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Chapter 5: Q9RP (page 306)

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

\({\bf{3y''' + 2y' - }}{{\bf{e}}^{\bf{t}}}{\bf{y = 5}}\)

Short Answer

Expert verified

The solution for the given is:

\(\begin{array}{c}{\bf{x}}_{\bf{1}}^{}{\bf{'(t) = }}{{\bf{x}}_{\bf{2}}}{\bf{(t)}}\\{\bf{x}}_{\bf{2}}^{}{\bf{'(t) = }}{{\bf{x}}_{\bf{3}}}{\bf{(t)}}\\{\bf{x}}_{\bf{2}}^{}{\bf{'(t) = }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{\bf{3}}}{{\bf{x}}_{\bf{1}}}{\bf{(t) - }}\frac{{\bf{2}}}{{\bf{3}}}{{\bf{x}}_{\bf{2}}}{\bf{(t) + }}\frac{{\bf{5}}}{{\bf{3}}}\end{array}\)

Step by step solution

01

Step 1:Rewrite the given equation

Rewrite the given equation as\({\bf{y'''(t) = }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{\bf{3}}}{\bf{y(t) - }}\frac{{\bf{2}}}{{\bf{3}}}{\bf{y'(t) + }}\frac{{\bf{5}}}{{\bf{3}}}\)and setting \({\bf{y(t) = }}{{\bf{x}}_{\bf{1}}}{\bf{(t),y'(t) = }}{{\bf{x}}_{\bf{2}}}{\bf{(t)}}\) and \({\bf{y''(t) = }}{{\bf{x}}_{\bf{3}}}.\)

02

Step 2:Finding the equivalent form

Therefore, the equivalent normal form is:

\(\begin{array}{c}{\bf{x}}_{\bf{1}}^{}{\bf{'(t) = }}{{\bf{x}}_{\bf{2}}}{\bf{(t)}}\\{\bf{x}}_{\bf{2}}^{}{\bf{'(t) = }}{{\bf{x}}_{\bf{3}}}{\bf{(t)}}\\{\bf{x}}_{\bf{2}}^{}{\bf{'(t) = }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{\bf{3}}}{{\bf{x}}_{\bf{1}}}{\bf{(t) - }}\frac{{\bf{2}}}{{\bf{3}}}{{\bf{x}}_{\bf{2}}}{\bf{(t) + }}\frac{{\bf{5}}}{{\bf{3}}}\end{array}\)

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Most popular questions from this chapter

In Problems 1鈥�7, convert the given initial value problem into an initial value problem for a system in normal form.

y''(t)+ty'(t)-3y(t)=t2;y(0)=3,y'(0)=-6

In Problems 3 鈥� 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

dxdt+y=t2,-x+dydt=1

In Problems 3鈥�6, find the critical point set for the given system.

dxdt=x2-2xy,dydt=3xy-y2

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In Problems 25 鈥� 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), z(t).

x'=3x+y-z,y'=x+2y-z,z'=3x+3y-z
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