Q5.3-13E - 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.

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

Step-by-Step Solution

Answer

The required solution is;

$$\begin{gathered} \mathrm{y}(0.25)=0.25 \\ \mathrm{y}(0.5)=0.5 \\ \mathrm{y}(0.75)=0.75 \\ \mathrm{y}\left(1\right)=1 \end{gathered}$$

Transform the equation

Here h=0.25 on [0,1].

 

The equations can be written as;

 

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

 

The transformation of the equation is;

 

$$\begin{gathered} \mathrm{x}{\text{'}}_1\left(\mathrm{t}\right)={\mathrm{x}}_2\left(\mathrm{t}\right) \\ \mathrm{x}{\text{'}}_2\left(\mathrm{t}\right)={\mathrm{t}}^2-{{\mathrm{x}}_1}^2 \end{gathered}$$

 

The initial conditions are;

 

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

Apply Euler’s method

Now, 

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

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

And 

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

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

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

This is the required result.