Q1E - 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.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{6}$

Step-by-Step Solution

Answer

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

Step 1: Express the equation in form of x.

Here given;

 $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}$

Denote,

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

 The equation transforms as;

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

Step 2: the initial conditions

The given initial conditions are $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{6}$

 

Initial conditions after transformations;

 $$\begin{gathered} {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{6} \end{gathered}$$

This is the required result.