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### 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 1–7, convert the given initial value problem into an initial value problem for a system in normal form.${{\mathbf{y}}}^{{\mathbf{6}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\left[\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]}^{{\mathbf{3}}}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right)}\mathbf{+}{{\mathbf{e}}}^{\mathbf{2}\mathbf{t}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{.}{.}{.}{.}{.}{.}{\mathbf{=}}{{\mathbf{y}}}^{\mathbf{\left(}\mathbf{5}\mathbf{\right)}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

$\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{6}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\left[{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]}^{\mathbf{3}}\mathbf{-}\mathbf{sin}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

See the step by step solution

## Step 1: Express the equation in form of x

Here given ${\mathbf{y}}^{\mathbf{6}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\left[\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]}^{\mathbf{3}}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

Denote,

${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{y}}^{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{5}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{y}}^{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{6}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{y}}^{\mathbf{5}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

The equation transforms as;

$\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{5}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{5}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{6}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{6}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\left[{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]}^{\mathbf{3}}\mathbf{-}\mathbf{sin}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

## Step 2: The initial conditions

The given initial conditions are $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}......\mathbf{=}{\mathbf{y}}^{\mathbf{\left(}\mathbf{5}\mathbf{\right)}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}.$

Initial conditions after transformations ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{5}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{6}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}.$

This is the required result.

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