Q6E - 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.

$$\begin{gathered} \mathbf{3}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{5}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{3}} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{6}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{4}} \\ {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \end{gathered}$$

Step 1: express the equation in form of x

Here given $\mathbf{3}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$ and $\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{=}\mathbf{0}$

 

Rewrite the equations  $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}}{\mathbf{3}}$ and $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{6}\mathbf{x}}{\mathbf{4}}$

 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;

 $$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{5}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{3}} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{6}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{4}} \end{gathered}$$

 

Step 2: the initial conditions

The given initial conditions are $\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ and $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$

 

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

This is the required result.