Question

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

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

Short answer

$y_0=-5.93807,y_2=0.29551,y_3=0.2276,y_4=-0.031,y_5=-0.4109,y_6=-0.8778,y_7=-1.41156$

Step-by-step solution

Consider the function

The function is,

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

The standard formula is given as,

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

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,

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

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.

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

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.

$\begin{array}{l}i=1\Rightarrow 0.5y_2-y_1+0.375y_0=-2.079\\ i=2\Rightarrow 0.75y_3-1.75y_2+1.25y_1=-0.3465\\ i=3\Rightarrow 0.833y_4-1.88y_3+1.167y_2=-0.1089\\ i=4\Rightarrow 0.875y_5-1.9375y_4+1.125y_3=-0.0433\\ i=5\Rightarrow 0.9y_6-1.96y_5+1.1y_4=-0.0188\\ i=6\Rightarrow 0.9167y_7-1.972y_6+1.083y_5=-0.0079\\ i=7\Rightarrow 0.928y_8-1.979y_7+1.071y_6=-0.0027\end{array}$

Substitute the boundary values, $y_1=0,y_8=-2$ in the equations.

$\begin{array}{l}0.5y_2+0.375y_0=-2.079\\ 0.75y_3-1.75y_2=-0.3465\\ 0.833y_4-1.88y_3+1.167y_2=-0.1089\\ 0.875y_5-1.9375y_4+1.125y_3=-0.0433\\ 0.9y_6-1.96y_5+1.1y_4=-0.0188\\ 0.9167y_7-1.972y_6+1.083y_5=-0.0079\\ -1.856-1.979y_7+1.071y_6=-0.0027\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_0=-5.93807,y_2=0.29551,y_3=0.2276,y_4=-0.031,y_5=-0.4109,y_6=-0.8778,y_7=-1.41156$