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Expert-verified Found in: Page 259 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{on}\mathbf{}\mathbf{\left[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{\right]}$

$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{\right)}\mathbf{=}\mathbf{1}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9375}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8164}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{651855}$

See the step by step solution

## Step 1: Transform equation

Here h=0.25 0n [0,1]

The equations can be written as;

${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

The transformation of the equation is;

$\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}{\mathbf{tx}}_{\mathbf{2}}$

The initial conditions are;

${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{0}}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{0}}$

## Step 2: Apply Euler’s method.

Now,

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

${\mathbf{t}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{t}}_{\mathbf{n}}\mathbf{+}\mathbf{h}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{1}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{1}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{25}$

And

${\mathbf{t}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{t}}_{\mathbf{n}}\mathbf{+}\mathbf{h}\phantom{\rule{0ex}{0ex}}{\mathbf{t}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9375}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{484375}$

${\mathbf{t}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}\mathbf{,}\mathbf{3}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{816406}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}\mathbf{,}\mathbf{3}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{658203}$

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

This is the required result. ### Want to see more solutions like these? 