20.20.11 problem 11

Internal problem ID [3901]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 9, First order linear systems. Section 9.11 (Chapter review), page 665
Problem number : 11
Date solved : Monday, January 27, 2025 at 08:04:15 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=-x_{1} \left (t \right )-5 x_{2} \left (t \right )+x_{3} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=4 x_{1} \left (t \right )-9 x_{2} \left (t \right )-x_{3} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=3 x_{3} \left (t \right ) \end{align*}

✓ Solution by Maple

Time used: 0.206 (sec). Leaf size: 78

dsolve([diff(x__1(t),t)=-1*x__1(t)-5*x__2(t)+x__3(t),diff(x__2(t),t)=4*x__1(t)-9*x__2(t)-x__3(t),diff(x__3(t),t)=0*x__1(t)+0*x__2(t)+3*x__3(t)],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= \frac {c_3 \,{\mathrm e}^{3 t}}{4}+{\mathrm e}^{-5 t} \cos \left (2 t \right ) c_{1} +{\mathrm e}^{-5 t} \sin \left (2 t \right ) c_{2} \\ x_{2} \left (t \right ) &= \frac {2 \,{\mathrm e}^{-5 t} \left (2 c_{1} \cos \left (2 t \right )-c_{2} \cos \left (2 t \right )+c_{1} \sin \left (2 t \right )+2 c_{2} \sin \left (2 t \right )\right )}{5} \\ x_{3} \left (t \right ) &= c_3 \,{\mathrm e}^{3 t} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.009 (sec). Leaf size: 106

DSolve[{D[x1[t],t]==-1*x1[t]-5*x2[t]+x3[t],D[x2[t],t]==4*x1[t]-9*x2[t]-x3[t],D[x3[t],t]==0*x1[t]+0*x2[t]+3*x3[t]},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{4} e^{-5 t} \left (c_3 e^{8 t}+(4 c_1-c_3) \cos (2 t)+2 (4 c_1-5 c_2-c_3) \sin (2 t)\right ) \\ \text {x2}(t)\to \frac {1}{2} e^{-5 t} (2 c_2 \cos (2 t)+(4 c_1-4 c_2-c_3) \sin (2 t)) \\ \text {x3}(t)\to c_3 e^{3 t} \\ \end{align*}