Q8RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

write the given higher-order equation or system in an equivalent normal form (compare Section\({\bf{5}}.{\bf{3}}\)).

\(2y'' - ty' + 8y = \sin t\)

Step-by-Step Solution

Answer

The solution for the given is:

 

\(\begin{array}{l}{{\bf{x}}_{\bf{1}}}{\bf{'(t) = }}{{\bf{x}}_{\bf{2}}}{\bf{(t)}}\\{{\bf{x}}_{\bf{2}}}{\bf{'(t) = }}\frac{{{\bf{sint}}}}{{\bf{2}}}{\bf{ + }}\frac{{\bf{t}}}{{\bf{2}}}{{\bf{x}}_{\bf{2}}}{\bf{ - 4}}{{\bf{x}}_{\bf{1}}}\end{array}\)

Step 1: Rewrite the given equation

Rewrite the given equation as\(y'' = \frac{{\sin t}}{2} + \frac{t}{2}y' - 4y\)and setting \(y(t) = {x_1}(t)\) and \(y'(t) = {x_2}(t)\)

Step 2: Finding the equivalent form

 

Therefore, the equivalent normal form is;

 

\(\begin{array}{l}{{\bf{x}}_{\bf{1}}}{\bf{'(t) = }}{{\bf{x}}_{\bf{2}}}{\bf{(t)}}\\{{\bf{x}}_{\bf{2}}}{\bf{'(t) = }}\frac{{{\bf{sint}}}}{{\bf{2}}}{\bf{ + }}\frac{{\bf{t}}}{{\bf{2}}}{{\bf{x}}_{\bf{2}}}{\bf{ - 4}}{{\bf{x}}_{\bf{1}}}\end{array}\)