20.17.12 problem 12

Internal problem ID [3866]
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 : 12
Date solved : Monday, January 27, 2025 at 08:03:36 AM
CAS classification : system_of_ODEs

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

✓ Solution by Maple

Time used: 0.260 (sec). Leaf size: 93

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

\begin{align*} x_{1} \left (t \right ) &= \cos \left (t \right ) c_3 -c_4 \sin \left (t \right ) \\ x_{2} \left (t \right ) &= c_3 \sin \left (t \right )+c_4 \cos \left (t \right ) \\ x_{3} \left (t \right ) &= \frac {8 c_3 \sin \left (t \right )}{25}+\frac {6 c_4 \sin \left (t \right )}{25}-\frac {6 \cos \left (t \right ) c_3}{25}+\frac {8 c_4 \cos \left (t \right )}{25}+c_{2} {\mathrm e}^{2 t} t +{\mathrm e}^{2 t} c_{1} \\ x_{4} \left (t \right ) &= -\frac {\cos \left (t \right ) c_3}{5}-\frac {2 c_3 \sin \left (t \right )}{5}-\frac {2 c_4 \cos \left (t \right )}{5}+\frac {c_4 \sin \left (t \right )}{5}+c_{2} {\mathrm e}^{2 t} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.211 (sec). Leaf size: 260

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

\begin{align*} \text {x1}(t)\to \frac {1}{2} e^{-i t} \left (c_1 \left (1+e^{2 i t}\right )+i c_2 \left (-1+e^{2 i t}\right )\right ) \\ \text {x2}(t)\to \frac {1}{2} e^{-i t} \left (c_2 \left (1+e^{2 i t}\right )-i c_1 \left (-1+e^{2 i t}\right )\right ) \\ \text {x3}(t)\to \frac {1}{25} e^{-i t} \left (c_1 \left (e^{(2+i) t} (5 t+6)-(3+4 i) e^{2 i t}-(3-4 i)\right )+c_2 \left (2 e^{(2+i) t} (5 t-4)+(4-3 i) e^{2 i t}+(4+3 i)\right )+25 e^{(2+i) t} (c_4 t+c_3)\right ) \\ \text {x4}(t)\to \frac {1}{10} e^{-i t} \left (c_1 \left (-(1-2 i) e^{2 i t}+2 e^{(2+i) t}+(-1-2 i)\right )+c_2 \left (-(2+i) e^{2 i t}+4 e^{(2+i) t}+(-2+i)\right )+10 c_4 e^{(2+i) t}\right ) \\ \end{align*}