Internal problem ID [386]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 7.6, Multiple Eigenvalue Solutions. Page 451
Problem number: problem 29.
ODE order: 1.
ODE degree: 1.
Solve \begin {align*} x_{1}^{\prime }\relax (t )&=-x_{1} \relax (t )+x_{2} \relax (t )+x_{3} \relax (t )-2 x_{4} \relax (t )\\ x_{2}^{\prime }\relax (t )&=7 x_{1} \relax (t )-4 x_{2} \relax (t )-6 x_{3} \relax (t )+11 x_{4} \relax (t )\\ x_{3}^{\prime }\relax (t )&=5 x_{1} \relax (t )-x_{2} \relax (t )+x_{3} \relax (t )+3 x_{4} \relax (t )\\ x_{4}^{\prime }\relax (t )&=6 x_{1} \relax (t )-2 x_{2} \relax (t )-2 x_{3} \relax (t )+6 x_{4} \relax (t ) \end {align*}
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 119
dsolve([diff(x__1(t),t)=-1*x__1(t)+1*x__2(t)+1*x__3(t)-2*x__4(t),diff(x__2(t),t)=7*x__1(t)-4*x__2(t)-6*x__3(t)+11*x__4(t),diff(x__3(t),t)=5*x__1(t)-1*x__2(t)+1*x__3(t)+3*x__4(t),diff(x__4(t),t)=6*x__1(t)-2*x__2(t)-2*x__3(t)+6*x__4(t)],[x__1(t), x__2(t), x__3(t), x__4(t)], singsol=all)
\[ x_{1} \relax (t ) = -\frac {{\mathrm e}^{-t} \left (t c_{4}+c_{3}\right )}{2} \] \[ x_{2} \relax (t ) = -c_{2} {\mathrm e}^{2 t} t +\frac {3 c_{4} {\mathrm e}^{-t} t}{2}-c_{1} {\mathrm e}^{2 t}+2 c_{2} {\mathrm e}^{2 t}+\frac {3 c_{3} {\mathrm e}^{-t}}{2}-\frac {{\mathrm e}^{-t} c_{4}}{2} \] \[ x_{3} \relax (t ) = c_{2} {\mathrm e}^{2 t} t +c_{1} {\mathrm e}^{2 t}+\frac {c_{4} {\mathrm e}^{-t} t}{2}+\frac {c_{3} {\mathrm e}^{-t}}{2} \] \[ x_{4} \relax (t ) = c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{-t}+c_{4} {\mathrm e}^{-t} t \]
✓ Solution by Mathematica
Time used: 0.011 (sec). Leaf size: 166
DSolve[{x1'[t]==-1*x1[t]+1*x2[t]+1*x3[t]-2*x4[t],x2'[t]==7*x1[t]-4*x2[t]-6*x3[t]+11*x4[t],x3'[t]==5*x1[t]-1*x2[t]+1*x3[t]+3*x4[t],x4'[t]==6*x1[t]-2*x2[t]-2*x3[t]+6*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {x1}(t)\to e^{-t} ((c_2+c_3-2 c_4) t+c_1) \\ \text {x2}(t)\to e^{-t} \left (-3 ((c_2+c_3-2 c_4) t+c_1)-e^{3 t} (c_1 (2 t-3)+c_4 (t-2)+c_3)+c_2+c_3-2 c_4\right ) \\ \text {x3}(t)\to e^{2 t} (2 c_1 t+c_4 t+c_1+c_3)-e^{-t} ((c_2+c_3-2 c_4) t+c_1) \\ \text {x4}(t)\to (2 c_1+c_4) e^{2 t}-2 e^{-t} ((c_2+c_3-2 c_4) t+c_1) \\ \end{align*}