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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.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{y}\mathbf{\right)}\mathbf{+}{{\mathbf{y}}}^{{\mathbf{2}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

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

See the step by step solution

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

Here given $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{y}\mathbf{\right)}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Denote,

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

The equation transforms as;

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

## Step 2: the initial conditions

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

Initial conditions after transformations;

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

This is the required result.