Question

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

$\mathit{y}\mathbf{"}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\sqrt{\mathbf{x}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{;}\mathit{n}\mathbf{=}\mathbf{6}$

Short answer

$y_0=4.32317,y_2=-0.336279,y_3=-0.841221,y_4=-1.04362,y_5=-1.06342$

Step-by-step solution

Consider the function

The function is,

$y"+5y\text{'}=4\sqrt{x},y\left(1\right)=1,y\left(2\right)=-1;n=6$

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,

$\begin{array}{l}P\left(x\right)=5,Q\left(x\right)=0,F\left(x\right)=4\sqrt{x}\\ h=\frac{2-1}{6}\Rightarrow h=\frac{1}{6}\end{array}$

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.

$\begin{array}{l}\left(1+\frac{1}{6}\times \frac{1}{2}\times 5\right)y_{i+1}+\left(-2+{\left(\frac{1}{6}\right)}^2\times 0\right)y_i+\left(1-\frac{1}{6}\times \frac{1}{2}\times 5\right)y_{i-1}=4\sqrt{x_i}\\ \Rightarrow 1.4167y_{i+1}-2y_i+0.5833y_{i-1}=0.111\sqrt{x_i}\end{array}$

As$n=6\Rightarrow i=1,2,3,4,5$.

For,

$\begin{array}{l}i=1\Rightarrow x_1=\frac{1}{6}=0.167\\ i=2\Rightarrow x_2=\frac{2}{6}=0.333\\ i=3\Rightarrow x_3=\frac{3}{6}=0.5\\ i=4\Rightarrow x_4=\frac{4}{6}=0.667\\ i=5\Rightarrow x_5=\frac{5}{6}=0.833\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}i=1\Rightarrow 1.4167y_2-2y_1+0.5833y_0=0.0453\\ i=2\Rightarrow 1.4167y_3-2y_2+0.5833y_1=0.0641\\ i=3\Rightarrow 1.4167y_4-2y_3+0.5833y_2=0.0078\\ i=4\Rightarrow 1.4167y_5-2y_4+0.5833y_3=0.090\\ i=5\Rightarrow 1.4167y_6-2y_5+0.5833y_4=0.1014\end{array}$

Substitute the boundary values, $y_1=1,y_6=-1$ in the equations.

$\begin{array}{l}1.4167y_2-2+0.5833y_0=0.0453\\ 1.4167y_3-2y_2+0.5833=0.0641\\ 1.4167y_4-2y_3+0.5833y_2=0.0078\\ 1.4167y_5-2y_4+0.5833y_3=0.090\\ -1.4167-2y_5+0.5833y_4=0.1014\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_0=4.32317,y_2=-0.336279,y_3=-0.841221,y_4=-1.04362,y_5=-1.06342$