Internal problem ID [388]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 7.6, Multiple Eigenvalue Solutions. Page 451
Problem number: problem 31.
ODE order: 1.
ODE degree: 1.
Solve \begin {align*} x_{1}^{\prime }\left (t \right )&=35 x_{1} \left (t \right )-12 x_{2} \left (t \right )+4 x_{3} \left (t \right )+30 x_{4} \left (t \right )\\ x_{2}^{\prime }\left (t \right )&=22 x_{1} \left (t \right )-8 x_{2} \left (t \right )+3 x_{3} \left (t \right )+19 x_{4} \left (t \right )\\ x_{3}^{\prime }\left (t \right )&=-10 x_{1} \left (t \right )+3 x_{2} \left (t \right )-9 x_{4} \left (t \right )\\ x_{4}^{\prime }\left (t \right )&=-27 x_{1} \left (t \right )+9 x_{2} \left (t \right )-3 x_{3} \left (t \right )-23 x_{4} \left (t \right ) \end {align*}
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 112
dsolve([diff(x__1(t),t)=35*x__1(t)-12*x__2(t)+4*x__3(t)+30*x__4(t),diff(x__2(t),t)=22*x__1(t)-8*x__2(t)+3*x__3(t)+19*x__4(t),diff(x__3(t),t)=-10*x__1(t)+3*x__2(t)+0*x__3(t)-9*x__4(t),diff(x__4(t),t)=-27*x__1(t)+9*x__2(t)-3*x__3(t)-23*x__4(t)],[x__1(t), x__2(t), x__3(t), x__4(t)], singsol=all)
\[ x_{1} \left (t \right ) = -\frac {{\mathrm e}^{t} \left (6 c_{4} t^{2}+6 c_{3} t +2 c_{4} t +6 c_{2} +c_{3} -c_{4} \right )}{6} \] \[ x_{2} \left (t \right ) = \frac {{\mathrm e}^{t} \left (-2 c_{4} t^{2}-2 c_{3} t -10 c_{4} t +4 c_{1} -2 c_{2} -5 c_{3} +6 c_{4} \right )}{12} \] \[ x_{3} \left (t \right ) = \frac {{\mathrm e}^{t} \left (6 c_{4} t^{2}+6 c_{3} t -2 c_{4} t +12 c_{1} +6 c_{2} -c_{3} \right )}{12} \] \[ x_{4} \left (t \right ) = {\mathrm e}^{t} \left (c_{4} t^{2}+c_{3} t +c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 187
DSolve[{x1'[t]==35*x1[t]-12*x2[t]+4*x3[t]+30*x4[t],x2'[t]==22*x1[t]-8*x2[t]+3*x3[t]+19*x4[t],x3'[t]==-10*x1[t]+3*x2[t]+0*x3[t]-9*x4[t],x4'[t]==-27*x1[t]+9*x2[t]-3*x3[t]-23*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {x1}(t)\to e^t (c_1 t (21 t+34)-(3 c_2-c_3) t (3 t+4)+6 c_4 t (3 t+5)+c_1) \\ \text {x2}(t)\to \frac {1}{2} e^t (t (c_1 (7 t+44)+(c_3-3 c_2) (t+6)+2 c_4 (3 t+19))+2 c_2) \\ \text {x3}(t)\to \frac {1}{2} e^t (2 c_3-t (c_1 (21 t+20)-3 c_2 (3 t+2)+c_3 (3 t+2)+18 c_4 (t+1))) \\ \text {x4}(t)\to e^t (c_4-3 t (c_1 (7 t+9)+(c_3-3 c_2) (t+1)+2 c_4 (3 t+4))) \\ \end{align*}