1} = y_n + h f(x_n, y_n).$$

Example: Solve $y' = x + y, y(0) = 1$, $h = 0.1$, $x = 0.1$:

$$y_1 = y_0 + h f(x_0, y_0) = 1 + 0.1 (0 + 1) = 1.1.$$

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4. Picard’s Method (7 Marks)

Picard’s method generates successive approximations for solving ODEs.

Solve:

$$\frac{dy}{dx} = xy^2, \, y(0) = 1.$$

1. First approximation: $y_1 = 1$.
2. Second approximation:

$$y_2 = 1 + \int_0^x x (1)^2 dx = 1 + \frac{x^2}{2}.$$

3. Continue iterating for better approximations.

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5. Newton-Raphson Method for $\sqrt[3]{N}$ (7 Marks)

The formula for the Newton-Raphson method is:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$

For $f(x) = x^3 - N$:

$$f'(x) = 3x^2, \quad x_{n+1} = x_n - \frac{x_n^3 - N}{3x_n^2}.$$

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6. Gauss Elimination Method (Example) (7 Marks)

Solve:

$$2x + y + 4z = 12, \, 8x - 3y + 2z = 23, \, 4x + 11y - z = 33.$$

Follow steps as explained in Question 1. Final solution:

$$x = 1, y = 2, z = 3.$$

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7. Regula-Falsi Method (7 Marks)

The formula is:

$$x = b - \frac{f(b) (b - a)}{f(b) - f(a)}.$$

Derive this from the equation of a straight line between $(a, f(a))$ and $(b, f(b))$. Use iterations like in the bisection method.

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8. Numerical Integration (Examples) (7 Marks)

Gauss Forward Formula:

Interpolate a polynomial using equally spaced points.

Simpson’s 1/3 Rule:

$$\int_a^b f(x) dx \approx \frac{h}{3} [f(a) + 4f(a+h) + f(b)].$$

Lagrange’s Interpolation:

$$P(x) = \sum_{i=0}^n y_i \prod_{j=0, j \neq i}^n \frac{x - x_j}{x_i - x_j}.$$

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9. Gauss-Seidel Method (7 Marks)

Iterative method to solve a system of linear equations:

1. Rewrite equations with diagonally dominant coefficients.
2. Solve iteratively: $$x_1^{(k+1)} = \frac{1 - y_k - z_k}{3}, \, y^{(k+1)} = \ldots$$

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10. False Position Method (7 Marks)

Solve $x^3 - 2x - 5 = 0$ using:

$$x = b - \frac{f(b) (b - a)}{f(b) - f(a)}.$$

Similar steps as bisection.

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11. Runge-Kutta Method (15 Marks)

Given:

$$\frac{dy}{dx} = x + y^2, \, y(0) = 1, \, h = 0.1.$$

Steps:

1. Compute: $$k_1 = f(x_n, y_n), \, k_2 = f(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_1).$$
2. Continue to $k_3, k_4$ and find $y_{n+1}$: $$y_{n+1}=y_n+\frac{h}{6}(k_1+2k_2+2k_3+k_4)\mathrm{.}$$

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12. Solve Using Runge-Kutta Method (4th Order)

Problem:

Solve the initial value problem (IVP):

$$\frac{dy}{dx} = x + y^2, \, y(0) = 1, \, h = 0.2, \, \text{find } y(0.2).$$

Solution:

The Runge-Kutta 4th-order formula is:

$$y_{n+1} = y_n + \frac{h}{6} (k_1 + 2k_2 + 2k_3 + k_4),$$

where:

$$k_1 = f(x_n, y_n),$$ $$k_2 = f(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_1),$$ $$k_3 = f(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_2),$$ $$k_4 = f(x_n + h, y_n + h k_3).$$

Steps:

Initial Values: $x_0 = 0, y_0 = 1, h = 0.2$.

1. Compute $k_1$:

   $$k_1 = f(x_0, y_0) = 0 + 1^2 = 1.$$

2. Compute $k_2$:

   $$k_2 = f(x_0 + \frac{h}{2}, y_0 + \frac{h}{2}k_1) = f(0.1, 1 + 0.1 \times 1) = f(0.1, 1.1).$$ $$k_2 = 0.1 + (1.1)^2 = 0.1 + 1.21 = 1.31.$$

3. Compute $k_3$:

   $$k_3 = f(x_0 + \frac{h}{2}, y_0 + \frac{h}{2}k_2) = f(0.1, 1 + 0.1 \times 1.31) = f(0.1, 1.131).$$ $$k_3 = 0.1 + (1.131)^2 = 0.1 + 1.279 = 1.379.$$

4. Compute $k_4$:

   $$k_4 = f(x_0 + h, y_0 + h k_3) = f(0.2, 1 + 0.2 \times 1.379) = f(0.2, 1.276).$$ $$k_4 = 0.2 + (1.276)^2 = 0.2 + 1.629 = 1.829.$$

5. Compute $y_{n+1}$:

   $$y_1 = y_0 + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4),$$ $$y_1 = 1 + \frac{0.2}{6}(1 + 2 \times 1.31 + 2 \times 1.379 + 1.829),$$ $$y_1 = 1 + \frac{0.2}{6}(8.207) = 1 + 0.2736 = 1.2736.$$

Final Answer:

At $x = 0.2, y(0.2) \approx 1.274$.

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13. Lagrange’s Interpolation

Problem:

Given points $(x_0, y_0) = (1, 1)$, $(x_1, y_1) = (2, 4)$, and $(x_2, y_2) = (3, 9)$, find $y$ when $x = 1.5$.

Solution:

The Lagrange interpolation formula is:

$$P(x) = \sum_{i=0}^n y_i \prod_{j=0, j \neq i}^n \frac{x - x_j}{x_i - x_j}.$$

Steps:

1. Compute $P(x)$:

   $$P(x) = y_0 \frac{(x - x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)} + y_1 \frac{(x - x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)} + y_2 \frac{(x - x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)}.$$

2. Substitute Values:
   For $x = 1.5$,

   $$P(1.5) = 1 \cdot \frac{(1.5 - 2)(1.5 - 3)}{(1 - 2)(1 - 3)} + 4 \cdot \frac{(1.5 - 1)(1.5 - 3)}{(2 - 1)(2 - 3)} + 9 \cdot \frac{(1.5 - 1)(1.5 - 2)}{(3 - 1)(3 - 2)}.$$

   Simplify each term:

   $$P(1.5) = 1 \cdot \frac{(-0.5)(-1.5)}{(-1)(-2)} + 4 \cdot \frac{(0.5)(-1.5)}{(1)(-1)} + 9 \cdot \frac{(0.5)(-0.5)}{(2)(1)}.$$ $$P(1.5) = 1 \cdot \frac{0.75}{2} + 4 \cdot \frac{-0.75}{-1} + 9 \cdot \frac{-0.25}{2}.$$ $$P(1.5) = 0.375 + 3 - 1.125 = 2.25.$$

Final Answer:

At $x = 1.5, y \approx 2.25$.

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14. Simpson’s 1/3 Rule

Problem:

Evaluate:

$$\int_0^6 \frac{1}{1 + x^2} dx,$$

using Simpson’s 1/3 rule with $n = 6$.

Solution:

Simpson’s 1/3 rule is:

$$\int_a^b f(x) dx \approx \frac{h}{3} \left[f(x_0) + 4 \sum_{i=1,3,5} f(x_i) + 2 \sum_{i=2,4} f(x_i) + f(x_n)\right],$$

where $h = \frac{b - a}{n}$.

1. Compute $h$:

   $$h = \frac{6 - 0}{6} = 1.$$

2. Compute $f(x)$:

   $$f(x) = \frac{1}{1 + x^2}.$$

   At $x = 0, 1, 2, 3, 4, 5, 6$:

   $$f(0) = 1, f(1) = 0.5, f(2) = 0.2, f(3) = 0.1, f(4) = 0.0588, f(5) = 0.0385, f(6) = 0.027.$$

3. Apply the Formula:

   $$\int_0^6 \frac{1}{1 + x^2} dx \approx \frac{1}{3} \left[1 + 4(0.5 + 0.1 + 0.0385) + 2(0.2 + 0.0588) + 0.027\right].$$

   Simplify:

   $$\int_0^6 \approx \frac{1}{3} \left[1 + 2.548 + 0.5176 + 0.027\right].$$ $$\int_0^6 \approx \frac{1}{3}(4.0926) = 1.3642.$$

Final Answer:

\int_0^6 \frac