Internal problem ID [357]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 7.6, Multiple Eigenvalue Solutions. Examples. Page 437
Problem number: Example 6.
ODE order: 1.
ODE degree: 1.
Solve \begin {align*} x_{1}^{\prime }\left (t \right )&=x_{3} \left (t \right )\\ x_{2}^{\prime }\left (t \right )&=x_{4} \left (t \right )\\ x_{3}^{\prime }\left (t \right )&=-2 x_{1} \left (t \right )+2 x_{2} \left (t \right )-3 x_{3} \left (t \right )+x_{4} \left (t \right )\\ x_{4}^{\prime }\left (t \right )&=2 x_{1} \left (t \right )-2 x_{2} \left (t \right )+x_{3} \left (t \right )-3 x_{4} \left (t \right ) \end {align*}
✓ Solution by Maple
Time used: 0.046 (sec). Leaf size: 95
dsolve([diff(x__1(t),t)=0*x__1(t)+0*x__2(t)+1*x__3(t)+0*x__4(t),diff(x__2(t),t)=0*x__1(t)+0*x__2(t)+0*x__3(t)+1*x__4(t),diff(x__3(t),t)=-2*x__1(t)+2*x__2(t)-3*x__3(t)+1*x__4(t),diff(x__4(t),t)=2*x__1(t)-2*x__2(t)+1*x__3(t)-3*x__4(t)],singsol=all)
\begin{align*} x_{1} \left (t \right ) &= c_{2} +c_{3} {\mathrm e}^{-2 t}+c_{4} {\mathrm e}^{-2 t} t \\ x_{2} \left (t \right ) &= -c_{3} {\mathrm e}^{-2 t}-c_{4} {\mathrm e}^{-2 t} t +c_{4} {\mathrm e}^{-2 t}+c_{2} +c_{1} {\mathrm e}^{-2 t} \\ x_{3} \left (t \right ) &= -{\mathrm e}^{-2 t} \left (2 c_{4} t +2 c_{3} -c_{4} \right ) \\ x_{4} \left (t \right ) &= -{\mathrm e}^{-2 t} \left (-2 c_{4} t +2 c_{1} -2 c_{3} +3 c_{4} \right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.06 (sec). Leaf size: 210
DSolve[{x1'[t]==0*x1[t]+0*x2[t]+1*x3[t]+0*x4[t],x2'[t]==0*x1[t]+0*x2[t]+0*x3[t]+1*x4[t],x3'[t]==-2*x1[t]+2*x2[t]-3*x3[t]+1*x4[t],x4'[t]==2*x1[t]-2*x2[t]+1*x3[t]-3*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {x1}(t)\to \frac {1}{4} e^{-2 t} \left (2 c_1 \left (2 t+e^{2 t}+1\right )+2 c_2 \left (-2 t+e^{2 t}-1\right )+c_3 e^{2 t}+2 c_3 t+c_4 e^{2 t}-2 c_4 t-c_3-c_4\right ) \\ \text {x2}(t)\to \frac {1}{4} e^{-2 t} \left (2 c_1 \left (-2 t+e^{2 t}-1\right )+2 c_2 \left (2 t+e^{2 t}+1\right )+c_3 e^{2 t}-2 c_3 t+c_4 e^{2 t}+2 c_4 t-c_3-c_4\right ) \\ \text {x3}(t)\to e^{-2 t} ((-2 c_1+2 c_2-c_3+c_4) t+c_3) \\ \text {x4}(t)\to e^{-2 t} ((2 c_1-2 c_2+c_3-c_4) t+c_4) \\ \end{align*}