Q11RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Write the given higher-order equation or system in an equivalent normal form (compare Section \(5.3\)).

\(\begin{array}{l}{\bf{x''' + y' + y'' = t }}\\{\bf{x'' - x' + y''' = 0}}\end{array}\)

 

Step-by-Step Solution

Answer

 

The equivalent normal form of the given is:

 

\(\begin{array}{c}{{\bf{x}}_{\bf{1}}}{\bf{' = }}{{\bf{x}}_{\bf{2}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{' = }}{{\bf{x}}_{\bf{3}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{' = t - }}{{\bf{x}}_{\bf{5}}}{\bf{ - }}{{\bf{x}}_{\bf{6}}}{\bf{,}}\\{{\bf{x}}_{\bf{4}}}{\bf{' = }}{{\bf{x}}_{\bf{5}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{5}}}{\bf{' = }}{{\bf{x}}_{\bf{6}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{6}}}{\bf{' = }}{{\bf{x}}_{\bf{2}}}{\bf{ - }}{{\bf{x}}_{\bf{3}}}{\bf{.}}\end{array}\)

Step 1: Rewrite the given equation

Rewrite the given system as:

 

\(\begin{array}{c}{\bf{x''' = t - y' - y'' }}\\{\bf{y''' = x' - x''}}\end{array}\)

 

Setting \({\bf{x = }}{{\bf{x}}_{\bf{1}}}{\bf{,x}}'{\bf{ = }}{{\bf{x}}_{\bf{2}}}{\bf{,x}}''{\bf{ = }}{{\bf{x}}_{\bf{3}}}{\bf{,y = }}{{\bf{x}}_{\bf{4}}}{\bf{,y}}'{\bf{ = }}{{\bf{x}}_{\bf{5}}}\) and \({\bf{y'' = }}{{\bf{x}}_{\bf{6}}}.\)

Step 2: Finding the derivatives of \({\bf{x}}\)

Therefore, the equivalent normal form;

 

\(\begin{array}{c}{{\bf{x}}_{\bf{1}}}{\bf{' = }}{{\bf{x}}_{\bf{2}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{' = }}{{\bf{x}}_{\bf{3}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{' = t - }}{{\bf{x}}_{\bf{5}}}{\bf{ - }}{{\bf{x}}_{\bf{6}}}{\bf{,}}\\{{\bf{x}}_{\bf{4}}}{\bf{' = }}{{\bf{x}}_{\bf{5}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{5}}}{\bf{' = }}{{\bf{x}}_{\bf{6}}}{\bf{,}}\;\;\;{{\bf{x}}_{\bf{6}}}{\bf{' = }}{{\bf{x}}_{\bf{2}}}{\bf{ - }}{{\bf{x}}_{\bf{3}}}{\bf{.}}\end{array}\)