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Chapter 5: Q8RP (page 306)

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

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

Short Answer

Expert verified

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 by step solution

01

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 \(y'(t) = {x_2}(t)\)

02

Finding the equivalent form

Therefore, the equivalent normal form is;

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

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