Q5.3-12E - 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.

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

Step-by-Step Solution

Answer

The solution is:

$$\begin{gathered} \mathrm{y}(1.25)=0.75 \\ \mathrm{y}(1.5)=0.625 \\ \mathrm{y}(1.75)=0.6 \\ \mathrm{y}\left(2\right)=0.6548 \end{gathered}$$

Transform the equation

Here h=0.25 on [0,2].

 

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:

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

The initial conditions are:

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

Apply Euler’s method

Now,

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

$$\begin{gathered} {\mathrm{t}}_{\mathrm{n}+1}={\mathrm{t}}_{\mathrm{n}}+\mathrm{h} \\ {\mathrm{t}}_1=1+0.25 \\ =1.25 \\ {\mathrm{x}}_1(1.25)={\mathrm{x}}_{1,1} \\ =0.75 \\ {\mathrm{x}}_2(1.25)={\mathrm{x}}_{2,1} \\ =-0.5 \end{gathered}$$

And 

$$\begin{gathered} {\mathrm{t}}_{\mathrm{n}+1}={\mathrm{t}}_{\mathrm{n}}+\mathrm{h} \\ {\mathrm{t}}_2=1.25+0.25 \\ =1.5 \\ {\mathrm{x}}_1(1.5)={\mathrm{x}}_{1,2} \\ =0.625 \\ {\mathrm{x}}_2(1.5)={\mathrm{x}}_{2,2} \\ =-0.1 \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_3=1.5+0.25 \\ =1.75 \\ {\mathrm{x}}_1(1.75)={\mathrm{x}}_{1,3} \\ =0.6 \\ {\mathrm{x}}_2(1.75)={\mathrm{x}}_{2,3} \\ =-0.219 \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_4=1.75+0.25 \\ =2 \\ {\mathrm{x}}_1\left(2\right)={\mathrm{x}}_{1,4} \\ =0.6548 \\ {\mathrm{x}}_2\left(2\right)={\mathrm{x}}_{2,4} \\ =0.4765 \end{gathered}$$

Thus, this is the required result