Question

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

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

Short answer

$y_1=-0.2259,y_2=-0.3356,y_3=-0.3308,y_4=-0.2167$

Step-by-step solution

Consider the function

The function is,

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

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,

$\begin{array}{l}P\left(x\right)=2,Q\left(x\right)=1,F\left(x\right)=5x\\ h=\frac{1-0}{5}\Rightarrow h=\frac{1}{5}\end{array}$

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.

$\begin{array}{l}\left(1+\frac{1}{5}\times \frac{1}{2}\times 2\right)y_{i+1}+\left(-2+{\left(\frac{1}{5}\right)}^2\times 1\right)y_i+\left(1-\frac{1}{5}\times \frac{1}{2}\times 2\right)y_{i-1}={\left(\frac{1}{5}\right)}^2\times 5x_i\\ \Rightarrow 1.2y_{i+1}-1.96y_i+0.8y_{i-1}=0.2x_i\end{array}$

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

For,

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

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}i=1\Rightarrow 1.2y_2-1.96y_1+0.8y_0=0.04\\ i=2\Rightarrow 1.2y_3-1.96y_2+0.8y_1=0.08\\ i=3\Rightarrow 1.2y_4-1.96y_3+0.8y_2=0.12\\ i=4\Rightarrow 1.2y_5-1.96y_4+0.8y_3=0.16\end{array}$

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

$\begin{array}{l}1.2y_2-1.96y_1=0.04\\ 1.2y_3-1.96y_2+0.8y_1=0.08\\ 1.2y_4-1.96y_3+0.8y_2=0.12\\ -1.96y_4+0.8y_3=0.16\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_1=-0.2259,y_2=-0.3356,y_3=-0.3308,y_4=-0.2167$