20.20.29 problem 29

Internal problem ID [3919]
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 : 29
Date solved : Monday, January 27, 2025 at 08:04:32 AM
CAS classification : system_of_ODEs

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

✓ Solution by Maple

Time used: 0.071 (sec). Leaf size: 78

dsolve([diff(x__1(t),t)=2*x__1(t)-2*x__2(t)+1*x__3(t)+t,diff(x__2(t),t)=1*x__1(t)-4*x__2(t)+1*x__3(t),diff(x__3(t),t)=2*x__1(t)+2*x__2(t)-3*x__3(t)+1],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= c_3 \,{\mathrm e}^{-5 t}+c_{2} {\mathrm e}^{2 t}-\frac {t}{2}-\frac {7}{20} \\ x_{2} \left (t \right ) &= \frac {c_{2} {\mathrm e}^{2 t}}{4}-\frac {t}{4}+\frac {1}{20}+c_{1} {\mathrm e}^{-2 t}+2 c_3 \,{\mathrm e}^{-5 t} \\ x_{3} \left (t \right ) &= \frac {c_{2} {\mathrm e}^{2 t}}{2}-\frac {t}{2}+\frac {3}{10}+2 c_{1} {\mathrm e}^{-2 t}-3 c_3 \,{\mathrm e}^{-5 t} \\ \end{align*}

✓ Solution by Mathematica

Time used: 4.724 (sec). Leaf size: 183

DSolve[{D[x1[t],t]==2*x1[t]-2*x2[t]+1*x3[t]+t,D[x2[t],t]==1*x1[t]-4*x2[t]+1*x3[t],D[x3[t],t]==2*x1[t]+2*x2[t]-3*x3[t]+1},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{140} e^{-5 t} \left (-7 e^{5 t} (10 t+7)+20 (7 c_1-2 c_2+c_3) e^{7 t}+40 c_2-20 c_3\right ) \\ \text {x2}(t)\to \frac {1}{140} e^{-5 t} \left (e^{5 t} (7-35 t)-35 (c_1-2 c_2-c_3) e^{3 t}+5 (7 c_1-2 c_2+c_3) e^{7 t}+80 c_2-40 c_3\right ) \\ \text {x3}(t)\to \frac {1}{70} \left (-35 t-35 (c_1-2 c_2-c_3) e^{-2 t}+30 (c_3-2 c_2) e^{-5 t}+5 (7 c_1-2 c_2+c_3) e^{2 t}+21\right ) \\ \end{align*}