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.

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

Short answer

$y_1=-0.0172,y_2=-0.0315,y_3=-0.0323$

Step-by-step solution

Consider the function

The function is,

$y"-y=x^2,y\left(0\right)=0,y\left(1\right)=0;n=4$

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,

$\begin{array}{l}P\left(x\right)=0,Q\left(x\right)=-1,F\left(x\right)=x^2\\ h=\frac{1-0}{4}\Rightarrow h=\frac{1}{4}\end{array}$

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.

$\begin{array}{l}\left(1+\frac{1}{4}\times \frac{1}{2}\times 0\right)y_{i+1}+\left(-2+{\left(\frac{1}{4}\right)}^2\times 9\right)y_i+\left(1-\frac{1}{4}\times \frac{1}{2}\times 0\right)y_{i-1}={\left(\frac{1}{4}\right)}^2\times x_i^2\\ \Rightarrow y_{i+1}-2.0625y_i+y_{i-1}=0.0625x_i^2\end{array}$

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

For

$\begin{array}{l}i=1\Rightarrow x_1=\frac{1}{4}=0.25\\ i=2\Rightarrow x_2=\frac{2}{4}=0.5\\ i=3\Rightarrow x_3=\frac{3}{4}=0.75\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}i=1\Rightarrow y_2-2.0625y_1+y_0=0.0039\\ i=2\Rightarrow y_3-2.0625y_2+y_1=0.0156\\ i=3\Rightarrow y_4-2.0625y_3+y_2=0.0351\end{array}$

Substitute the boundary values,$y_0=0,y_4=0$ in the equations.

$\begin{array}{l}{y}_2-2.0625y_1=0.0039\\ y_3-2.0625y_2+y_1=0.0156\\ -2.0625y_3+y_2=0.0351\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_1=-0.0172,y_2=-0.0315,y_3=-0.0323$