Q4RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution  $\mathbf{x}\left(\mathbf{t}\right)\mathbf{,}\mathbf{y}\left(\mathbf{t}\right)$ for the given system.

 $$\begin{gathered} \mathbf{x}\text{'}\text{'}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{y}\text{'}\text{'}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ \mathbf{x}\text{'}\text{'}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\text{'}\text{'}\mathbf{=}\mathbf{0} \end{gathered}$$

Step-by-Step Solution

Answer

The solution to the given system is:

 

 $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{,} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{\mathbf{6}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\frac{{\mathbf{c}}_{\mathbf{2}}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{4}} \end{gathered}$$

Step 1: Using the elimination method

One will solve the given system using the elimination method. One will first rewrite the system in operator form:

 

 $$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\right)\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{-}{\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}{\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathbf{0} \end{gathered}$$

 

One will eliminate $y$  from the system by adding those two equations together:

 

 $$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ \mathbf{2}{\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ {\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \end{gathered}$$

 

So, one has that $\mathbf{x}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$. One will solve for  $x$ integrating twice the previous equation:

 

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

Step 2: Integrating the equation

The first equation of the given system gives us that  $\mathbf{y}\text{'}\text{'}\mathbf{=}\mathbf{x}\text{'}\text{'}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$ and since one already has that $\mathbf{x}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$ one can calculate a solution of $y$. 

 

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

 

Integrating the previous equation twice one will get:

 

 $$\begin{gathered} \mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\int \mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}\mathbf{=}\int \left({\mathbf{c}}_{\mathbf{1}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right)\mathbf{dt}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{3}} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\int \mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}\mathbf{=}\int \left(\frac{{\mathbf{c}}_{\mathbf{1}}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\right)\mathbf{dt} \\ \mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{\mathbf{6}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\frac{{\mathbf{c}}_{\mathbf{2}}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{4}} \end{gathered}$$

 

So, the solution to the given system is:

 $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{,} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{\mathbf{6}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\frac{{\mathbf{c}}_{\mathbf{2}}}{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{4}} \end{gathered}$$