Question

Rework example 1 using $\mathbf{n}\mathbf{=}\mathbf{8}$.

Short answer

So, the approximation of the boundary problem is${\text{y}}_1=0.3492,{\text{y}}_2=0.7202$

${\text{y}}_3=1.1363,{\text{y}}_4=1.6233,{\text{y}}_5=2.2118,{\text{y}}_6=2.9386and{\text{y}}_7=3.8490.$

Step-by-step solution

Definition of finite difference method

Finite difference methods are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences.

Solve for finite difference

Given boundary value problem is compared with (7) we get

$P\left(x\right)=0,Q\left(x\right)=9,f\left(x\right)=0,a=0,b=2$

Since $\text{n}=8$it has,

$\begin{array}{rcl}\text{h}& =& \frac{\text{b}-\text{a}}{\text{n}}\\ & =& \frac{2-0}{8}\\ & =& \frac{1}{4}\\ & =& 0.25\end{array}$

From (8) it gives the finite difference equation as:

$$\begin{gathered} \left(1+\frac{\text{h}}{2}.0\right){\text{y}}_{\text{i}+1}+\left(-2+{\text{h}}^2·\text{a}\right){\text{y}}_{\text{i}}+\left(1-\frac{\text{h}}{2}.0\right){\text{y}}_{\text{i}-1}=0 \\ {\text{y}}_{\text{i}+1}-1.4375{\text{y}}_{\text{i}}+{\text{y}}_{\text{i}-1}=0......\left(1\right) \end{gathered}$$

Solve for interior points

Now the interior points are found as:

$$\begin{gathered} x_1=0+0.25=0.25,x_2=0+2\times 0.25=0.5,x_3=0+3\times 0.25=0.75 \\ {x}_4=0+4\times 0.25=1,x_5=0+5\times 0.25=1.25,x_6=0+6\times 0.25=1.5 \\ {x}_7=0+7\times 0.25=1.75,x_8=0+8\times 0.25=2 \end{gathered}$$

For $\text{i}=1,2,3,4,5,6,7\dots \dots \left(1\right)$ yields;

$$\begin{gathered} y_3-1.4375y_2+y_1=0 \\ {y}_4-1.4375y_3+y_2=0 \\ {y}_5-1.4375y_4+y_3=0 \\ {y}_6-1.4375y_5+y_4=0 \\ {y}_7-1.4375y_6+y_5=0 \\ {y}_8-1.4375y_7+y_6=0 \end{gathered}$$

Use initial condition

But from initial conditions ${\text{y}}_0=4$and ${\text{y}}_8=1$substituting and solving it gives:

$$\begin{gathered} {\text{y}}_1=0.3492 \\ {\text{y}}_2=0.7202 \\ {\text{y}}_3=1.1363 \\ {\text{y}}_4=1.6233 \\ {\text{y}}_5=2.2118 \\ {\text{y}}_6=2.9386 \\ {\text{y}}_7=3.8490 \end{gathered}$$

Thus, the required values are ${\text{y}}_1=0.3492,{\text{y}}_2=0.7202,{\text{y}}_3=1.1363,{\text{y}}_4=1.6233$

${\text{y}}_5=2.2118,{\text{y}}_6=2.9386and{\text{y}}_7=3.8490$