Q5.3-11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{on}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{]}$

Step-by-Step Solution

Answer

The solution is:

$\begin{array}{rcl}& & \mathrm{y}(0.25)=0.75\\ & & \mathrm{y}(0.5)=0.625\\ & & \mathrm{y}(0.75)=0.57353\\ & & \mathrm{y}\left(1\right)=0.563603\end{array}$

Transform the equation

Here h=0.25 0n [0,1]

 

The equations can be written as,

 

$$\begin{gathered} {\mathrm{x}}_1\left(\mathrm{t}\right)=\mathrm{y}\left(\mathrm{t}\right) \\ {\mathrm{x}}_2\left(\mathrm{t}\right)=\mathrm{x}\text{'}\left(\mathrm{t}\right) \end{gathered}$$

 

The transformation of the equation is;

 $$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{x}}_{\mathbf{2}} \end{gathered}$$

The initial conditions are;

$$\begin{gathered} {\mathrm{x}}_1\left(0\right)={\mathrm{y}}_1\left(0\right)=1={\mathrm{x}}_{1,0} \\ {\mathrm{x}}_2\left(0\right)=\mathrm{y}\text{'}\left(0\right)=-1={\mathrm{x}}_{2,0} \end{gathered}$$

 

Apply Euler’s method.

Now, ${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{hf}({\mathrm{t}}_{\mathrm{n}},{\mathrm{x}}_{\mathrm{n}})$

$$\begin{gathered} {\mathrm{t}}_{\mathrm{n}+1}={\mathrm{t}}_{\mathrm{n}}+\mathrm{h} \\ {\mathrm{t}}_1=0+0.25 \\ {\mathrm{x}}_1(0.25)={\mathrm{x}}_{1,1}=0.75 \\ {\mathrm{x}}_2(0.25)={\mathrm{x}}_{2,1}=-0.5 \end{gathered}$$

And 

$$\begin{gathered} {\mathrm{t}}_{\mathrm{n}+1}={\mathrm{t}}_{\mathrm{n}}+\mathrm{h} \\ {\mathrm{t}}_2=0.25+0.25=0.5 \\ {\mathrm{x}}_1(0.5)={\mathrm{x}}_{1,2}=0.625 \\ {\mathrm{x}}_2(0.5)={\mathrm{x}}_{2,2}=-0.20588 \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_3=0.5+0.25=0.75 \\ {\mathrm{x}}_1(0.75)={\mathrm{x}}_{1,3}=0.57353 \\ {\mathrm{x}}_2(0.75)={\mathrm{x}}_{2,3}=-0.039704 \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_4=0.75+0.25=1 \\ {\mathrm{x}}_1\left(1\right)={\mathrm{x}}_{1,4}=0.563604 \\ {\mathrm{x}}_2\left(1\right)={\mathrm{x}}_{2,4}=0.058413 \end{gathered}$$

This is the required result.