76.26.14 problem 18

Internal problem ID [17846]
Book : Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015
Section : Chapter 6. Systems of First Order Linear Equations. Section 6.4 (Nondefective Matrices with Complex Eigenvalues). Problems at page 419
Problem number : 18
Date solved : Tuesday, January 28, 2025 at 11:03:58 AM
CAS classification : system_of_ODEs

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

✓ Solution by Maple

Time used: 0.154 (sec). Leaf size: 130

dsolve([diff(x__1(t),t)=0*x__1(t)+3*x__2(t)+0*x__3(t)-2*x__4(t),diff(x__2(t),t)=-1/2*x__1(t)+1*x__2(t)-3*x__3(t)-5/2*x__4(t),diff(x__3(t),t)=0*x__1(t)+3*x__2(t)-5*x__3(t)-3*x__4(t),diff(x__4(t),t)=1*x__1(t)+3*x__2(t)+0*x__3(t)-3*x__4(t)],singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.015 (sec). Leaf size: 261

DSolve[{D[x1[t],t]==0*x1[t]+3*x2[t]+0*x3[t]-2*x4[t],D[x2[t],t]==-1/2*x1[t]+1*x2[t]-3*x3[t]-5/2*x4[t],D[x3[t],t]==0*x1[t]+3*x2[t]-5*x3[t]-3*x4[t],D[x4[t],t]==1*x1[t]+3*x2[t]+0*x3[t]-3*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{2} e^{-2 t} \left (c_1 e^t-c_4 e^t+2 (-c_2+c_3+c_4) \cos (3 t)+(c_1+2 c_2-c_4) \sin (3 t)+c_1+2 c_2-2 c_3-c_4\right ) \\ \text {x2}(t)\to \frac {1}{2} e^{-2 t} \left ((c_4-c_1) e^t+(c_1+2 c_2-c_4) \cos (3 t)+2 (c_2-c_3-c_4) \sin (3 t)\right ) \\ \text {x3}(t)\to \frac {1}{2} e^{-2 t} ((c_1+2 c_2-c_4) \cos (3 t)+2 (c_2-c_3-c_4) \sin (3 t)-c_1-2 c_2+2 c_3+c_4) \\ \text {x4}(t)\to \frac {1}{2} e^{-2 t} \left (c_1 \left (-e^t\right )+c_4 e^t+2 (-c_2+c_3+c_4) \cos (3 t)+(c_1+2 c_2-c_4) \sin (3 t)+c_1+2 c_2-2 c_3-c_4\right ) \\ \end{align*}