Question

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

$\mathit{y}\mathbf{"}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}\mathit{n}\mathbf{=}\mathbf{4}$

Short answer

$y_1=-5.6774,y_2=-2.5807,y_3=6.3226$

Step-by-step solution

Consider the function

The function is,

$y"+9y=0,y\left(0\right)=4,y\left(2\right)=1;n=4$

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,

$\begin{array}{l}P\left(x\right)=0,Q\left(x\right)=9,F\left(x\right)=0\\ h=\frac{1-0}{4}\Rightarrow h=\frac{1}{4}\end{array}$

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.

$\begin{array}{l}\left(1+\frac{1}{4}\times \frac{1}{2}\times 0\right)y_{i+1}+\left(-2+{\left(\frac{1}{4}\right)}^2\times 9\right)y_i+\left(1-\frac{1}{4}\times \frac{1}{2}\times 0\right)y_{i-1}={\left(\frac{1}{4}\right)}^2\times 0\\ \Rightarrow y_{i+1}+0.25y_i+y_{i-1}=0\end{array}$

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

For

$\begin{array}{l}i=1\Rightarrow x_1=\frac{1}{4}\\ i=2\Rightarrow x_2=\frac{2}{4}\\ i=3\Rightarrow x_3=\frac{3}{4}\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}i=1\Rightarrow y_2+0.25y_1+y_0=0\\ i=2\Rightarrow y_3+0.25y_2+y_1=0\\ i=3\Rightarrow y_4+0.25y_3+y_2=0\end{array}$

Substitute the boundary values, $y_0=4,y_4=1$ in the equations.

$\begin{array}{l}{y}_2+0.25y_1+4=0\\ y_3+0.25y_2+y_1=0\\ 1+0.25y_3+y_2=0\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_1=-5.6774,y_2=-2.5807,y_3=6.3226$