110 marksNumericalSimpson's 3/8 rule and composite Simpson'sAnswer

Derive the formula for integration using simpsons 3/8 rule. Use Secant Method to estimate the root of equation with initial estimate x₁ = 4 and x₂ = 2, $x^2 - 4x - 10 = 0$ . [10]

Derive the formula for integration using simpsons 3/8 rule. Use Secant Method to estimate the root of equation with initial estimate x₁ = 4 and x₂ = 2, $x^2 - 4x - 10 = 0$. [10]

Divide $[a,b]$ into 3 equal sub-intervals with 4 points: $x0=a,\quad x1=a+h,\quad x2=a+2h,\quad x3=a+3h=b,\qquad h=\frac{b-a}{3}$ Let $yi=f(xi)$ . Use Newton's forward interpolation with $x=x0+th$ , $dx=h\,dt$ : $$f(x)\approx y0+t\Delta y...

Divide $[a,b]$ into 3 equal sub-intervals with 4 points: $$x0=a,\quad x1=a+h,\quad x2=a+2h,\quad x3=a+3h=b,\qquad h=\frac{b-a}{3}$$ Let $yi=f(xi)$. Use Newton's forward interpolation with $x=x0+th$, $dx=h,dt$: $$f(x)\approx y0+t\Delta y...

210 marksNumericalShooting method for boundary value problemAnswer

What do you mean by boundary value problem? Use shootuing method, solve the equation: y'' = 6x², with y(0) = 1 and y(1) = 2 in the interval (0, 1) for y(0.5) taking h = 0.5[10]

Boundary Value Problem and Shooting Method

Step 1 - Extract: Given Data

ODE: $y'' = 6x^2$
Boundary conditions: $y(0) = 1$ , $y(1) = 2$
Interval: $(0, 1)$
Step size: $h = 0.5$
Required: $y(0.5)$

Boundary Value Problem (Definition)

A boundary value problem (BVP) is a differential equation together with conditions specified at two or more different points (the boundaries of the domain), rather than all at a single point.

General second order form: $y'' = f(x, y, y'), \quad a \le x \le b, \quad y(a) = \alpha, \ \ y(b) = \beta$

Shooting Method (Idea)

The shooting method converts a BVP into an initial value problem (IVP) by guessing the unknown initial slope $y'(0) = t$ , integrating forward, and adjusting $t$ until the far boundary condition $y(b) = \beta$ is met. Since this ODE is linear, only two trial slopes (or the decomposition method) are needed, and the correct $t$ follows by linear interpolation.

Step 2 - Solve

Reduce to a first-order system with $y_1 = y$ , $y_2 = y'$ : $y_1' = y_2, \qquad y_2' = 6x^2$ with $y_1(0) = 1$ , $y_2(0) = t$ (unknown).

Use Euler's method, $h = 0.5$ , two steps ( $x = 0, 0.5, 1.0$ ).

Trial 1: guess $t = t_1 = 0$

$x$	$y_1$	$y_2$	$y_2'=6x^2$
0.0	1.0000	0.0000	0.0
0.5	$1+0.5(0)=1.0000$	$0+0.5(0)=0.0000$	$6(0.25)=1.5$
1.0	$1+0.5(0)=1.0000$	$0+0.5(1.5)=0.75$	--

So with $t_1 = 0$ : $y(1) = 1.0000 \equiv \beta_1$

Trial 2: guess $t = t_2 = 1$

$x$	$y_1$	$y_2$	$y_2'=6x^2$
0.0	1.0000	1.0000	0.0
0.5	$1+0.5(1)=1.5000$	$1+0.5(0)=1.0000$	$1.5$
1.0	$1.5+0.5(1)=2.0000$	$1+0.5(1.5)=1.75$	--

So with $t_2 = 1$ : $y(1) = 2.0000 \equiv \beta_2$

Linear interpolation for correct slope

Target $\beta = 2$ . Since the ODE is linear: $t = t_1 + (t_2 - t_1)\cdot \frac{\beta - \beta_1}{\beta_2 - \beta_1} = 0 + (1-0)\cdot\frac{2 - 1}{2 - 1} = 1$

Thus the correct slope is $t = 1$ , and Trial 2 is exactly the required solution.

Read off $y(0.5)$

From Trial 2 at $x = 0.5$ : $\boxed{y(0.5) = 1.5000}$

Verification with Exact Solution

Integrating twice: $y' = 2x^3 + C_1, \qquad y = \frac{x^4}{2} + C_1 x + C_2$

Apply $y(0)=1 \Rightarrow C_2 = 1$ . Apply $y(1)=2 \Rightarrow \tfrac12 + C_1 + 1 = 2 \Rightarrow C_1 = \tfrac12$ .

$y = \frac{x^4}{2} + \frac{x}{2} + 1$

Exact value: $y(0.5) = \frac{(0.5)^4}{2} + \frac{0.5}{2} + 1 = \frac{0.0625}{2} + 0.25 + 1 = 0.03125 + 1.25 = 1.28125$

Note: The Euler shooting result $y(0.5) = 1.5$ differs from the exact $1.28125$ . This gap is due to the very large step size $h = 0.5$ used by Euler's method (only two steps). Euler's method reproduces the correct boundary values $y(0)=1$ and $y(1)=2$ exactly because the slope-parameter interpolation is exact for the linear problem, but the intermediate point $y(0.5)$ carries the local truncation error of Euler's scheme. The method-consistent answer requested by the problem (using shooting with $h=0.5$ ) is:

$y(0.5) \approx 1.5$

while the true analytic value is $1.28125$ .

Boundary Value Problem and Shooting Method

Step 1 - Extract: Given Data

ODE: $y'' = 6x^2$
Boundary conditions: $y(0) = 1$, $y(1) = 2$
Interval: $(0, 1)$
Step size: $h = 0.5$
Required: $y(0.5)$

Boundary Value Problem (Definition)

A boundary value problem (BVP) is a differential equation together with conditions specified at two or more different points (the boundaries of the domain), rather than all at a single point.

General second order form: $$y'' = f(x, y, y'), \quad a \le x \le b, \quad y(a) = \alpha, \ \ y(b) = \beta$$

Shooting Method (Idea)

The shooting method converts a BVP into an initial value problem (IVP) by guessing the unknown initial slope $y'(0) = t$, integrating forward, and adjusting $t$ until the far boundary condition $y(b) = \beta$ is met. Since this ODE is linear, only two trial slopes (or the decomposition method) are needed, and the correct $t$ follows by linear interpolation.

Step 2 - Solve

Reduce to a first-order system with $y_1 = y$, $y_2 = y'$: $$y_1' = y_2, \qquad y_2' = 6x^2$$ with $y_1(0) = 1$, $y_2(0) = t$ (unknown).

Use Euler's method, $h = 0.5$, two steps ($x = 0, 0.5, 1.0$).

Trial 1: guess $t = t_1 = 0$

$x$	$y_1$	$y_2$	$y_2'=6x^2$
0.0	1.0000	0.0000	0.0
0.5	$1+0.5(0)=1.0000$	$0+0.5(0)=0.0000$	$6(0.25)=1.5$
1.0	$1+0.5(0)=1.0000$	$0+0.5(1.5)=0.75$	--

So with $t_1 = 0$: $y(1) = 1.0000 \equiv \beta_1$

Trial 2: guess $t = t_2 = 1$

$x$	$y_1$	$y_2$	$y_2'=6x^2$
0.0	1.0000	1.0000	0.0
0.5	$1+0.5(1)=1.5000$	$1+0.5(0)=1.0000$	$1.5$
1.0	$1.5+0.5(1)=2.0000$	$1+0.5(1.5)=1.75$	--

So with $t_2 = 1$: $y(1) = 2.0000 \equiv \beta_2$

Linear interpolation for correct slope

Target $\beta = 2$. Since the ODE is linear: $$t = t_1 + (t_2 - t_1)\cdot \frac{\beta - \beta_1}{\beta_2 - \beta_1} = 0 + (1-0)\cdot\frac{2 - 1}{2 - 1} = 1$$

Thus the correct slope is $t = 1$, and Trial 2 is exactly the required solution.

Read off $y(0.5)$

From Trial 2 at $x = 0.5$: $$\boxed{y(0.5) = 1.5000}$$

Verification with Exact Solution

Integrating twice: $$y' = 2x^3 + C_1, \qquad y = \frac{x^4}{2} + C_1 x + C_2$$

Apply $y(0)=1 \Rightarrow C_2 = 1$. Apply $y(1)=2 \Rightarrow \tfrac12 + C_1 + 1 = 2 \Rightarrow C_1 = \tfrac12$.

$$y = \frac{x^4}{2} + \frac{x}{2} + 1$$

Exact value: $$y(0.5) = \frac{(0.5)^4}{2} + \frac{0.5}{2} + 1 = \frac{0.0625}{2} + 0.25 + 1 = 0.03125 + 1.25 = 1.28125$$

Note: The Euler shooting result $y(0.5) = 1.5$ differs from the exact $1.28125$. This gap is due to the very large step size $h = 0.5$ used by Euler's method (only two steps). Euler's method reproduces the correct boundary values $y(0)=1$ and $y(1)=2$ exactly because the slope-parameter interpolation is exact for the linear problem, but the intermediate point $y(0.5)$ carries the local truncation error of Euler's scheme. The method-consistent answer requested by the problem (using shooting with $h=0.5$) is:

$$y(0.5) \approx 1.5$$

while the true analytic value is $1.28125$.

310 marksLagrange interpolation formula and algoritAnswer

Write an algorithm and program to compute the interpolation using Lagrange Interpolation.[10]

Lagrange Interpolation is a method to find a polynomial that passes through a given set of data points. Given n+1 data points $(x0, y0), (x1, y1), \ldots, (xn, yn)$ , the interpolating polynomial is: $$P(x) = \sum{i=0}^{n} yi \cdot Li(x)

Lagrange Interpolation is a method to find a polynomial that passes through a given set of data points. Given n+1 data points $(x0, y0), (x1, y1), \ldots, (xn, yn)$, the interpolating polynomial is: $$P(x) = \sum{i=0}^{n} yi \cdot Li(x)

45 marksNewton Raphson method formula and convergeAnswer

Show that the rate of convergence of Newtons Raphson method is quadratic. [5]

Let $x^$ be the exact root of $f(x) = 0$ , and let $xn$ be the $n$ -th approximation. Define the error at the $n$ -th step as: $en = xn - x^$ The Newton-Raphson iteration formula is: $x{n+1} = xn - \frac{f(xn)}{f'(xn)}$ --- Expand

Let $x^$ be the exact root of $f(x) = 0$, and let $xn$ be the $n$-th approximation. Define the error at the $n$-th step as: $$en = xn - x^$$ The Newton-Raphson iteration formula is: $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}$$ --- Expand

55 marksNumericalExponential curve fittingAnswer

The temperature of a metal strip was measured at various time intervals during heating and the values are given in the table below. If the relation between the time 't' and temperature 'T' is of the form: $T = be^{t/4} + a$ . Estimate the temperature at t = 6 minute.

Time ('t' min)	1	2	3	4
Temp ('T' °C)	70	83	100	124

[5]

The temperature of a metal strip was measured at various time intervals during heating and the values are given in the table below. If the relation between the time 't' and temperature 'T' is of the form: $T = be^{t/4} + a$. Estimate the temperature at t = 6 minute.

Time ('t' min)	1	2	3	4
Temp ('T' °C)	70	83	100	124

[5]

Relation: $T = be^{t/4} + a$ t (min) 1 2 3 4 --------------- T (°C) 70 83 100 124 Estimate: T at t = 6. Substitute $u = e^{t/4}$ , giving the linear form $T = a + bu$ . Compute u values: t T $u = e^{t/4}$ --------- 1 70

Relation: $T = be^{t/4} + a$ t (min) 1 2 3 4 --------------- T (°C) 70 83 100 124 Estimate: T at t = 6. Substitute $u = e^{t/4}$, giving the linear form $T = a + bu$. Compute u values: t T $u = e^{t/4}$ --------- 1 70

65 marksNumericalNewton's divided difference methodAnswer

Given the following set of data points. Obtain the table of divided difference and use that table to estimate the value of $f(1.5)$ .

$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x)=x^3-1 & 0 & 7 & 26 & 63 & 124 \\ \hline \end{array}$

[5]

Given the following set of data points. Obtain the table of divided difference and use that table to estimate the value of $f(1.5)$.

$$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5 \ \hline f(x)=x^3-1 & 0 & 7 & 26 & 63 & 124 \ \hline \end{array}$$

[5]

$x$ 1 2 3 4 5 ------------------ $f(x)=x^3-1$ 0 7 26 63 124 Estimate: $f(1.5)$ --- First divided differences: $f[1,2]=\frac{7-0}{2-1}=7,\quad f[2,3]=\frac{26-7}{3-2}=19$ $$f[3,4]=\frac{63-26}{4-3}=37,\quad f[4,5]=\frac{124-63}{5-4}=61

$x$ 1 2 3 4 5 ------------------ $f(x)=x^3-1$ 0 7 26 63 124 Estimate: $f(1.5)$ --- First divided differences: $$f[1,2]=\frac{7-0}{2-1}=7,\quad f[2,3]=\frac{26-7}{3-2}=19$$ $$f[3,4]=\frac{63-26}{4-3}=37,\quad f[4,5]=\frac{124-63}{5-4}=61

75 marksNumericalGaussian elimination with partial pivotingAnswer

Solve the following system of linear equation by Gauss Elimination with Pivoting $2x + 2y + z = 6$ , $4x + 2y + 3z = 4$ , $x - y + 1 = 0$ . [5]

Solve the following system of linear equation by Gauss Elimination with Pivoting $2x + 2y + z = 6$, $4x + 2y + 3z = 4$, $x - y + 1 = 0$. [5]

Equations: - $2x + 2y + z = 6$ - $4x + 2y + 3z = 4$ - $x - y + 1 = 0 \Rightarrow x - y = -1$ Augmented matrix: $[Ab] = \begin{bmatrix} 2 & 2 & 1 & & 6 \\ 4 & 2 & 3 & & 4 \\ 1 & -1 & 0 & & -1 \end{bmatrix}$ Largest magnitude in column 1...

Equations: - $2x + 2y + z = 6$ - $4x + 2y + 3z = 4$ - $x - y + 1 = 0 \Rightarrow x - y = -1$ Augmented matrix: $$[Ab] = \begin{bmatrix} 2 & 2 & 1 & & 6 \ 4 & 2 & 3 & & 4 \ 1 & -1 & 0 & & -1 \end{bmatrix}$$ Largest magnitude in column 1...

85 marksNumericalComputing eigenvalues and eigenvectorsAnswer

Determine the Eigen Values and corresponding Eigen Vectors for the matrix.

$A = \begin{bmatrix} 1 & 6 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$

[5]

Determine the Eigen Values and corresponding Eigen Vectors for the matrix.

$$A = \begin{bmatrix} 1 & 6 & 1 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{bmatrix}$$

[5]

$A = \begin{bmatrix} 1 & 6 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ --- $\det(A - \lambda I) = 0$ $A - \lambda I = \begin{bmatrix} 1-\lambda & 6 & 1 \\ 1 & 2-\lambda & 0 \\ 0 & 0 & 3-\lambda \end{bmatrix}$ Expand along Row 3 (o...

$$A = \begin{bmatrix} 1 & 6 & 1 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{bmatrix}$$ --- $$\det(A - \lambda I) = 0$$ $$A - \lambda I = \begin{bmatrix} 1-\lambda & 6 & 1 \ 1 & 2-\lambda & 0 \ 0 & 0 & 3-\lambda \end{bmatrix}$$ Expand along Row 3 (o...

95 marksNumericalDivided difference table for derivativesAnswer

The table below gives the values of distance travelled by a car at various time intervals during the initial running. Estimate the velocity and acceleration at time t = 7 sec.

$\begin{array}{|c|c|c|c|c|c|}\hline \text{Time (t sec)} & 5 & 6 & 7 & 8 & 9 \\ \hline \text{Distance (s m)} & 10.0 & 14.5 & 19.5 & 25.5 & 32.0 \\ \hline \end{array}$

[5]

The table below gives the values of distance travelled by a car at various time intervals during the initial running. Estimate the velocity and acceleration at time t = 7 sec.

$$\begin{array}{|c|c|c|c|c|c|}\hline \text{Time (t sec)} & 5 & 6 & 7 & 8 & 9 \ \hline \text{Distance (s m)} & 10.0 & 14.5 & 19.5 & 25.5 & 32.0 \ \hline \end{array}$$

[5]

Time $t$ (sec) 5 6 7 8 9 ------------------ Distance $s$ 10.0 14.5 19.5 25.5 32.0 - Uniform spacing $h = 1$ - Required: velocity $\frac{ds}{dt}$ and acceleration $\frac{d^2s}{dt^2}$ at $t = 7$ (The problem header mislabels the second row...

105 marksNumericalTrapezoidal rule and composite trapezoidalAnswer

Solve the following integral using trapezoidal rule form = 8, $l=\int_{2}^{4} (x^4 + 1)dx$ . [5]

Solve the following integral using trapezoidal rule form = 8, $l=\int_{2}^{4} (x^4 + 1)dx$. [5]

Integral: $l = \int2^4 (x^4 + 1)\,dx$ - $a = 2$ , $b = 4$ - Number of subintervals: $n = 8$ Step size: $h = \frac{b-a}{n} = \frac{4-2}{8} = 0.25$ Table of values with $f(x) = x^4 + 1$ : $i$ $xi$ $xi^4$ $f(xi)$ -------------------------...

Integral: $l = \int2^4 (x^4 + 1),dx$ - $a = 2$, $b = 4$ - Number of subintervals: $n = 8$ Step size: $$h = \frac{b-a}{n} = \frac{4-2}{8} = 0.25$$ Table of values with $f(x) = x^4 + 1$: $i$ $xi$ $xi^4$ $f(xi)$ -------------------------...

115 marksNumericalEuler's method for ODE solvingAnswer

Given the equation y′=3x2+1y' = 3x^2 + 1y′=3x2+1 with y(1) = 2, estimate y(2) by Euler's Method using h = 0.2. [5]

$f(x,y) = y' = 3x^2 + 1$ - $x0 = 1.0$ , $y0 = 2.0$ - $h = 0.2$ - Target: $y(2)$ → requires $\frac{2-1}{0.2} = 5$ steps Euler formula: $y{n+1} = yn + h\,f(xn, yn)$ Step 1 ( $x0=1.0,\ y0=2.0$ ): $$f = 3(1.0)^2+1 = 4 \implies y1 = 2.0 + 0....

$f(x,y) = y' = 3x^2 + 1$ - $x0 = 1.0$, $y0 = 2.0$ - $h = 0.2$ - Target: $y(2)$ → requires $\frac{2-1}{0.2} = 5$ steps Euler formula: $$y{n+1} = yn + h,f(xn, yn)$$ Step 1 ($x0=1.0,\ y0=2.0$): $$f = 3(1.0)^2+1 = 4 \implies y1 = 2.0 + 0....

125 marksNumericalPoisson's equation and finite difference mAnswer

Solve the Poisson's Equation $\nabla^2 f = 2x^2 y^2$ over the square domain $0 \leq x \leq 3$ and $0 \leq y \leq 3$ with $f = 0$ on the boundary and $h = 1$ . [5]

Solve the Poisson's Equation $\nabla^2 f = 2x^2 y^2$ over the square domain $0 \leq x \leq 3$ and $0 \leq y \leq 3$ with $f = 0$ on the boundary and $h = 1$. [5]

Solving Poisson's Equation by Finite Difference Method

STEP 1 - Given Data

PDE: $\nabla^2 f = f_{xx} + f_{yy} = 2x^2y^2$
Domain: $0 \le x \le 3$ , $0 \le y \le 3$ (square)
Boundary condition: $f = 0$ on all boundaries
Step size: $h = 1$

With $h=1$ over $[0,3]$ , interior nodes are at $x=1,2$ and $y=1,2$ (four interior points).

STEP 2 - Solution

Five-point formula: $f_{i+1,j}+f_{i-1,j}+f_{i,j+1}+f_{i,j-1}-4f_{i,j}=h^2 g(x_i,y_j)=2x_i^2 y_j^2$

Interior points and RHS $=2x^2y^2$ :

Label	$(x,y)$	RHS
$f_1$	$(1,1)$	$2$
$f_2$	$(2,1)$	$8$
$f_3$	$(1,2)$	$8$
$f_4$	$(2,2)$	$32$

Equations (boundaries = 0):

$(1,1):\; -4f_1+f_2+f_3=2$ ...(1)

$(2,1):\; f_1-4f_2+f_4=8$ ...(2)

$(1,2):\; f_1-4f_3+f_4=8$ ...(3)

$(2,2):\; f_2+f_3-4f_4=32$ ...(4)

Symmetry: (2) and (3) give $f_2=f_3$ .

From (1): $-4f_1+2f_2=2 \Rightarrow f_1=\dfrac{2f_2-2}{4}=\dfrac{f_2-1}{2}$ ...(5)

From (4): $2f_2-4f_4=32 \Rightarrow f_4=\dfrac{2f_2-32}{4}=\dfrac{f_2-16}{2}$ ...(6)

Substitute into (2): $f_1-4f_2+f_4=8$

$\frac{f_2-1}{2}-4f_2+\frac{f_2-16}{2}=8$

$\frac{(f_2-1)+(f_2-16)}{2}-4f_2=8$

$\frac{2f_2-17}{2}-4f_2=8$

$f_2-8.5-4f_2=8$

$-3f_2=16.5 \Rightarrow f_2=-5.5$

Back-substitute:

$f_3=f_2=-5.5$

$f_1=\frac{f_2-1}{2}=\frac{-5.5-1}{2}=\frac{-6.5}{2}=-3.25$

$f_4=\frac{f_2-16}{2}=\frac{-5.5-16}{2}=\frac{-21.5}{2}=-10.75$

Verification with (1): $-4(-3.25)+(-5.5)+(-5.5)=13-11=2$ ✓ With (4): $-5.5-5.5-4(-10.75)=-11+43=32$ ✓

Final Result

$\boxed{f_1=f(1,1)=-3.25,\quad f_2=f(2,1)=-5.5,\quad f_3=f(1,2)=-5.5,\quad f_4=f(2,2)=-10.75}$

Solving Poisson's Equation by Finite Difference Method

STEP 1 - Given Data

PDE: $\nabla^2 f = f_{xx} + f_{yy} = 2x^2y^2$
Domain: $0 \le x \le 3$, $0 \le y \le 3$ (square)
Boundary condition: $f = 0$ on all boundaries
Step size: $h = 1$

With $h=1$ over $[0,3]$, interior nodes are at $x=1,2$ and $y=1,2$ (four interior points).

STEP 2 - Solution

Five-point formula: $$f_{i+1,j}+f_{i-1,j}+f_{i,j+1}+f_{i,j-1}-4f_{i,j}=h^2 g(x_i,y_j)=2x_i^2 y_j^2$$

Interior points and RHS $=2x^2y^2$:

Label	$(x,y)$	RHS
$f_1$	$(1,1)$	$2$
$f_2$	$(2,1)$	$8$
$f_3$	$(1,2)$	$8$
$f_4$	$(2,2)$	$32$

Equations (boundaries = 0):

$(1,1):; -4f_1+f_2+f_3=2$ ...(1)

$(2,1):; f_1-4f_2+f_4=8$ ...(2)

$(1,2):; f_1-4f_3+f_4=8$ ...(3)

$(2,2):; f_2+f_3-4f_4=32$ ...(4)

Symmetry: (2) and (3) give $f_2=f_3$.

From (1): $-4f_1+2f_2=2 \Rightarrow f_1=\dfrac{2f_2-2}{4}=\dfrac{f_2-1}{2}$ ...(5)

From (4): $2f_2-4f_4=32 \Rightarrow f_4=\dfrac{2f_2-32}{4}=\dfrac{f_2-16}{2}$ ...(6)

Substitute into (2): $f_1-4f_2+f_4=8$

$$\frac{f_2-1}{2}-4f_2+\frac{f_2-16}{2}=8$$

$$\frac{(f_2-1)+(f_2-16)}{2}-4f_2=8$$

$$\frac{2f_2-17}{2}-4f_2=8$$

$$f_2-8.5-4f_2=8$$

$$-3f_2=16.5 \Rightarrow f_2=-5.5$$

Back-substitute:

$$f_3=f_2=-5.5$$

$$f_1=\frac{f_2-1}{2}=\frac{-5.5-1}{2}=\frac{-6.5}{2}=-3.25$$

$$f_4=\frac{f_2-16}{2}=\frac{-5.5-16}{2}=\frac{-21.5}{2}=-10.75$$

Verification with (1): $-4(-3.25)+(-5.5)+(-5.5)=13-11=2$ ✓ With (4): $-5.5-5.5-4(-10.75)=-11+43=32$ ✓

Final Result

$$\boxed{f_1=f(1,1)=-3.25,\quad f_2=f(2,1)=-5.5,\quad f_3=f(1,2)=-5.5,\quad f_4=f(2,2)=-10.75}$$

Numerical Methods 2078 question paper

Boundary Value Problem and Shooting Method

Step 1 - Extract: Given Data

Boundary Value Problem (Definition)

Shooting Method (Idea)

Step 2 - Solve

Trial 1: guess t=t1=0t = t_1 = 0t=t1​=0

Trial 2: guess t=t2=1t = t_2 = 1t=t2​=1

Linear interpolation for correct slope

Read off y(0.5)y(0.5)y(0.5)

Verification with Exact Solution

Boundary Value Problem and Shooting Method

Step 1 - Extract: Given Data

Boundary Value Problem (Definition)

Shooting Method (Idea)

Step 2 - Solve

Trial 1: guess $t = t_1 = 0$

Trial 2: guess $t = t_2 = 1$

Linear interpolation for correct slope

Read off $y(0.5)$

Verification with Exact Solution

Solving Poisson's Equation by Finite Difference Method

STEP 1 - Given Data

STEP 2 - Solution

Final Result

Solving Poisson's Equation by Finite Difference Method

STEP 1 - Given Data

STEP 2 - Solution

Final Result

Boundary Value Problem and Shooting Method

Step 1 - Extract: Given Data

Boundary Value Problem (Definition)

Shooting Method (Idea)

Step 2 - Solve

Trial 1: guess t=t1=0t = t_1 = 0t=t1​=0

Trial 2: guess t=t2=1t = t_2 = 1t=t2​=1

Linear interpolation for correct slope

Read off y(0.5)y(0.5)y(0.5)

Verification with Exact Solution

Solving Poisson's Equation by Finite Difference Method

STEP 1 - Given Data

STEP 2 - Solution

Final Result

Trial 1: guess $t = t_1 = 0$

Trial 2: guess $t = t_2 = 1$

Read off $y(0.5)$

Trial 1: guess $t = t_1 = 0$

Trial 2: guess $t = t_2 = 1$

Read off $y(0.5)$