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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 1–7, convert the given initial value problem into an initial value problem for a system in normal form.$\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\phantom{\rule{0ex}{0ex}}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

$\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{t}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{5}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{5}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{1}}{\mathbf{5}}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{5}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

See the step by step solution

## Step 1: Express equation in form of x

Here given $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{t}$ and $\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{1}.$

Rewrite the equations $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{+}\mathbf{t}$ and $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{1}}{\mathbf{2}}$

Denote,

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

The equation transforms as:

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

## Step 2: The initial conditions

The given initial conditions are $\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$ and $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}{\mathbf{.}}$

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

This is the required result.