Question

Question: Use the finite difference method with n = 10 to approximate the solution of the boundary-value problem

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{55}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}$

Short answer

So, the approximate solutions of the boundary-value problem are

$\begin{array}{ccc}{\text{y}}_1=4.1987,& {\text{y}}_2=8.1049,& {\text{y}}_3=11.384,\\ {\text{y}}_4=13.7038,& {\text{y}}_5=14.777,& {\text{y}}_6=14.4083,\\ {\text{y}}_7=12.5396,& {\text{y}}_8=9.2847,& {\text{y}}_9=4.9450\end{array}$

Step-by-step solution

Definition of finite difference

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 $y^{\text{'}}+6.55(1+x)y=1,y\left(0\right)=0,y\left(1\right)=0$.

Comparing with (7) we get

$P\left(x\right)=0,Q\left(x\right)=6.55(1+x),f\left(x\right)=1,a=0,b=1$

Since n = 10 we have$\text{h}=\frac{\text{b}-\text{a}}{\text{n}}=\frac{1-0}{10}=0.1$

Find interior points

From (8) we get the finite difference equation as

$$\begin{gathered} \left(1+\frac{\text{h}}{2}.0\right){\text{y}}_{\text{i}+1}+\left(-2+{\text{h}}^2\times 6.55\left(1+{\text{x}}_{\text{i}}\right)\right){\text{y}}_{\text{i}}+\left(1-\frac{\text{h}}{2}.0\right){\text{y}}_{\text{i}-1}={\text{h}}^2.1 \\ \Rightarrow {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}}_{\text{i}+1}+\left(0.0655{\text{x}}_{\text{i}}-1.9345\right){\text{y}}_{\text{i}}+{\text{y}}_{\text{i}-1}=0.001\dots \dots \mathrm{..}(*) \end{gathered}$$

Now the interior points are

$$\begin{gathered} x_1=0+0.1=0.1, \\ {x}_2=0+2\times 0.1=0.2, \\ {x}_3=0+3\times 0.1=0.3, \\ {x}_4=0+4\times 0.1=0.4, \\ {x}_5=0+5\times 0.1=0.5, \\ {x}_6=0+6\times 0.1=0.6, \\ {x}_7=0+7\times 1.0=0.7, \\ {x}_8=0+8\times 0.1=0.8, \\ {x}_9=0+9\times 0.1=0.9, \\ {x}_{10}=0+10\times 0.1=1 \end{gathered}$$

Substitution

For $\text{i}=1,2,3,4,5,6,7,8,9\dots \dots (*)$yields

$$\begin{gathered} {\text{y}}_2+(-1.92795){\text{y}}_1+{\text{y}}_0=0.001 \\ {\text{y}}_3+(-1.9214){\text{y}}_2+{\text{y}}_1=0.001 \\ {\text{y}}_4+(-1.91485){\text{y}}_3+{\text{y}}_2=0.001 \\ {\text{y}}_5+(-1.9083){\text{y}}_4+{\text{y}}_3=0.001 \\ {\text{y}}_6+(-1.90175){\text{y}}_5+{\text{y}}_4=0.001 \\ {\text{y}}_7+(-1.8952){\text{y}}_6+{\text{y}}_5=0.001 \\ {\text{y}}_8+(-1.88865){\text{y}}_7+{\text{y}}_6=0.001 \\ {\text{y}}_9+(-1.8821){\text{y}}_8+{\text{y}}_7=0.001 \\ {\text{y}}_{10}+(-1.87555){\text{y}}_9+{\text{y}}_8=0.001 \end{gathered}$$

Use initial value

But from initial conditions ${\text{y}}_0=0,{\text{y}}_{10}=0$substituting and solving we get

$$\begin{gathered} {\text{y}}_1=4.1987 \\ {\text{y}}_2=8.1049 \\ {\text{y}}_3=11.384 \\ {\text{y}}_4=13.7038 \\ {\text{y}}_5=14.777 \\ {\text{y}}_6=14.4083 \\ {\text{y}}_7=12.5396 \\ {\text{y}}_8=9.2847 \\ {\text{y}}_9=4.9450 \end{gathered}$$

Thus, the required solutions are

$\begin{array}{ccc}{\text{y}}_1=4.1987,& {\text{y}}_2=8.1049,& {\text{y}}_3=11.384,\\ {\text{y}}_4=13.7038,& {\text{y}}_5=14.777,& {\text{y}}_6=14.4083,\\ {\text{y}}_7=12.5396,& {\text{y}}_8=9.2847,& {\text{y}}_9=4.9450\end{array}$