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.

${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{"}\mathbf{+}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{n}\mathbf{=}\mathbf{8}$

Short answer

$y_1=3.8842,y_2=2.9640,y_3=2.2064,y_4=1.5826,y_5=1.0681,y_6=0.6430,y_7=0.2913$

Step-by-step solution

Consider the function

The function is,

$x^{2}y"+3xy\text{'}+3y=0,y\left(1\right)=5,y\left(2\right)=0;n=8$

$\Rightarrow y"+\frac{3}{x}y\text{'}+\frac{3}{x^2}y=0,y\left(1\right)=5,y\left(2\right)=0;n=8$

The standard formula is given as,

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

Compute the constants

The values are,

$\begin{array}{l}P\left(x\right)=\frac{3}{x},Q\left(x\right)=\frac{3}{x^2},F\left(x\right)=0\\ h=\frac{2-1}{8}\Rightarrow h=\frac{1}{8}\end{array}$

The standard formula is given as,

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

Substitute the values.

$\begin{array}{l}\left(1+\frac{1}{8}\times \frac{1}{2}\times \frac{3}{x_i}\right)y_{i+1}+\left(-2+{\left(\frac{1}{8}\right)}^2\times \frac{3}{{x^2}_i}\right)y_i+\left(1-\frac{1}{8}\times \frac{1}{2}\times \frac{3}{x_i}\right)y_{i-1}=0\\ \Rightarrow \left(1+\frac{3}{16x_i}\right)y_{i+1}+\left(-2+\frac{3}{64{x^2}_i}\right)y_i+\left(1-\frac{3}{16x_i}\right)y_{i-1}=0\end{array}$

As$n=8\Rightarrow i=1,2,3,4,5,6,7$.

For,

$\begin{array}{l}i=1\Rightarrow x_1=\frac{1}{8}\\ i=2\Rightarrow x_2=\frac{2}{8}\\ i=3\Rightarrow x_3=\frac{3}{8}\\ i=4\Rightarrow x_4=\frac{4}{8}\\ i=5\Rightarrow x_5=\frac{5}{8}\\ i=6\Rightarrow x_6=\frac{6}{8}\\ i=7\Rightarrow x_7=\frac{7}{8}\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}i=1\Rightarrow \frac{5}{2}{y}_2+y_1-\frac{1}{2}{y}_0=0\\ i=2\Rightarrow \frac{7}{4}{y}_3-\frac{5}{4}{y}_2+\frac{1}{4}{y}_1=0\\ i=3\Rightarrow \frac{3}{2}{y}_4-\frac{5}{3}{y}_3+\frac{1}{2}{y}_2=0\\ i=4\Rightarrow \frac{11}{8}{y}_5-\frac{29}{16}{y}_4+\frac{5}{8}{y}_3=0\\ i=5\Rightarrow \frac{13}{10}{y}_6-\frac{47}{25}{y}_5+\frac{7}{10}{y}_4=0\\ i=6\Rightarrow \frac{5}{4}{y}_7-\frac{23}{12}{y}_6+\frac{3}{4}{y}_5=0\\ i=7\Rightarrow \frac{17}{14}{y}_8-\frac{95}{49}{y}_7+\frac{11}{14}{y}_6=0\end{array}$

Substitute the boundary values, $y_0=5,y_8=0$ in the equations.

$\begin{array}{l}\frac{5}{2}{y}_2+y_1=\frac{5}{2}\\ \frac{7}{4}{y}_3-\frac{5}{4}{y}_2+\frac{1}{4}{y}_1=0\\ \frac{3}{2}{y}_4-\frac{5}{3}{y}_3+\frac{1}{2}{y}_2=0\\ \frac{11}{8}{y}_5-\frac{29}{16}{y}_4+\frac{5}{8}{y}_3=0\\ \frac{13}{10}{y}_6-\frac{47}{25}{y}_5+\frac{7}{10}{y}_4=0\\ \frac{5}{4}{y}_7-\frac{23}{12}{y}_6+\frac{3}{4}{y}_5=0\\ -\frac{95}{49}{y}_7+\frac{11}{14}{y}_6=0\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_1=3.8842,y_2=2.9640,y_3=2.2064,y_4=1.5826,y_5=1.0681,y_6=0.6430,y_7=0.2913$