Question

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

$\mathit{y}\mathbf{"}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{n}\mathbf{=}\mathbf{6}$

Short answer

$y_1=3.3751,y_2=3.6306,y_3=3.6448,y_4=3.2355,y_5=2.1411$

Step-by-step solution

Consider the function

The function is,

$y"-4y\text{'}+4y=(x+1)e^{2x},y\left(0\right)=3,y\left(1\right)=0;n=6$

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,

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

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.

$\begin{array}{l}\left(1+\frac{1}{6}\times \frac{1}{2}\times (-4)\right)y_{i+1}+\left(-2+{\left(\frac{1}{6}\right)}^2\times 4\right)y_i+\left(1-\frac{1}{6}\times \frac{1}{2}\times (-4)\right)y_{i-1}={\left(\frac{1}{6}\right)}^2\left(x_i+1\right)e^{2x_i}\\ \Rightarrow 0.667y_{i+1}-1.889y_i+1.333y_{i-1}=0.027\left(x_i+1\right)e^{2x_i}\end{array}$

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.

$\begin{array}{l}i=1\Rightarrow 0.667y_2-1.889y_1+1.333y_0=0.04396\\ i=2\Rightarrow 0.667y_3-1.889y_2+1.333y_1=0.0701\\ i=3\Rightarrow 0.667y_4-1.889y_3+1.333y_2=0.1100\\ i=4\Rightarrow 0.667y_5-1.889y_4+1.333y_3=0.1707\\ i=5\Rightarrow 0.667y_6-1.889y_5+1.333y_4=0.2620\end{array}$

Substitute the boundary values, $y_0=3,y_6=0$ in the equations.

$\begin{array}{l}0.667y_2-1.889y_1+3.999y_0=0.04396\\ 0.667y_3-1.889y_2+1.333y_1=0.0701\\ 0.667y_4-1.889y_3+1.333y_2=0.1100\\ 0.667y_5-1.889y_4+1.333y_3=0.1707\\ -1.889y_5+1.333y_4=0.2620\end{array}$

Final answer

Use computer algebra system to compute the values.

$\therefore y_1=3.3751,y_2=3.6306,y_3=3.6448,y_4=3.2355,y_5=2.1411$