Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{]}$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathbf{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}\mathbf{=}\mathbf{1} \\ \mathbf{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9375} \\ \mathbf{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8164} \\ \mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{651855} \end{gathered}$$

Step 1: Transform equation

Here h=0.25 0n [0,1]

 

The equations can be written as;

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

The transformation of the equation is;

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

 

The initial conditions are;

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

Step 2: Apply Euler’s method.

Now,

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

$$\begin{gathered} {\mathbf{t}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{t}}_{\mathbf{n}}\mathbf{+}\mathbf{h} \\ \mathbf{=}\mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25} \\ {\mathbf{x}}_{\mathbf{1}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{1}} \\ \mathbf{=}\mathbf{1} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{1}} \\ \mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{25} \end{gathered}$$ 

And

$$\begin{gathered} {\mathbf{t}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{t}}_{\mathbf{n}}\mathbf{+}\mathbf{h} \\ {\mathbf{t}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25} \\ {\mathbf{x}}_{\mathbf{1}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{2}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9375} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{2}} \\ \mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{484375} \end{gathered}$$

 $$\begin{gathered} {\mathbf{t}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75} \\ {\mathbf{x}}_{\mathbf{1}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{3}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{816406} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{3}} \\ \mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{658203} \end{gathered}$$

$$\begin{gathered} {\mathbf{t}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25} \\ \mathbf{=}\mathbf{1} \\ {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{4}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{651855} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{4}} \\ \mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{738891} \end{gathered}$$

 

 

This is the required result.