Question

In Problems 1-10 use the finite difference method and the indicated value of n to approximate the solution of the given boundary-value problem.

$\mathit{y}\mathbf{"}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{x}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathit{n}\mathbf{=}\mathbf{10}$

Short answer

$$\begin{gathered} y_1=-0.968,y_2=-2.8077,y_3=-4.3850,y_4=-5.5769, \\ {y}_5=-6.2753,y_6=-6.3905,y_7=-5.8544,y_8=-4.6217,y_9=-2.670 \end{gathered}$$

Step-by-step solution

Consider the function

The function is,

$y"+xy\text{'}+y=x,y\left(0\right)=1,y\left(1\right)=0;n=10$

The standard formula is given as,

$y"+P\left(x\right)y\text{'}+Q\left(x\right)y=F\left(x\right)$

Compute the constants

The values are,

$$\begin{gathered} p\left(x\right)=x,Q\left(x\right)=1,F\left(x\right)=x \\ h=\frac{1-0}{10}\Rightarrow h=\frac{1}{10} \end{gathered}$$

The standard formula is given as,

$\left(1+\frac{h}{2}{P}_i\right)y_{i+1}+(-2+h^2{Q}_i)y_i+\left(1-\frac{h}{2}{p}_i\right)y_{i-1}=h^2{f}_i$

Substitute the values.

$$\begin{gathered} \left(1+\frac{1}{10}\times \frac{1}{2}\times x_i\right)y_{i+1}+\left(-2+{\left(\frac{1}{10}\right)}^2\times 1\right)y_i+\left(1-\frac{1}{10}\times \frac{1}{2}\times x_i\right)y_{i-1}=x_i \\ \Rightarrow \left(1+\frac{x_i}{20}\right)y_{i+1}-1.99y_i+\left(1-\frac{x_i}{20}\right)y_{i-1}=x_i \end{gathered}$$As $n=10\Rightarrow i=1,2,3,4,5,6,7,8,9$.

For,

$$\begin{gathered} i=1\Rightarrow x_1=\frac{1}{10} \\ i=2\Rightarrow x_2=\frac{2}{10} \\ i=3\Rightarrow x_3=\frac{3}{10} \\ i=4\Rightarrow x_4=\frac{4}{10} \\ i=5\Rightarrow x_5=\frac{5}{10} \\ i=6\Rightarrow x_6=\frac{6}{10} \\ i=7\Rightarrow x_7=\frac{7}{10} \\ i=8\Rightarrow x_8=\frac{8}{10} \\ i=9\Rightarrow x_9=\frac{9}{10} \end{gathered}$$

Compute the equations

Substitute the values of variables in the equation.

$$\begin{gathered} i=1\Rightarrow 1.005y_2-1.99y_1+0.995y_0=0.1 \\ i=2\Rightarrow 1.01y_3-1.99y_2+0.99y_1=0.2 \\ i=3\Rightarrow 1.015y_4-1.99y_3+0.985y_2=0.3 \\ i=4\Rightarrow 1.02y_5-1.99y_4+0.98y_3=0.4 \\ i=5\Rightarrow 1.025y_6-1.99y_5+0.975y_4=0.5 \\ i=6\Rightarrow 1.03y_7-1.99y_6+0.97y_5=0.6 \\ i=7\Rightarrow 1.035y_8-1.99y_7+0.965y_6=0.7 \\ i=8\Rightarrow 1.04y_9-1.99y_8+0.960y_7=0.8 \\ i=9\Rightarrow 1.055y_{10}-1.99y_9+0.955y_8=0.9 \end{gathered}$$

Substitute the boundary values $y_0=1,y_{10}=0$, in the equations.

$$\begin{gathered} 1.005y_2-1.99y_1+0.995=0.1 \\ 1.01y_3-1.99y_2+0.99y_1=0.2 \\ 1.015y_4-1.99y_3+0.985y_2=0.3 \\ 1.02y_5-1.99y_4+0.98y_3=0.4 \\ 1.025y_6-1.99y_5+0.975y_4=0.5 \\ 1.03y_7-1.99y_6+0.97y_5=0.6 \\ 1.035y_8-1.99y_7+0.965y_6=0.7 \\ 1.04y_9-1.99y_8+0.960y_7=0.8 \\ -1.99y_9+0.955y_8=0.9 \end{gathered}$$

Final answer

Use computer algebra system to compute the values.

$$\begin{gathered} y_1=-0.968,y_2=-2.8077,y_3=-4.3850,y_4=-5.5769,y_5=-6.2753, \\ {y}_6=-6.3905,y_7=-5.8544,y_8=-4.6217,y_9=-2.670 \end{gathered}$$