20.17.13 problem 13

Internal problem ID [3867]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 9, First order linear systems. Section 9.5 (Defective coefficient matrix), page 619
Problem number : 13
Date solved : Monday, January 27, 2025 at 08:03:38 AM
CAS classification : system_of_ODEs

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

✓ Solution by Maple

Time used: 0.079 (sec). Leaf size: 78

dsolve([diff(x__1(t),t)=-2*x__1(t)+3*x__2(t)+0*x__3(t)+0*x__4(t),diff(x__2(t),t)=3*x__1(t)-2*x__2(t)+0*x__3(t)+0*x__4(t),diff(x__3(t),t)=1*x__1(t)+0*x__2(t)+1*x__3(t)+1*x__4(t),diff(x__4(t),t)=0*x__1(t)+1*x__2(t)+0*x__3(t)+1*x__4(t)],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= c_3 \,{\mathrm e}^{t}-c_4 \,{\mathrm e}^{-5 t} \\ x_{2} \left (t \right ) &= c_3 \,{\mathrm e}^{t}+c_4 \,{\mathrm e}^{-5 t} \\ x_{3} \left (t \right ) &= \left (\left (\frac {1}{2} t^{2}+t \right ) c_3 +c_{2} t +c_{1} \right ) {\mathrm e}^{t}+\frac {7 c_4 \,{\mathrm e}^{-5 t}}{36} \\ x_{4} \left (t \right ) &= c_3 \,{\mathrm e}^{t} t +c_{2} {\mathrm e}^{t}-\frac {c_4 \,{\mathrm e}^{-5 t}}{6} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.009 (sec). Leaf size: 192

DSolve[{D[x1[t],t]==-2*x1[t]+3*x2[t]+0*x3[t]+0*x4[t],D[x2[t],t]==3*x1[t]-2*x2[t]+0*x3[t]+0*x4[t],D[x3[t],t]==1*x1[t]+0*x2[t]+1*x3[t]-1*x4[t],D[x4[t],t]==0*x1[t]+1*x2[t]+0*x3[t]+1*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{2} e^{-5 t} \left (c_1 \left (e^{6 t}+1\right )+c_2 \left (e^{6 t}-1\right )\right ) \\ \text {x2}(t)\to \frac {1}{2} e^{-5 t} \left (c_1 \left (e^{6 t}-1\right )+c_2 \left (e^{6 t}+1\right )\right ) \\ \text {x3}(t)\to \frac {1}{72} e^{-5 t} \left (c_1 \left (e^{6 t} \left (-18 t^2+42 t+5\right )-5\right )+c_2 \left (e^{6 t} \left (-18 t^2+30 t-5\right )+5\right )+72 e^{6 t} (c_3-c_4 t)\right ) \\ \text {x4}(t)\to \frac {1}{12} e^{-5 t} \left (c_1 \left (e^{6 t} (6 t-1)+1\right )+c_2 \left (e^{6 t} (6 t+1)-1\right )+12 c_4 e^{6 t}\right ) \\ \end{align*}