Question

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

$\mathit{y}\mathbf{"}\mathbf{-}\mathbf{10}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{n}\mathbf{=}\mathbf{5}$

Short answer

$y_1=1.96,y_2=3.88,y_3=7.72,y_4=15.4$

Step-by-step solution

Consider the function

The function is,

$y"-10y\text{'}+25y=1,y\left(0\right)=1,y\left(1\right)=0;n=5$

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,

$\begin{array}{l}P\left(x\right)=-10,Q\left(x\right)=25,F\left(x\right)=1\\ h=\frac{1-0}{5}\Rightarrow h=\frac{1}{5}\end{array}$

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.

$\begin{array}{l}\left(1+\frac{1}{5}\times \frac{1}{2}\times (-10)\right)y_{i+1}+\left(-2+{\left(\frac{1}{5}\right)}^2\times 25\right)y_i+\left(1-\frac{1}{5}\times \frac{1}{2}\times (-10)\right)y_{i-1}={\left(\frac{1}{5}\right)}^2\times 1\\ \Rightarrow -y_i+2y_{i-1}=0.04\end{array}$

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

For,

$\begin{array}{l}i=1\Rightarrow x_1=\frac{1}{5}=0.2\\ i=2\Rightarrow x_2=\frac{2}{5}=0.4\\ i=3\Rightarrow x_3=\frac{3}{5}=0.6\\ i=4\Rightarrow x_4=\frac{4}{5}=0.8\end{array}$

Compute the equations 

Substitute the values of variables in the equation.

$\begin{array}{l}i=1\Rightarrow -y_1+2y_0=0.04\\ i=2\Rightarrow -y_2+2y_1=0.04\\ i=3\Rightarrow -y_3+2y_2=0.04\\ i=4\Rightarrow -y_4+2y_3=0.04\end{array}$

Substitute the boundary values,$y_0=1,y_5=0$ in the equations.

$\begin{array}{l}-y_1+2=0.04\\ -y_2+2y_1=0.04\\ -y_3+2y_2=0.04\\ -y_4+2y_3=0.04\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_1=1.96,y_2=3.88,y_3=7.72,y_4=15.4$