Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

[hint] ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{x}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{y}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{y}\mathbf{\text{'}}$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{t} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{1} \\ {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Step 1: Express the equation in form of x

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

 

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

 

Denote,

 $$\begin{gathered} {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ {\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ {\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{(}\mathbf{t}\mathbf{)} \end{gathered}$$

The equation transforms as;

 $$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{t} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{1} \end{gathered}$$

Step 2: the initial conditions

The given initial conditions are $\mathbf{x}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{2}$ and $\mathbf{y}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}.$

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

 This is the required result.